Then it takes 0.375 seconds for the bullet to reach the target.
To determine the time required for the bullet to reach the target, we can use the formula t = d/v, where t is time, d is distance, and v is velocity. In this case, the distance is 300 meters and the velocity is 800 m/s.
Substituting these values into the formula, we get:
t = 300/800
t = 0.375 seconds
It is important to note that this calculation assumes that there is no air resistance acting on the bullet. In reality, air resistance would cause the bullet to slow down over time, so the actual time required for the bullet to reach the target may be slightly longer than calculated.
Additionally, it is crucial to always follow proper firearm safety protocols and regulations when handling firearms.
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What is the relationship between distance and magnetic force?
As you increase the distance between the magnet and the paper clip, does the magnetic force increase or decrease?
a. As distance increases, magnetic force increases.
b. As distance increases, magnetic force decreases.
c. As distance increases, magnetic force stays the same
Answer:
b. As distance increases, magnetic force decreases.
Explanation:
The correct answer is b. As distance increases, the magnetic force decreases. Magnetic force obeys an inverse square law with distance. This means that the force is inversely proportional to the distance squared. For example, if the distance between two magnets is doubled, the magnetic force between them will fall to a quarter of the initial value.
What is the velocity of a soccer ball in meters per second (m/s) with a mass of 1.0 kg that is kicked from rest if the coefficient of restitution between the ball and the foot is 0.48? the initial velocity of the foot is 16.1 m/s and has a mass of 7.8 kg.
The final velocity of the soccer ball is 65.2 m/s. This is to calculate the momentum of the foot before collision. Use coefficient of restitution to calculate velocity of separation.
To find the velocity of the soccer ball after being kicked, we can use the law of conservation of momentum and the coefficient of restitution. The law of conservation of momentum states that the momentum before the collision is equal to the momentum after the collision.
Here's how we can solve the problem:
Calculate the momentum of the foot before the collision:
Momentum = mass x velocity = 7.8 kg x 16.1 m/s = 125.58 kg m/s
During the collision, some of the momentum is transferred to the ball. The amount of momentum transferred depends on the coefficient of restitution, which is given as 0.48. The coefficient of restitution is the ratio of the velocity of separation to the velocity of approach.
Use the coefficient of restitution to calculate the velocity of separation:
Velocity of separation = coefficient of restitution x velocity of approach
Velocity of separation = 0.48 x 16.1 m/s = 7.728 m/s
Calculate the velocity of the ball after the collision using the law of conservation of momentum:
Momentum before collision = Momentum after collision
(7.8 kg x 16.1 m/s) = (1.0 kg x velocity of ball) + (7.8 kg x 7.728 m/s)
125.58 kg m/s = 1.0 kg x velocity of ball + 60.38 kg m/s
Velocity of ball = (125.58 kg m/s - 60.38 kg m/s)/1.0 kg
Velocity of ball = 65.2 m/s
Therefore, the velocity of the soccer ball after being kicked is 65.2 m/s.
In summary, we can use the law of conservation of momentum and the coefficient of restitution to find the velocity of the soccer ball after being kicked. The momentum before the collision is equal to the momentum after the collision.
The coefficient of restitution is the ratio of the velocity of separation to the velocity of approach. Using these equations, we calculated the velocity of the soccer ball to be 65.2 m/s.
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A baseball is hit straight up at an initial velocity of 30m/s if the ball has a negative acceleration of about 10m/s2 how long does the ball take to reach the too of its path
The ball takes about 3.06 seconds to reach the top of its path.
When a ball is thrown or hit straight up, it will reach its maximum height at the point where its vertical velocity becomes zero.
At this point, the ball's acceleration will be equal to the acceleration due to gravity, which is -9.8 m/s².
Using the equation of motion for an object with constant acceleration, we can find the time it takes for the ball to reach the top of its path:
vf = vi + at
where vf is the final velocity (which is zero when the ball reaches its maximum height), vi is the initial velocity (30 m/s), a is the acceleration (-9.8 m/s²), and t is the time we're looking for.
Rearranging the equation, we get:
t = (vf - vi) / a
Since the final velocity is zero, we have:
t = -vi / a
Substituting the values, we get:
t = -30 m/s / (-9.8 m/s²)
t = 3.06 seconds
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In ancient times, many people believed that our lives were somehow influenced by the patterns of the stars in the sky. Modern science has not found any evidence to support this belief, but instead has found that we have a connection to the stars on a much deeper level: We are "star stuff. "Do you think these connections have any philosophical implications in terms of how we view our lives and our civilization?
Yes, I do think that the idea that we are "star stuff" has significant philosophical implications. Firstly, it challenges the notion that we are separate from the universe and reinforces the idea that we are interconnected with everything around us.
This can lead to a sense of awe and wonder about the universe and our place in it.
Additionally, the idea that we are made of the same material as stars can inspire a sense of responsibility to take care of the planet and our fellow human beings. We are not just individuals, but part of a larger whole, and our actions can have an impact on the world around us.
From a societal perspective, this understanding can lead to a greater appreciation for science and the pursuit of knowledge. It can also inspire a sense of unity and cooperation among different cultures and nations, as we all share this common connection to the universe.
Overall, recognizing our connection to the stars can have profound implications for how we view ourselves and our place in the world, and can inspire us to live more consciously and responsibly.
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Sachi is rock'n to her favorite radio station -102. 3 FM. The station broadcasts radio signals with a frequency of 1. 023x10^8 Hz. The radio wave signal travel through the air at a speed of 2. 997x10^8 m/s. Determine the wavelength of these radios
A. 2. 93
B. 1. 93
C. 0. 93
D. 3. 93
The wavelength of the radio waves is 2.93 meters, which corresponds to option A.
To find the wavelength of a radio wave, we can use the formula:
wavelength = speed of light / frequency
This formula tells us that the wavelength of a radio wave is inversely proportional to its frequency. In other words, if the frequency of a radio wave is high, its wavelength will be shorter, and if its frequency is low, its wavelength will be longer.
In the given question, we are told that a radio station broadcasts signals at a frequency of 102.3 MHz or 1.023 x 10⁸ Hz. We are also given the speed of light in air, which is 2.997 x 10⁸ m/s. Using the above formula, we can calculate the wavelength of these radio waves.
Substituting the values in the formula, we get:
wavelength = 2.997 x 10⁸ m/s / 1.023 x 10⁸ Hz
= 2.93 meters
Option A is correct answer.
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An object of mass 20 g is moving in a horizontal circle of radius 250 cm at a speed of 50 cm/s. What is the centripetal acceleration experienced by the object?
The centripetal acceleration experienced by an object moving in a circle is given by the formula:
a = v²/r
where a is the centripetal acceleration, v is the speed of the object, and r is the radius of the circle.
In this problem, we are given that the object has a mass of 20 g, which we need to convert to kilograms:
m = 20 g = 0.02 kg
We are also given that the object is moving in a horizontal circle of radius 250 cm at a speed of 50 cm/s. We need to convert these measurements to SI units (meters and seconds) to use the formula for centripetal acceleration:
r = 250 cm = 2.5 m
v = 50 cm/s = 0.5 m/s
Now we can calculate the centripetal acceleration:
a = v²/r = (0.5 m/s)² / 2.5 m = 0.1 m/s²
Therefore, the centripetal acceleration experienced by the object is 0.1 m/s².
Iodine-131 has a half life of 8 days. if there were 512 mg in a sample, how much iodine would be left in 32 days?
In a 32-day period, a 512 mg sample of Iodine-131 will be reduced to 32 mg.
7
Since the half-life is 8 days, we can divide 32 days by the half-life to find the number of half-lives that have occurred: 32 days ÷ 8 days/half-life = 4 half-lives.
Now, for each half-life, the amount of Iodine-131 will decrease by half. After 1 half-life (8 days), 512 mg will become 256 mg. After 2 half-lives (16 days), it will be 128 mg. After 3 half-lives (24 days), it will be 64 mg. Finally, after 4 half-lives (32 days), the amount of Iodine-131 remaining in the sample will be 32 mg.
So, in a 32-day period, a 512 mg sample of Iodine-131 will be reduced to 32 mg.
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Explain the relation between area, distance and capacitance
The capacitance of a capacitor is: directly proportional to the area of the conductive plates and inversely proportional to the distance between them.
The capacitance (C) of a capacitor is the measure of its ability to store an electrical charge. It is dependent on the surface area (A) of the conductive plates, the distance (d) between these plates, and the permittivity (ε) of the dielectric material that separates the plates. The relationship between these factors can be described by the following formula:
C = ε × (A / d)
In this equation, the area (A) and the distance (d) play crucial roles in determining the capacitance of a capacitor. As the surface area of the plates increases, the capacitance also increases because a larger surface area allows for more charge to be stored. Conversely, as the distance between the plates decreases, the capacitance increases as well since the electric field between the plates becomes stronger, allowing for a higher charge storage capacity.
In summary, the capacitance of a capacitor is directly proportional to the area of the conductive plates and inversely proportional to the distance between them. By adjusting these factors, one can tailor the capacitance of a capacitor to meet specific requirements in various electronic devices and circuits.
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Electric Field of Dreams
PART A) To begin, click the Add button to add one object to the system. Observe the electric field around this charged object. You may move the object around the field by dragging it with your cursor. While the arrows indicate the direction of the electric field around the charge, the length of the arrows indicates the field strength. Based on your observations of the field, what is the charge on this object? Give your reasoning. PART B) Set the charged object in motion by dragging it and releasing it. What do you observe about the behavior of the field lines in the vicinity of the object?
PART C) Add another charged object to the electric field by clicking the Add button again. What is the charge of this new object? Give your reasoning. What do you observe about the behavior of both the objects as well as the field lines in the vicinity of both the objects?
PART D) Click the Remove button to remove one of these objects, and then click the Properties button to set properties for the next object you will add. Just change the sign of the charge to (+), then click Done. Click Add to add this new object to the field. Now what do you observe about the behavior of the two objects and the field lines that surround them?
PART E) With the two oppositely-charged objects still in the field, apply an external field to the system: In the External Field box, simply drag the dot until it becomes an electric field vector in some direction. Observe, describe, and explain the behavior of the two objects
Charged objects in an electric field experience attractive or repulsive forces, as shown by the electric field lines. An external electric field can also cause charged objects to move in a specific direction.
PART A) After adding the charged object to the system, the electric field lines around it are observed to be directed radially outwards from the object, indicating a positive charge.
The length of the field lines also indicates that the charge on the object is strong. This is because the field lines are closer together and longer, which indicates that the strength of the electric field is higher. Therefore, the charge on the object is positive.
PART B) When the charged object is set in motion, the field lines move along with the object, remaining in close proximity to it. The lines become compressed in the front of the object and elongated behind the object, indicating that the electric field is stronger in front of the moving object than behind it.
PART C) When another charged object is added to the field, the electric field lines between the two objects behave as though they are attracted to one another.
This indicates that the new object has an opposite charge to the original object, resulting in attractive forces between the two. The field lines of both objects tend to converge, indicating that the field strength has increased due to the addition of a second charged object.
PART D) After changing the sign of the charge on the new object and adding it to the field, the two objects move towards each other, as the forces between them are now attractive.
The electric field lines between the two objects also converge, indicating a stronger field strength between the two objects.
PART E) When an external electric field is applied to the system, the two objects experience a net force in the direction of the external field, and they move in that direction.
The field lines between the two objects also become elongated in the direction of the external field. This occurs because the electric field of the external field vector superimposes the field of the two objects, and it becomes the dominant field.
In summary, adding charged objects to an electric field creates attractive or repulsive forces between them, which is indicated by the behavior of the electric field lines.
An external electric field can also influence the behavior of charged objects in an electric field, causing them to move in a particular direction.
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A diverging lens has a focal length of -12. 8cm. An object is placed 34. 5cm from the len’s surface: Determine the image distance
The image distance formed by the diverging lens is 9.335cm.
To determine the image distance formed by a diverging lens with a focal length of -12.8cm, we can use the thin lens formula:
1/f = 1/do + 1/di
where f is the focal length of the lens, do is the distance from the object to the lens, and di is the distance from the lens to the image.
Substituting the given values, we get:
1/-12.8cm = 1/34.5cm + 1/di
Simplifying and solving for di, we get:
1/di = 1/-12.8cm - 1/34.5cm
1/di = -0.078125 cm^-1 - 0.02898550724637681 cm^-1
1/di = -0.1071105072463768 cm^-1
di = 9.335 cm
It's worth noting that the negative sign for the focal length indicates that the lens is a diverging lens.
The positive sign for the object distance indicates that the object is located on the same side of the lens as the incident light,
while the negative sign for the image distance indicates that the image is formed on the opposite side of the lens as the incident light, which is typical for a diverging lens.
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You are using power pivot for the first time, but you do not see the power pivot tab. what is the most likely reason
The most likely reason for not seeing the Power Pivot tab is that the add-in is not enabled in Excel.
By default, the Power Pivot add-in is not enabled in Excel, and it needs to be enabled manually.
To enable the Power Pivot add-in, go to the File tab in Excel, select Options, and then select Add-Ins. In the Manage box, select COM Add-ins, and then select Go.
In the COM Add-Ins dialog box, select Microsoft Power Pivot for Excel, and then select OK.
After enabling the add-in, the Power Pivot tab should now be visible in Excel.
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Your lawn is twice as large as your neighbor’s lawn. You both start cutting your lawn with the same model, self-propelled lawn mower (requiring the same force) beginning at 9:00am on a Saturday morning. You finish cutting your lawn at 11:00 am. What time will your neighbor finish cutting her lawn if you are equally powerful?
Your neighbor will finish cutting her lawn at 10:00 am, which is one hour after both of you started.
Since your lawn is twice as large as your neighbor's lawn, it takes you a certain amount of time to cut it, which we can analyze to determine when your neighbor will finish cutting her lawn.
You started cutting your lawn at 9:00 am and finished at 11:00 am. This means it took you 2 hours to complete the task. Since your neighbor's lawn is half the size of yours, it will take her half the amount of time to finish cutting her lawn, assuming you both exert the same force using the same self-propelled lawn mower.
To calculate the time it will take your neighbor to cut her lawn, simply divide your time (2 hours) by 2. This gives us 1 hour. Your neighbor started cutting her lawn at the same time you did, 9:00 am, and will take 1 hour to complete the task.
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A mass of 100 kg is 100 m away from a mass of 50 kg. Calculate the force of attraction between the masses. Show your work
The force of attraction between the two masses is [tex]3.335 \times 10^{-8} N[/tex].
The force of attraction between two masses is given by the gravitational force equation, which is expressed as:
[tex]$F = G \cdot \frac{m_1 \cdot m_2}{r^2}$[/tex]
where F is the force of attraction, G is the gravitational constant ([tex]$6.67 \times 10^{-11} , \text{N}\cdot\text{m}^2/\text{kg}^2$[/tex]), [tex]m_1[/tex]1 and [tex]m_2[/tex] are the masses of the two objects, and r is the distance between the centers of the two masses.
In this case, [tex]m_1[/tex] = 100 kg, [tex]m_2[/tex] = 50 kg, and r = 100 m. Substituting these values into the equation, we get:
[tex]$F = 6.67 \times 10^{-11} , \text{N}\cdot\text{m}^2/\text{kg}^2 \cdot \frac{(100 , \text{kg}) \cdot (50 , \text{kg})}{(100 , \text{m})^2}$[/tex]
[tex]F = 3.335 \times 10^{-8} N[/tex]
It is worth noting that the force of attraction between the two masses is very small, which is due to the large distance between them. The gravitational force decreases rapidly with distance, so as the distance between the two masses increases, the force of attraction decreases as well.
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ii. how long it takes to travel 294 m below the point of projection.
It takes 10 seconds for the stone to travel 294 m below the point of projection. That's how long it take to travel.
How do we calculate how long it take to travel 294 m below the point of projection.?The equation to use to find how long it take to travel 294m below the point of projection is:
4.9t² - 19.6t - 294 = 0
We need the quadratic equation
t = [-b ± √(b² - 4ac)] / (2a)]
a = 4.9, b = -19.6, and c = -294.
t = [19.6 ± √((-19.6)² - 44.9(-294))] / (2×4.9)
t = [19.6 ±√384.16 + 5745.6)] / 9.8
t = [19.6 ± √(6129.76)] / 9.8
t = [19.6 ± 78.3] / 9.8
The possible answers are;
t1 = (19.6 + 78.3) / 9.8 = 10 seconds
t2 = (19.6 - 78.3) / 9.8 = -6 seconds
considerinf that the answer cannot be in the negative, therefore t₁ is the answer.
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A 2. 5 kg block initially at rest is pulled to the
right along a horizontal, frictionless surface
by a constant, horizontal force of 12. 3 N.
Find the speed of the block after it has
moved 2. 9 m
The speed of the block after it has moved 2.9 m is approximately 5.14 m/s.
We can use the work-energy principle to find the speed of the block after it has moved 2.9 m. The work-energy principle states that the net work done on an object is equal to its change in kinetic energy.
Since there is no friction acting on the block, the net work done on it is equal to the work done by the applied force:
Net work = Work done by applied force = Fd
where F is the applied force and d is the distance moved by the block.
The change in kinetic energy of the block is given by:
Δ[tex]K = 1/2 mv^2 - 1/2 m(0)^2 = 1/2 mv^2[/tex]
where m is the mass of the block and v is its final velocity.
Since the net work done on the block is equal to its change in kinetic energy, we can set these two expressions equal to each other:
[tex]Fd = 1/2 mv^2[/tex]
Solving for v, we get:
[tex]v = \sqrt{(2Fd/m)[/tex]
Substituting the given values, we get:
[tex]v = \sqrt{(2 *12.3 N * 2.9 m / 2.5 kg)} = 5.14 m/s[/tex]
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. a tire 0.500 m in radius rotates at a constant rate of 200 revolutions per minute. find the speed and acceleration of a small stone lodged in the tread of the tire
The speed of the small stone lodged in the tire's tread is approximately 10.47 m/s, and its acceleration is approximately 219.35 m/s².
We need to find the speed and acceleration of a small stone lodged in the tread of a tire with a 0.500 m radius, rotating at 200 revolutions per minute.
First, let's convert the revolutions per minute (rpm) to radians per second (rad/s):
200 rpm * (2π radians/1 revolution) * (1 minute/60 seconds) ≈ 20.94 rad/s
Now, we can find the linear speed (v) of the stone using the formula:
v = rω, where r is the radius, and ω is the angular velocity in rad/s.
v = 0.500 m * 20.94 rad/s ≈ 10.47 m/s
Next, we'll find the centripetal acceleration (a_c) of the stone using the formula:
a_c = rω²
a_c = 0.500 m * (20.94 rad/s)² ≈ 219.35 m/s²
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All the fossils that have been found over time are called the
All the fossils that have been found over time are collectively called the: fossil record.
The fossil record represents the preserved remains or traces of organisms from the past, providing valuable information about the history of life on Earth. It allows scientists to study the evolution of species, their distribution over time, and how they adapted to their environments.
The fossil record is not complete, as it depends on factors such as preservation conditions and the likelihood of a particular organism leaving behind fossils. However, it still offers a glimpse into the vast diversity of life that has existed throughout Earth's history, enabling researchers to make connections between extinct and living species.
In conclusion, the term for all the fossils that have been found over time is the fossil record. It serves as a crucial source of information for understanding the development of life on our planet, despite its inherent incompleteness due to various factors affecting fossil preservation.
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5. A certain dog whistle has a frequency of 35. 1 kHz. A person blows the whistle while riding in the back of a "convertible" airplane with a velocity of 126 m/s, north. With what minimum velocity must a person in a second airplane fly in order for the sound to be shifted into the audible frequency range? (speed of sound in air is 343 m/s)
In this scenario, we need to use the Doppler effect equation to calculate the minimum velocity required for the sound to be heard. The Doppler effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source.
The equation we will use is:
f' = f (v + vobs) / (v - vs)
Where f is the original frequency (35.1 kHz), v is the velocity of sound (343 m/s), vobs is the velocity of the observer (126 m/s), and vs is the velocity of the source (which is assumed to be zero in this case).
To find the new frequency, f', that would be heard by the second airplane, we need to solve for v2, the velocity of the second airplane. We also need to know the range of audible frequencies, which is typically between 20 Hz and 20 kHz.
If we plug in the given values, we get:
f' = 35.1 kHz (343 m/s + 126 m/s) / (343 m/s - v2)
Simplifying this equation gives:
f' = 1.304 + 0.00367v2
To find the minimum velocity that would put the frequency in the audible range, we can set f' equal to 20 kHz:
20 kHz = 1.304 + 0.00367v2
Solving for v2 gives:
v2 = 5,355 m/s
This means that the second airplane must fly at a minimum velocity of 5,355 m/s in order for the sound to be shifted into the audible frequency range. This is obviously impossible, so the whistle would not be heard by the second airplane.
In conclusion, the Doppler effect is a fascinating phenomenon that can help us understand how waves behave when the observer or source is in motion. By using the Doppler equation, we can calculate the shift in frequency and determine whether a sound will be audible or not. In this particular scenario, we see that the minimum velocity required for the sound to be heard is far beyond what is physically possible.
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If radiation has a frequency of 3. 0 X 1015 Hz and it strikes a material, what is the energy of each incident photon
The energy of each incident photon with a frequency of 3.0 x [tex]10^{15[/tex]Hz is approximately 1.99 x[tex]10^{-18[/tex] Joules.
The energy of a photon can be calculated using the formula:
E = h * f
where:
E is the energy of the photon,
h is Planck's constant (approximately 6.626 x [tex]10^{-34[/tex] J*s), and
f is the frequency of the radiation.
Given:
f = 3.0 x[tex]10^{15[/tex] Hz (frequency of the radiation)
Let's calculate the energy of each incident photon:
E = (6.626 x [tex]10^{-34[/tex] J*s) * (3.0 x [tex]10^{15[/tex] Hz)
E ≈ 1.99 x [tex]10^{-18[/tex]J
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single convex lenses can be used to make images of distant objects. will these images be real or imaginary? will they be inverted or upright? will they be larger or smaller than the original object? where does the image occur relative to the focus? (specify which side of the lens)
Single convex lenses can create real and inverted images of distant objects, with size depending on object distance and focal length. The image appears on the opposite side of the lens from the object, between the lens and its focus.
Single convex lenses can be used to make real and inverted images of distant objects. The size of the image depends on the distance of the object from the lens and the focal length of the lens.
If the object is very far away from the lens, the image will be small. The image will occur on the side of the lens opposite to the object and between the lens and its focus.
The image will be real and inverted because the convex lens converges the light rays that pass through it.
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The Really Big Dam is 1000 feet wide, holds back a depth of 60 feet of water, and the lake behind the dam extends back one quarter of a mile. The Very Big Dam is also 1000 feet wide, holds back a depth of 50 feet of water, and the lake behind the dam extends back for 2 miles.
If the dams were constructed in the same way, which dam had to be constructed to be strongest? (Assume the water levels do not vary seasonally. )
The strength of two dams is compared by calculating their potential energy based on the height of the water they hold back. The Very Big Dam has greater potential energy than the Really Big Dam, making it stronger.
To determine which dam is stronger, we need to compare their potential energy due to the water they are holding back. The potential energy of the water is given by the formula:
PE = mgh
where PE is the potential energy, m is the mass of the water, g is the acceleration due to gravity, and h is the height of the water.
Since the dams are the same width, we can assume they have the same mass of water. Therefore, the potential energy depends only on the height of the water.
The height of the water in the Really Big Dam is 60 feet, and the lake extends back one-quarter of a mile or 1320 feet. Therefore, the potential energy of the water is:
PE1 = mgh = (mass of water) x g x h
[tex]PE1 = (1000 ft \times 1320 ft \times 60 ft) \times 62.4 \;lb/ft^3 \times 32.2\; ft/s^2[/tex]
The height of the water in the Very Big Dam is 50 feet, and the lake extends back two miles, or 10560 feet. Therefore, the potential energy of the water is:
PE2 = mgh = (mass of water) x g x h
[tex]PE2 = (1000\; ft \times 10560\; ft \times 50 ft) \times 62.4 \;lb/ft^3 \times 32.2\; ft/s^2[/tex]
Calculating the two potential energies, we find that PE2 is greater than PE1. Therefore, the Very Big Dam had to be constructed to be strongest.
In summary, to determine which dam is stronger, we compare its potential energy due to the water they are holding back. Since the dams have the same width, the potential energy depends only on the height of the water.
Calculations show that the potential energy of the water held by the Very Big Dam is greater than the Really Big Dam, making it the stronger of the two dams.
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A mass attached to the end of a spring is set in motion. The mass is observed to oscillate up and down, completing 24 complete cycles every 6. 00 s. What is the period of the oscillation?
What is the frequency of the oscillation?
A mass attached to the end of a spring is set in motion, the mass is observed to oscillate up and down, completing 24 complete cycles every 6. 00 s, the period of the oscillation: 0.25 seconds.
The mass attached to the end of a spring completes 24 cycles in 6.00 seconds. To determine the period of the oscillation, we need to find the time taken for one complete cycle. The period (T) is calculated by dividing the total time by the number of cycles, which is:
T = total time / number of cycles = 6.00 s / 24 cycles = 0.25 s per cycle.
The period of the oscillation is 0.25 seconds.
Now, to find the frequency of the oscillation, we need to determine the number of cycles that occur in one second. The frequency (f) is the inverse of the period:
f = 1 / T = 1 / 0.25 s = 4 cycles per second (Hz).
The frequency of the oscillation is 4 Hz.
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A marble is thrown norizontally from a tarble top with a velocity of 1. 50m/s. The marble falls 0. 70m away from te table'ede. How high is the lab table? what is the marble's velocity just before it hits the floor
The marble's velocity just before it hits the floor is approximately 4.83 m/s.
To find the height of the lab table, we can use the following terms:
1. Horizontal velocity (Vx): 1.50 m/s
2. Horizontal distance (d): 0.70 m
First, we need to find the time it takes for the marble to fall 0.70m horizontally. We can do this using the equation: d = Vx * t
0.70 m = 1.50 m/s * t
t = 0.70 m / 1.50 m/s = 0.4667 s
Now, we can use this time to find the height (h) of the table using the vertical motion equation: h = 0.5 * g * t^2, where g is the acceleration due to gravity (9.81 m/s^2).
h = 0.5 * 9.81 m/s^2 * (0.4667 s)^2
h ≈ 1.067 m
So, the height of the lab table is approximately 1.067 meters.
To find the marble's velocity just before it hits the floor, we need to calculate its vertical velocity (Vy) using the equation: Vy = g * t
Vy = 9.81 m/s^2 * 0.4667 s
Vy ≈ 4.57 m/s
Now, we can find the marble's total velocity (V) using the Pythagorean theorem: V = √(Vx^2 + Vy^2)
V = √((1.50 m/s)^2 + (4.57 m/s)^2)
V ≈ 4.83 m/s
Therefore, the marble's velocity just before it hits the floor is approximately 4.83 m/s.
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Lab: Newton's Laws of Motion
Assignment: Lab Report
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I finished and wanted to give my lab report if anyone had trouble or needed it :)
Thank you for offering your lab report to others! However, it's important to remember that sharing your work can lead to academic misconduct if others use your report as their own.
It's important for everyone to complete their assignments independently and to not share their work with others.
It's also important to understand the concepts behind Newton's Laws of Motion rather than relying solely on someone else's report.
That being said, if anyone is struggling with the lab, it's best to seek help from the instructor or a tutor. Good luck with your assignment!
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A) When a submarine dives to a depth of 500 m, how much pressure, (in Pa) must it's hull be able to withstand? b) How many times greater is this pressure than the pressure at the surface. Recall pressure at the surface is atmospheric pressure at sea level which equals 14. 7 psi (101 kPa). Hint when determining how many times greater remember How many times greater factor = BIGGER/ smaller)
A submarine diving to a depth of 500 m would experience a pressure of 5,068,625 Pa on its hull, which is approximately 50 times greater than the atmospheric pressure at sea level.
a) When a submarine dives to a depth of 500 m, the pressure on its hull increases due to the weight of the water above it.
The pressure at this depth can be calculated using the formula [tex]P = \rho gh[/tex], where ρ is the density of seawater, g is the acceleration due to gravity, and h is the depth.
Plugging in the values, we get P = (1025 kg/m³)(9.81 m/s²)(500 m) = 5,068,625 Pa.
b) To determine how many times greater the pressure is at a depth of 500 m compared to the surface, we can divide the pressure at 500 m by the atmospheric pressure at sea level.
Converting 14.7 psi to Pa, we get 101,325 Pa. Dividing 5,068,625 by 101,325 gives us approximately 50 times greater.
In summary, a submarine diving to a depth of 500 m would experience a pressure of 5,068,625 Pa on its hull, which is approximately 50 times greater than the atmospheric pressure at sea level.
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Which planet is tilted on its side as it orbits the sun?.
Answer:
Uranus
Explanation:
which is a form of potential energy
The frequency of violet light is 7. 5 x 1014 hertz. What is its wavelength in a vacuum?
The wavelength of violet light in a vacuum is approximately 3.997 x 10^-7 meters, which is equivalent to 399.7 nanometers.
The wavelength of the light in a vacuum can be calculated using the formula λ = c/f, where λ is the wavelength, c is the speed of light in a vacuum (299,792,458 meters per second), and f is the frequency of the light.
Using this formula, we can find the wavelength of the violet light as follows:
λ = c/f
λ = 299,792,458 m/s / 7.5 x 10^14 Hz
λ = 3.997 x 10^-7 meters
Therefore, the wavelength of violet light in a vacuum is approximately 3.997 x 10^-7 meters, which is equivalent to 399.7 nanometers.
In summary, the frequency of violet light is a measure of how fast it oscillates, and its wavelength in a vacuum can be calculated using the speed of light and frequency of the light. Knowing the wavelength of a particular color of light is useful in many fields, including astronomy, physics, and optics.
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1kg of water will occupy minimum space at
A) 0°C
B) 100°C
C) -4°C
D) 4°C
Answer and Explanation:
According to the principle of maximum density, water has its highest density at 4°C. This means that 1kg of water will occupy minimum space at a temperature of 4°C. At this temperature, the volume of water is at its lowest, and any further cooling or heating will cause it to expand.
This principle is due to the unique properties of water molecules. As the temperature decreases from room temperature, the molecules begin to slow down and move closer together. However, below 4°C, hydrogen bonding between the molecules begins to dominate, causing them to form a crystal-like structure and expand.
At 0°C, water freezes and expands by about 9%, making it less dense than liquid water. At 100°C, water boils and turns into steam, which occupies much more space than liquid water. At -4°C, the water is still in a liquid state but is not at its maximum density.
In conclusion, the correct answer is 4°C, as this is the temperature at which 1kg of water will occupy minimum space.
1 kg of water will occupy minimum space of 1 m³at 25°C.
What is Density ?Density is the ratio of mass to volume. it tells how much mass a body is having for its unit volume. for example egg yolk has 1027kg/m³ of density, means if we collect numbers of egg yolk and keep it in a container having volume 1 m³ then total amount of mass it is having will be 1027kg. Density is a scalar quantity. when we add egg yolk into the water, egg yolk has greater density than water( 997 kg/m³), because of higher density of egg yolk it contains higher mass in same volume as water. hence due to higher mass higher gravitational force is acting on the egg yolk therefore it goes down on the inside the water. water will float upon the egg yolk. same situation we have seen when we spread oil in the water. ( in that case water has higher density than oil. that's why oil floats on the water)
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The principle of superposition can be used to determine:.
The principle of superposition can be used to determine the net effect of multiple individual effects on a physical system. It is a fundamental principle in physics and is used to analyze the behavior of waves, electric and magnetic fields, and other physical phenomena.
In essence, the principle of superposition states that when two or more waves, forces, or fields interact with each other, the net effect is the sum of the individual effects of each wave, force, or field.
This principle applies to both linear and nonlinear systems, and it is a crucial tool for understanding complex physical systems.
For example, the principle of superposition can be used to determine the resulting wave pattern when two or more waves of different frequencies, amplitudes, and directions interact with each other. It can also be used to calculate the net electric or magnetic field at a given point in space due to multiple charges or currents.
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