Answers
I think it’s 9ms2. Sorry if I’m wrong.
Related Questions
What is the ratio of the displacement amplitudes of two sound waves given that they are both5.0 kHz but have a 3.0 dB intensity level difference?
Answers
Answer:
The ratio of the displacement amplitudes of two sound waves is 1.16.
Explanation:
Given that,
Frequency = 5.0 kHz
Intensity level difference = 3.0 dB
We know that,
The sound intensity is inversely proportional to the square of distance.
[tex]I\propto\dfrac{1}{r^2}[/tex]
The sound intensity for first wave is
[tex]\beta_{1}=10\log\dfrac{I_{1}}{I_{0}}[/tex]...(I)
The sound intensity for second wave is
[tex]\beta_{2}=10\log\dfrac{I_{2}}{I_{0}}[/tex]...(II)
We need to calculate the ratio of intensity
From equation (I) and (II)
[tex]\beta_{2}-\beta_{1}=10\log\dfrac{I_{2}}{I_{0}}-10\log\dfrac{I_{1}}{I_{0}}[/tex]
[tex]\Delta \beta=10\log(\dfrac{I_{2}}{I_{1}})[/tex]
Put the value into the formula
[tex]3.0=10\log(\dfrac{I_{2}}{I_{1}})[/tex]
[tex]\dfrac{I_{2}}{I_{1}}=e^{\dfrac{3.0}{10}}[/tex]
[tex]\dfrac{I_{2}}{I_{1}}=1.34[/tex]
We need to calculate the ratio of the displacement
Using formula of displacement
[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{\dfrac{I_{2}}{I_{1}}}[/tex]
Put the value into the formula
[tex]\dfrac{r_{1}}{r_{2}}=\sqrt{1.34}[/tex]
[tex]\dfrac{r_{1}}{r_{2}}=1.16[/tex]
Hence, The ratio of the displacement amplitudes of two sound waves is 1.16.
At a distance of 10.0 m from a loudspeaker, the sound intensity level is measured to be 70 dB. At what distance from the source will the intensity be 40 dB?
Answers
Answer:
At a distance of 100 m from the source the intensity will be 40 dB.
Explanation:
Sound intensity is the acoustic power transferred by a sound wave per unit area normal to the direction of propagation.
The sound intensity depends on the power of the sound source, where the higher the power the greater the intensity, the distance to the sound source, the greater the distance being the lower the intensity, and the nature of the transmission medium.
The conversion between intensity and decibels corresponds to:
[tex]L=10*log\frac{I}{I0}[/tex]
where I0 = 10⁻¹² W/m² and corresponds to a level of 0 decibels therefore.
In this case, you can apply the following relationship between two intensities and distance, considering that the intensity of the sound level decreases with distance:
[tex]L1 - L2=10*log\frac{I1}{I0} - 10*log\frac{I2}{I0}[/tex]
[tex]L1 - L2=10*(log\frac{I1}{I0} - *log\frac{I2}{I0})[/tex]
[tex]L1 - L2=10*[log(\frac{I1}{I0}\frac{I0}{I2})][/tex]
[tex]L1 - L2=10*[log(\frac{I1}{I2})][/tex]
Being L1= 70 dB and L2= 40 dB
[tex]70 dB - 40 dB=10*[log(\frac{I1}{I2})][/tex]
[tex]30=10*[log(\frac{I1}{I2})][/tex]
[tex]\frac{30}{10} =log(\frac{I1}{I2})[/tex]
[tex]3=log(\frac{I1}{I2})[/tex]
[tex]10^{3} =\frac{I1}{I2}[/tex]
[tex]1,000=\frac{I1}{I2}[/tex]
The intensity is inversely proportional to the square of the distance to the source. The relationship between the intensities I1 and I2 at distances d1 and d2 respectively is:
[tex]\frac{I1}{I2} =\frac{d2^{2} }{d1^{2} }[/tex]
Then:
[tex]1,000=\frac{d2^{2} }{d1^{2} }[/tex]
Being d1= 10 m
[tex]1,000=\frac{d2^{2} }{10^{2} }[/tex]
[tex]1,000=\frac{d2^{2} }{100}[/tex]
1,000*100= d2²
10,000= d2²
√10,000= d2
100 m= d2
At a distance of 100 m from the source the intensity will be 40 dB.
Choose the best description of a magnet
O A. Something with magnetic domains
O B. Something that attracts iron
O C. Iron, cobalt, and nickel
O D. Something that becomes magnetic with the application of a
current
Answers
Answer:
Explanation:
Something that attracts iron...duuhhh
The smallest living organism discovered so far is called a mycoplasm. Its mass is estimated as
1.0 × 10–16 g. Express this mass in a. petagrams. b. femtograms. c. attograms.
Answers
Answer:
10-16= -4 so, 1.0x-4 is -4. So the answer is -4.
Explanation:
How would you measure and compare the size of the two balls ?
Answers
Answer:
you would use a measuring tape and compare your answers
1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal. The coefficient of friction between the crate and the floor is 0.12.
a) Calculate the net force and acceleration of the crate.
b) If the crate was moving at 1.0 m/s when it was pulled, what would be its velocity after pulling it for 5.0 s?
Answers
Answer:
(a) The net force is 80.394 N
The acceleration of the crate is 0.804 m/s²
(b) the final velocity of the crate is 5.02 m/s
Explanation:
Given;
mass of the crate, m = 100 kg
applied force, F = 250 N
angle of inclination, θ = 45°
coefficient of friction, μ = 0.12
Applied force in y-direction, [tex]F_y = Fsin \theta = 250sin45 = 176.78 \ N[/tex]
Applied force in x-direction, [tex]F_x = Fcos \theta = 250cos45 = 176.78 \ N[/tex]
The normal force is calculated as;
N + Fy -W = 0
N = W - Fy
N = (100 x 9.8) - 176.78
N = 980 - 176.78 = 803.22 N
The frictional force is given by;
Fk = μN
Fk = 0.12 x 803.22
Fk = 96.386 N
(a) The net force is given by;
[tex]F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N[/tex]
Apply Newton's second law of motion;
F = ma
[tex]a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2[/tex]
(b) the velocity of the crate after 5.0 s
[tex]F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s[/tex]
Which is a belief held by sociologists who work from a social-conflict
perspective?
O A. The best approach for a study is from a micro-level orientation.
O B. Personal background has little impact on how individuals react
with one another.
C. Some social patterns are helpful, while others are harmful.
D. Data are irrelevant to the study of sociology.
SUBMIT
Answers
Answer:
C. Some social patterns are helpful, while others are harmful.
Explanation:
Hope this was helpful, Have an amazing,spooky Halloween!!
A wire loop with 3030 turns is formed into a square with sides of length ss . The loop is in the presence of a 1.20 T1.20 T uniform magnetic field B⃗ B→ that points in the negative yy direction. The plane of the loop is tilted off the x-axisx-axis by θ=15∘θ=15∘ . If i=1.10 Ai=1.10 A of current flows through the loop and the loop experiences a torque of magnitude 0.0256 N⋅m0.0256 N⋅m , what are the lengths of the sides ss of the square loop, in centimeters?
Answers
Answer:
2.59 cm
Explanation:
The torque τ on a current carrying loop of wire is given by τ = NiABsinθ where N = number of turns of loop, i = current in loop, A = area of loop and B = magnetic field.
Now, given that τ = 0.0256 Nm, i = 1.10 A, B = 1.20 T,N = 30 and since the loop is tilted 15° off the x-axis and the magnetic field points in the negative y- direction, the angle between the normal to the loop and the magnetic field is thus 90° - 15° = 75°. So, θ = 75°.
We now find the area of the loop A from
τ = NiABsinθ
A = τ/NiBsinθ
substituting the values of the variables, we have
A = 0.0256 Nm/30 × 1.10 A × 1.20 T × sin75°
A = 0.0256 Nm/38.25
A = 6.69 × 10⁻⁴ m²
Since the loop is a square, with length of side L, its area A = L² and
L = √A
= √(6.69 × 10⁻⁴ m²)
= 2.59 × 10⁻² m
converting to cm, we have
L = 2.59 × 10⁻² m × 100 cm/m
L = 2.59 cm
So, the lengths of sides of the loop is 2.59 cm
To begin your activity, open this Tracker experiment: Car Round Trip. Click play to watch the video. The other video controls allow you to rewind the video or step forward or backward one frame at a time.
Answers
Answer:
The car appears to have a constant, positive acceleration for most of the video clip.
61) A bicycle wheel of radius 0.36 m and mass 3.2 kg is set spinning at 4.00 rev/s. A very light bolt is attached to extend the axle in length, and a string is attached to the axle at a distance of 0.10 m from the wheel. Initially the axle of the spinning wheel is horizontal, and the wheel is suspended only from the string. We can ignore the mass of the axle and spokes. At what rate will the wheel process about the vertical
Answers
Answer:
The rate the wheel will process about the vertical is 2.86 RPM
Explanation:
Given;
radius of the bicycle wheel, R = 0.36 m
mass of the wheel, m = 3.2 kg
angular velocity, ω = 4 rev/s
The rate at which the wheel will process about the vertical is given by;
Ф = τ/L
Where;
τ is the torque
L is the angular momentum
τ = Fr
τ = mgr = 3.2 x 9.8 x 0.1 = 3.126 N.m
L = Iω = MR²ω
L = 3.2 x (0.36)² x (4 x 2π)
L = 10.4244 kg.m²/s
Ф = τ/L
Ф = (3.126) / (10.4244)
Ф = 0.29987 rad/s
Ф = 0.29987 rad/s x (60 / 2π)
Ф = 2.86 RPM
Therefore, the rate the wheel will process about the vertical is 2.86 RPM
I need help with this answer
Answers
The gravitational force acting on a lead ball is much larger than that acting on a wooden ball of the same size. Which statement is true about when they are dropped
Answers
Complete question is;
The gravitational force acting on a lead ball is much larger than that acting on a wooden ball of the same size. Which statement is true about when they are dropped?
A) The acceleration due to gravity is the same for both balls.
B) The lead ball’s acceleration due to gravity is larger than the wooden ball’s.
C) The lead ball’s acceleration due to gravity is less than the wooden ball’s.
Answer:
Option A) The acceleration due to gravity is the same for both balls.
Explanation:
Density of Lead is 11.34 g/cm³ while density of hard wood is around 1.3 g/cm³
Thus, it means lead is the heavier object when both have same masses.
We are told that the wood and lead both have same size but from the density given above, it's clear that the mass of the lead ball is more than that of the wooden ball and thus, the force from would be more for the lead ball than the wooden ball.
Now, If we were to measure the position of both balls when falling, we will observe that they do not fall fall with a constant speed. This is because they fall with a constant acceleration and as such the speed increases.
Therefore both balls will hit the ground at the same time because they both start from a position of rest and both will have the same acceleration.
A physics student throws a softball straight up into the air. The ball was in the air for a total of 3.56 s before it was caught at its original position. What is the initial velocity of the ball? Consider upwards to be the positive direction.
Answers
Answer:
The initial velocity of the softball is 14.711 meters per second.
Explanation:
This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.
From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:
[tex]y = y_{o} + v_{o}\cdot t + \frac{1}{2}\cdot g \cdot t^{2}[/tex] (Eq. 1)
Where:
[tex]y_{o}[/tex] - Initial height of the softball, measured in meters.
[tex]y[/tex] - Final height of the softball, measured in meters.
[tex]v_{o}[/tex] - Initial velocity of the softball, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]y = y_{o}[/tex], [tex]t = 3.56\,s[/tex] and [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], the initial velocity of the softball is:
[tex]v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0[/tex]
[tex]3\cdot v_{o} -44.132\,m= 0[/tex]
[tex]v_{o} = 14.711\,\frac{m}{s}[/tex]
The initial velocity of the softball is 14.711 meters per second.
g A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 1.97 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 1680 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 4.34 V/m, (b) in the negative z direction and has a magnitude of 4.34 V/m, and (c) in the positive x direction and has a magnitude of 4.34 V/m
Answers
Answer:
a) 1.22*10^-18 N in the positive z direction
b) 1.65*10^-19 N in the negative z direction
c) (6.94*10^-19 N) in the positive x direction + (5.30*10^-19 N) in the positive z direction
Explanation:
See attachment for calculations
(a) The electromagnetic force on the proton is 1.224 × 10⁻¹⁸ [tex]\hat k[/tex] N
(b) The force is 1.65 × 10⁻¹⁹ [tex](-\hat k)[/tex] N
(c) The force is (6.94 [tex]\hat i[/tex] + 5.29 [tex]\hat k[/tex]) × 10⁻¹⁹ N
Electromagnetic force on the proton:Given a proton moving in the positive y-direction with a speed of :
v = 1680 m/s [tex]\hat j[/tex]
The magnetic field is in the negative x-direction with magnitude:
B = 1.97 mT [tex](-\hat i)[/tex]
(a) Electric field applied in positive z-direction :
E = 4.34 V/m [tex]\hat k[/tex]
The net force on the proton is iven by:
F = q (E + v×B)
where q is the charge on proton, given by:
q = 1.6×10⁻¹⁹ C
So,
F = 1.6×10⁻¹⁹( 4.34 [tex]\hat k[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )
F = 1.6×10⁻¹⁹ ( 4.34 [tex]\hat k[/tex] + 3.309 [tex]\hat k[/tex])
F = 1.224 × 10⁻¹⁸ [tex]\hat k[/tex] N
(b) Electric field applied in negative z-direction :
E = 4.34 V/m [tex](-\hat k)[/tex]
The net force on the proton is iven by:
F = q (E + v×B)
where q is the charge on proton, given by:
q = 1.6×10⁻¹⁹ C
So,
F = 1.6×10⁻¹⁹( 4.34 [tex](-\hat k)[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )
F = 1.6×10⁻¹⁹ ( 4.34 [tex](-\hat k)[/tex] + 3.309 [tex]\hat k[/tex])
F = 1.65 × 10⁻¹⁹ [tex](-\hat k)[/tex] N
(c) Electric field applied in positive x-direction :
E = 4.34 V/m [tex]\hat i[/tex]
The net force on the proton is iven by:
F = q (E + v×B)
where q is the charge on proton, given by:
q = 1.6×10⁻¹⁹ C
So,
F = 1.6×10⁻¹⁹( 4.34 [tex]\hat i[/tex] + 1680 [tex]\hat j[/tex] × 1.97×10⁻³ [tex](-\hat i)[/tex] )
F = 1.6×10⁻¹⁹ ( 4.34 [tex]\hat i[/tex] + 3.309 [tex]\hat k[/tex])
F = (6.94 [tex]\hat i[/tex] + 5.29 [tex]\hat k[/tex]) × 10⁻¹⁹ N
Learn more about electromagnetic force:
https://brainly.com/question/807785?referrer=searchResults
An element with 7 valence electrons will most likely-
be found in group 17 and be highly reactive
be found in group 7 and be highly reactive
be found in group 17 and be inert
be found in group 17 and be somewhat reactive
Answers
Answer: be found in group 17 and be highly reactive
Explanation:
Elements are distributed in groups and periods in a periodic table.
Elements that belong to same groups will show similar chemical properties because they have same number of valence electrons.
Flourine, chlorine, bromine and iodine are elements which belong to Group 17. All of them contain 7 valence electrons each and need one electron to complete their octet.
The chemical reactivity of elements is governed by the valence electrons present in the element and thus all of them are highly reactive.
A Labrador retriever runs 50 m in 7.2 s to retrieve a toy bird. The dog then runs half way
back in 3.85 s. Determine the average speed and velocity of the dog
Answers
Answer:
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
Explanation:
From Physics we must remember the definitions of average speed and average velocity, both measured in meters per second. Velocity is a vectorial quantity, that is, it has both magnitude and direction, whereas speed is an scalar quantity, which is a quantity that is represented solely by its magnitude. We assume that dog moves at constant speed.
For the case of the dog, we get that average speed and average velocity of the animal are, respectively:
Average velocity:
[tex]\vec v_{avg} = \frac{1}{\Delta t}\cdot (\vec r_{B}-\vec r_{A})[/tex] (Eq. 1)
Where:
[tex]\Delta t[/tex] - Travelling time of the dog, measured in seconds.
[tex]\vec r_{A}[/tex] - Initial vector position of the dog, measured in meters.
[tex]\vec r_{B}[/tex] - Final vector position of the dog, measured in meters.
Average speed:
[tex]v_{avg} = \frac{1}{\Delta t} \cdot (s_{A}+s_{B})[/tex] (Eq. 2)
Where [tex]s_{A}[/tex] and [tex]s_{B}[/tex] are the travelled distances of each stage, measured in meters.
If we know that [tex]\Delta t = 11.05\,s[/tex], [tex]\vec r_{A} = 0\,\hat{i}\,\,\,[m][/tex] and [tex]\vec r_{B} = 25\,\hat{i}\,\,\,[m][/tex], [tex]s_{A} = 50\,m[/tex] and [tex]s_{B} = 25\,m[/tex], average velocity and average speed are, respectively:
[tex]\vec v_{avg} = \frac{1}{11.05\,s}\cdot (25\,\hat{i})\,\,\,[m][/tex]
[tex]\vec v_{avg} = 2.262\,\hat{i}\,\,\,\left[\frac{m}{s} \right][/tex]
[tex]v_{avg} = \frac{75\,m}{11.05\,s}[/tex]
[tex]v_{avg} = 6.787\,\frac{m}{s}[/tex]
The average velocity and average speed of the dog are 2.262 meters per second and 6.787 meters per second, respectively.
What is the mass of an object that has a volume of 56 ml and a density of 1.24 g/ml
Answers
Answer:
The answer is 69.44 gExplanation:
The mass of a substance when given the density and volume can be found by using the formula
mass = Density × volumeFrom the question
volume = 56 ml
density = 1.24 g/ml
We have
mass = 56 × 1.24
We have the final answer as
69.44 gHope this helps you
If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
Answers
Complete Question
In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
Answer:
The speed of the helicopter is [tex]u = 7.73 \ m/s [/tex]
Explanation:
From the question we are told that
The height at which he let go of the brief case is h = 130 m
The time taken before the the brief case hits the water is t = 6 s
Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as
[tex]s = h+ u t + 0.5 gt^2[/tex]
Here s is the distance covered by the bag at sea level which is zero
[tex]0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2[/tex]
=> [tex]0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2[/tex]
=> [tex]u = \frac{-130 + (0.5 * 9.8 * 6^2) }{6}[/tex]
=> [tex]u = 7.73 \ m/s [/tex]
Predict how the total pressure varies during the gas-phase reaction 2 ICl(g)+H2 (g)→I 2 (g)+2 HCl(g) in a constant-volume container.
Answers
Answer:
Explanation:
First thing we should note, is remember Avogadro's law, which states that "equal no of moles of gas occupy equal volume". And thus, the number of moles on both the sides of above are the same, this means that the number of particles that are colliding with wall due to which pressure will be present. Also, if the number of moles are the same, which they are, then the pressure will also be the same.
The Initial number of moles is(see first attachment), without ammonia.
At a point in the reaction, α is the amount or number of N2 that has reacted. Therefore, the total number of gas moles is get by is(see second attachment)
The first two terms represents a change in number of moles of reactant while the last one stands for the number of ammonia produced.
A truck is speeding up as it travels on an interstate. The truck's momentum (in kg · m/s) is proportional to the truck's speed (in m/s).At some moment the truck's momentum is 50600 kg · m/s and the truck's speed is 23 m/s. At this moment, the truck's momentum (in kg · m/s) is how many times as large as the truck's speed (in m/s)?IncorrectThis means that as the truck travels, the truck's momentum (in kg · m/s) is always Incorrect times as large as the truck's speed (in m/s).If the truck is traveling at 16 m/s, what is the truck's momentum?
Answers
Answer:
At the moment, the truck's momentum (in kg · m/s) is 2200 times as large as the truck's speed (in m/s).
This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).
If the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.
Explanation:
From the question,
The truck's momentum (in kg · m/s) is proportional to the truck's speed (in m/s).
Let the truck's momentum be P and the truck's speed be v,
Then we can write that
P∝v
Then,
P = kv
Where k is the proportionality constant
From the question,
At some moment the truck's momentum is 50600 kg · m/s and the truck's speed is 23 m/s,
To determine how many times the truck's speed is as large as the truck's momentum at this moment, we will divide the truck's momentum by the speed, that is
50600 ÷ 23 = 2200
Hence, at the moment, the truck's momentum (in kg · m/s) is 2200 times as large as the truck's speed (in m/s).
Since, dividing the truck's momentum by the truck's speed gives the proportionality constant k (that is, P/v = k), then
This means that as the truck travels, the truck's momentum (in kg · m/s) is always 2200 times as large as the truck's speed (in m/s).
From
P = kv
Then, k = P/v
At a moment, P = 50600 kg · m/s and v = 23 m/s
∴ k = 50600 kg · m/s ÷ 23 m/s = 2200 kg
k = 2200 kg
To determine the truck's momentum if the truck is traveling at 16 m/s
From
P = kv
k = 2200 kg
v = 16 m/s
∴ P = 2200 kg × 16 m/s
P = 35200 kg · m/s
Hence, if the truck is travelling at 16 m/s, the truck's momentum is 35200 kg · m/s.
Which statements about potential energy are true?
▫ Gaining potential energy is always associated with a force field.
▫ A change in position always means that an object gains potential energy.
▫ There's only one kind of potential energy.
▫ Some kinds of potential energy are related to electric forces exerted by atoms and molecules.
Answers
Answer:
the answer is 1 and 4
Explanation
Plato users
For the potential energy, statement 1 and statement 4 are correct.
The potential energy of the object the energy of the object in its steady position. When the object is at rest, the energy of the object in that condition is called potential energy.
Let us consider an electron having charge [tex]e[/tex] is moving the distance [tex]d[/tex] in uniform electric field E.
Its potential energy can be written as,
[tex]P = eEd[/tex]
Where P is the potential energy and E is the electric field.
Hence, the potential energy of the electron is associated with the electric field.
The electric force can be written as,
[tex]F =eE[/tex]
Where [tex]F[/tex] is the electric force, [tex]E[/tex] is the electric field and [tex]e[/tex] is the charge on the electron.
So, the potential energy can be written as,
[tex]P=Fd[/tex]
Hence, the potential energy is related to electric force.
For more information, follow the link given below.
https://brainly.com/question/1413008.
The period of a wave is equal to the time it takes for one wavelength to pass by a fixed point. You stand on a pier watching water waves and see 10.9 waves pass by in a time of 28 seconds.What is the period of the water waves? Round to nearest .01 and do not include units in your answer.
Answers
Answer:
T = 2.57 s
Explanation:
The period of a wave is equal to the time it takes for one wavelength to pass by a fixed point.
No of waves observed = 10.9
Time taken, t = 28 s
We need to find the period of the water waves. Number of waves per unit time is called frequency. Let it is f. So,
[tex]f=\dfrac{n}{t}\\\\f=\dfrac{10.9}{28}\\\\f=0.389\ Hz[/tex]
If T is period of the wave. It is equal to the reciprocal of frequency.
[tex]T=\dfrac{1}{f}\\\\T=\dfrac{1}{0.389}\\\\T=2.57\ s[/tex]
So, the period of the water waves is 2.57 seconds.
How will the motion of an object that is moving to the right change, if it is pushed in the opposite direction with a greater force?
Оооо
The object will move at a constant speed in the same direction for a while and then slow down and stop.
The object will slow down for a while and then move at a slower constant speed in the same direction.
The object will slow down and then begin to move faster and faster in the opposite direction.
0 The object will speed up and then begin to move at a slower speed in the opposite direction.
Answers
A toy helicopter takes off and moves 6 m up and then 1 m back down. What
is the displacement of the helicopter?
A. 7 m
5 m up
C. 6 m
D. 1 m down
Answers
the answer is 5 m up AKA "B"
Derek is watching steam form swirling patterns above the boiling water in his beaker.
Which type of thermal energy transfer is he observing?
A.Conduction
B.Translation
C.Convection
D.Radiation
Answers
Answer:
Convection
Explanation:
Answer:
C. Convection
Explanation:
I just did it
The forces exerted on an object are shown. (3 points)
A box has an arrow pointing up labeled F and an arrow pointing down labeled 3 N.
If the net force on the object along the vertical plane is zero, which statement is correct?
F equals 3 N and the object moves up.
F equals 3 N and the object remains stationary.
F equals 0 N and the object moves down.
F equals 0 N and the object remains stationary.
Answers
Answer:
F equals 3 N and the object remains stationary. (second option in the list)
Explanation:
For sure to cancel acting forces, F must be 3N pointing up. But with regards to the object stationary or not, the question is tricky. We could have a ZERO net force applied, and the object moving at constant speed, which could still verify Newton's Laws. But considering the first answer option that refers to vertical motion upward where the object could be gaining potential energy, the most accurate response is that the force F has to be 3 N pointing up to make the object in equilibrium, and no motion in the vertical axis.
Answer: F equals 3 N and the object remains stationary.
Explanation:
Time it takes stone to fall from the height of 80 m is approximately equal to *
A. 1 s
B. 2 s
C. 4 s
D. 8 s
Answers
Answer:
D
Explanation:
Answer:
c.4s
Explanation:
A spring has natural length 16 cm. A force of 3 N is required to holdthe spring compressed compressed to 11 cm. Find the amount ofwork instretching the spring from 17 cm to 19 cm.
Answers
Answer:
W = 0.012 J
Explanation:
For this exercise let's use Hooke's law to find the spring constant
F = K Δx
K = F / Δx
K = 3 / (0.16 - 0.11)
K = 60 N / m
Work is defined by
W = F. x = F x cos θ
in this case the force and the displacement go in the same direction therefore the angle is zero and the cosine is equal to 1
W = ∫ F dx
W = k ∫ x dx
we integrate
W = k x² / 2
W = ½ k x²
let's calculate
W = ½ 60 (0.19 -0.17)²
W = 0.012 J
A cannon fires a shell straight upward; 1.6 s after it is launched, the shell is moving upward with a speed of 19 m/s. Assuming air resistance is negligible, find the speed (magnitude of velocity) of the shell at launch and 5.5 s after the launch.
Answers
Answer:
-19.259m/s
Explanation:
Given;
Final velocity = 19m/s
time t = 1.6s
u is the initial velocity
g is the acceleration due to gravity = 9.81m/s²
Using the equation of motion to first get the initial velocity of the shell:
v = u-gt
19 = u - (9.81)(1.6)
19 = u - 15.696
u = 19+15.696
u = 34.696m/s
The initial velocity of the shell is 34.696m/s
Next is to find the speed of the shell 5.5s after the launch
Using the equation of motion:
v = u-gt
v = 34.696-9.81(5.5)
v = 34.696 - 53.955
v = -19.259m/s
The negative value of the velocity shows that the velocity is travelling in the downward direction
a tiger leaps with an initial velocity of 55 km/hr at an angle of 13° with respect to the horizontal. what are the components of the tigers velocity?
Answers
Answer:
vₓ = 53.6 km/h
vy = 12.4 km/h
Explanation:
if we define two axis perpendicular each other with origin in the point represented by the tiger leaping (assuming we can treat it as a point mass) coincidently with the horizontal (x-axis) and vertical (y-axis) directions, we can obtain the components of the velocity in both independent directions.We can do it simply getting the projections of the velocity vector on both axes, using simple trigonometry, as follows:[tex]v_{x} = v_{o} * cos \theta = 55 km/h * cos 13 = 53.6 km/h[/tex]
[tex]v_{y} = v_{o} * sin\theta = 55 km/h * sin 13 = 12.4 km/h[/tex]
