Answer:
voltage= 17.88volts
current= 0.04 amps
Explanation:
Step one:
given data
resistance R=400 ohms
Power P= 0.8W
a. What is the maximum voltage that can be applied across this resistor without damaging it?
the expression relating power and voltage is
P=V^2/R
substituting we have
0.8=V^2/400
V^2=0.8*400
V^2=320
V=√320
V=17.88 volts
the maximum voltage is 17.88volts
b.What is the maximum current it can draw?
we know that from Ohm' law
V=IR
17.88=I*400
I=17.88/400
I=0.04amps
answer pls urgent pls
Answer:
I think its A: Magnesium
Explanation:
Basic oxide is formed by metal, and magnesium is the only metal in these options :)
Please inform me if this is incorrect
This is my question
Answer:
baby yoda
Explanation:
A train slows down as it rounds a sharp horizontal turn, going from 88.0 km/h to 52.0 km/h in the 18.0 s that it takes to round the bend. The radius of the curve is 120 m. Compute the acceleration at the moment the train speed reaches 52.0 km/h. Assume the train continues to slow down at this time at the same rate.
Answer:
The acceleration at the moment the train speed reaches 52 kilometers per hour is approximately 1.826 meters per square second.
Explanation:
According to Rotational Physics, the total acceleration of the train rounding the horizontal turn is a combination of tangential ([tex]a_{t}[/tex]) and radial accelerations ([tex]a_{r}[/tex]), measured in meters per square second. The former one represents the change in the magnitude of the velocity, whereas the latter one represents the change in its direction. By definition of magnitude and Pythagorean Theorem we get that magnitude of total acceleration ([tex]a[/tex]), measured in meters per square second, is:
[tex]a = \sqrt{a_{r}^{2}+a_{t}^{2}}[/tex] (Eq. 1)
Magnitudes of tangential and radial accelerations are determined by using the following formulas:
[tex]a_{t} = \frac{v_{f}-v_{o}}{t}[/tex] (Eq. 1)
[tex]a_{r} = \frac{v_{f}^{2}}{R}[/tex] (Eq. 2)
Where:
[tex]v_{o}[/tex], [tex]v_{f}[/tex] - Initial and final speeds, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]R[/tex] - Radius, measured in meters.
If we know that [tex]v_{o} = 24.444\,\frac{m}{s}[/tex], [tex]v_{f} = 14.444\,\frac{m}{s}[/tex], [tex]t = 18\,s[/tex] and [tex]R = 120\,m[/tex], then the magnitude of the total acceleration when the train speed reaches 52 kilometers per hour is:
[tex]a_{t} = \frac{14.444\,\frac{m}{s}-24.444\,\frac{m}{s} }{18\,s}[/tex]
[tex]a_{t} = -0.556\,\frac{m}{s^{2}}[/tex]
[tex]a_{r} = \frac{\left(14.444\,\frac{m}{s} \right)^{2}}{120\,m}[/tex]
[tex]a_{r} = 1.739\,\frac{m}{s^{2}}[/tex]
[tex]a = \sqrt{\left(-0.556\,\frac{m}{s^{2}} \right)^{2}+\left(1.739\,\frac{m}{s^{2}} \right)^{2}}[/tex]
[tex]a \approx 1.826\,\frac{m}{s^{2}}[/tex]
The acceleration at the moment the train speed reaches 52 kilometers per hour is approximately 1.826 meters per square second.
A rocket is fired in deep space, where gravity is negligible. In the first second it ejects 1/160 of its mass as exhaust gas and has an acceleration of 14.0 m/s^2.
Required:
What is the speed vgas of the exhaust gas relative to the rocket?
Answer:
840 m/s
Explanation:
Given that,
In the first second the rocket ejects 1/160 of its mass as exhaust gas and has an acceleration of 14.0 m/s².
We need to find the speed of the exhaust gas relative to the rocket.
The thrust of rocket is given by :
[tex]T=v_{gas}\dfrac{dm}{dt}\\\\ma=v_{gas}\dfrac{dm}{dt}\\\\v_{gas}=\dfrac{ma}{\dfrac{dm}{dt}}\\\\v_{gas}=\dfrac{14m}{\dfrac{1}{60}m}\\\\v_{gas}=840\ m/s[/tex]
So, the speed of the exhaust gas relative to the rocket is 840 m/s.
The speed ( Vgas) of the exhaust gas relative to rocket is : 840 m/s
Given data :
In first round Rocket ejects 1/60 of mass as exhaust gas
Acceleration of rocket ( a ) = 14.0 m/s²
Determine the speed of the exhaust gas relative to rocket
We will apply the equation for Rocket thrust
T = Vgas * [tex]\frac{dm}{dt}[/tex]
where : T = ma
∴ Vgas = ma / [tex]\frac{dm}{dt}[/tex]
= 14 m / [tex]\frac{1}{60}[/tex] m
therefore V gas = 840 m/s
Hence we can conclude that the speed ( Vgas) of the exhaust gas relative to rocket is : 840 m/s
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A physicist's left eye is myopic (i.e., nearsighted). This eye can see clearly only out to a distance of 35 cm.Find the focal length and the power of a lens that will correct this myopia when worn 2.0 cm in front of the eye.
Answer:
Explanation:
To get the focal length, we will use the lens formula;
1/f = 1/u + 1/v
f is the focal length
u is the object distance
v is the image distance
Given
since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.
u = 35cm
v = -2.0
1/f = 1/35-1/2
1/f = 2-35/70
1/f = -33/70
f = -70/33
f = -2.12 cm
f = -0.0212m
Power of a lend is the reciprocal of its focal length
Power of the lens = 1/f
P = 1/-0.0212
P = -47.17dioptres
The power of the lens is -47.17D
A 5-newton force directed east and a 5-newton force directed
north act concurrently on a point. Calculate the resultant of the two forces.
Answer:
[tex]5\sqrt{2} \ N[/tex]
Explanation:
It is given that,
A 5-newton force directed east and a 5-newton force directed north.
We need to find the resultant of the two forces.
As the two forces are acting in perpendicular to each other. The resultant of two such forces is given by :
[tex]F=\sqrt{5^2+5^2} \\\\=5\sqrt{2} \ N[/tex]
So, the resulatnt force is [tex]5\sqrt{2} \ N[/tex]
How do you calculate the net force when there are multiple forces in different directions?
To find [tex]F_{net}[/tex] we need to use vector addition and use the x and y components. First we subtract vector 2 from vector 5 which results in a vector with a length of 3 pointing directly east, then we use the distance formula to find the length of the net force [tex]F_{net} = \sqrt{(3)^2+(4)^2} \\[/tex] which gives [tex]F_{net} = 5[/tex]. We now have a magnitude but we also need a direction, since vector 4 and vector 5 are perpendicular. Using [tex]\theta = \tan^{-1} (\frac{4}{3})[/tex] where tan^-1(y/x) we get an angle of 53 degrees. The resultant force vector is 5 distance with an angle of 53 degrees north east.
During a soccer game, the goalie kicks the ball out toward the other end of the field. It travels 60 yards in 8 seconds before landing on the ground. What is the velocity of the ball?
Answer:
velocity = 7.5 yards / second
Explanation:
velocity = distance / time
where distance = 60 yards
time = 8 seconds
plug in values into the formula:
velocity = 60 yards
8 seconds
velocity = 7.5 yards / second
Answer:
the velocity of the ball is 7.5 yards/second
Explanation:
A 5 kg ball and a 2 kg ball are placed at opposite ends of a massless beam so that the system is in equilibrium. What is the ratio of the length of the beam to the distance from the heavier ball to the pivot
Answer:
L/x = 3.5 (The ratio of length of beam to the distance from heavier ball to pivot).
Explanation:
In order for the system to be in equilibrium, the moment created by both masses about the pivot point must be equal:
m₁x = m₂y
where,
m₁ = 5 kg
m₂ = 2 kg
x = distance of 5 kg ball from pivot
y = distance of 2 kg ball from pivot
Therefore,
(5 kg)x = (2 kg)y
y = (5kg/2kg)x
y = 2.5 x
but,
x + y = L
where,
L = length of beam
using the value of y from the previous equation:
x + 2.5 x = L
3.5 x = L
L/x = 3.5 (The ratio of length of beam to the distance from heavier ball to pivot).
If you can answer my last post ill give you 75 points pls its very important and please make sure it correct!!!!!!!
Answer:
I can't see the post :/
Explanation:
Answer:
ok
Explanation:
ok how many points is this?????
C4. A 50.0 kg boy runs at 10.0 m/s, jumps on a cart and rolls off at 2.50 m/s. What is the mass of the cart
Answer:
The mass of the cart is 150 kg.
Explanation:
Given that,
Mass of a boy, m₁ = 50 kg
Initial speed of boy, u₁ = 10 m/s
Initial speed of car, u₂ = 0 (at rest)
The speed of the cart with the boy on it is 2.50 m/s, V = 2.5 m/s
Let m₂ is the mass of the cart. Using the conservation of momentum as follows :
[tex]m_1u_1+m_2u_2=(m_1+m_2)V\\\\50(10)+m_2(0)=(50+m_2)(2.5)\\\\500=125+2.5m_2\\\\375=2.5m_2\\\\m_2=150\ kg[/tex]
So, the mass of the cart is 150 kg.
A 600 kg rocket sled can be accelerated at a constant rate from rest to 1400 km/h in 2.1 s. What is the magnitude of the required net force
Answer:
7.2 × 10^5 N
Explanation:
The first step is to convert 1400 km/hr to m/s
= 1,400×1000/3600
= 1,400,000/3600
= 388.88 m/s
The acceleration can be calculated as follows
a= v-u/t
= 388.88/2.1
= 185.18
Therefore the required net force can be calculated as follow
= 388.88 × 185.18
= 7.2 × 10^5 N
A rock climber is scaling rock wall at 10 m above the ground, and is moving downward at a speed of 0.25 m/s. If their phone falls out of their back pocket, how long before it shatters onto the ground?
Answer:
The time taken before the phone hits the ground is 1.4 s.
Explanation:
Given;
height of the rock wall, h = 10 m
speed of the climber, v = 0.25 m/s
The time taken before the phone hits the ground is given;
h = ut + ¹/₂gt²
10 = 0.25t + ¹/₂(9.8)t²
20 = 0.5t + 9.8t²
9.8t² + 0.5t - 20 = 0
solving this quadratic equation;
t = 1.4 s
Therefore, the time taken before the phone hits the ground is 1.4 s.
A 306-kg car moving at 16.5 m/s hits from behind a 810-kg car moving at 13.2 m/s in the same direction. If the new speed of the heavier car is 17.5 m/s, what is the velocity of the lighter car after the collision, assuming that any unbalanced forces on the system are negligibly small?
Answer:
v₂f = 5.1 m/s
Explanation:
Assuming no external forces acting during the collision, total momentum must be conserved.⇒ p₀ = pfThe initial momentum p₀, can be written as follows:[tex]p_{o} = m_{1} *v_{1o} + m_{2} *v_{2o} (1)[/tex]
where m₁ = 306 kg, m₂ = 810 kg, v₁₀ = 16.5 m/s, v₂₀ = 13.2 m/s
The final momentum, pf, can be written as follows:[tex]p_{f} = m_{1} *v_{1f} + m_{2} *v_{2f} (2)[/tex]
where v₂f = 17.5 m/s
Since p₀ = pf, which means that (1) is equal to (2),Replacing by the givens, and rearranging, we can solve for the only unknown that still remains, v₁f, as follows:[tex]v_{1f} = v_{1o} +\frac{m_{2} }{m_{1}} * (v_{2o} - v_{2f} ) \\= 16.5 m/s + \frac{810}{306} * (13.2 m/s - 17.5 m/s) \\= 16.5 m/s + \frac{810}{306} * (-4.3 m/s) \\= 16.5 m/s -11.4 m/s = 5.1 m/s[/tex]
The velocity of the lighter car after the collision is 5.1 m/s.A car travels at a speed of 55 km/hr and slows down to 10 km/hr in 20 seconds. What is the acceleration?
Answer:
a = 0.62 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics. But first, we must convert speeds from kilometers per hour to meters per second.
[tex]v_{f} =v_{o} -a*t[/tex]
[tex]55[\frac{km}{hr}]*\frac{1hr}{3600s}*\frac{1000m}{1km} =15.27[\frac{m}{s} ]\\10[\frac{km}{hr} ]*\frac{1hr}{3600s}*\frac{1000m}{1km} = 2.77[\frac{m}{s} ][/tex]
where:
Vf = final velocity = 2,77 [m/s]
Vo = initial velocity = 15.27 [m/s]
t = time = 20 [s]
a = acceleration [m/s²]
Now replacing:
[tex]2.77=15.27-a*(20)\\20*a=12.49\\a = 0.62[m/s^{2}][/tex]
Answer:
It would be 10.00
Explanation:
Hope this helps its different for everyone what was it for u it was D for me
1.) There was an earthquake in Salt Lake City, Utah, on March 18, 2020, in the morning at 9 hours, 9 minutes, and 45 seconds Mountain Standard Time (9:9:45 MST). If the velocity of the p-wave is 7.3 km/sec and the velocity of the s-wave is 5.1 km/sec and the s-p time lag is 16 seconds, what is the distance in kilometers from Salt Lake City to the focus of the earthquake? Explain how you calculated the answer.
Answer:
7 because salt lake and Southis weat
What property of a wave remains unchanged when a wave enters a different medium? 5 points amplitude frequency wavelength velocity
Answer:
frequency
Explanation:
When a wave enters a different medium, the property that remains unchanged is the frequency of the wave.
The frequency of a wave is defined as the number of wave cycles passing a given point per unit time. It represents the rate at which the wave oscillates or repeats its pattern.
The frequency of a wave remains constant when it enters a different medium, regardless of any changes in other properties such as amplitude, wavelength, or velocity.
Therefore, When a wave enters a different medium, the property that remains unchanged is the frequency of the wave.
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A ball has a diameter of 3.77 cm and average density of 0.0839 g/cm3. What force is required to hold it completely submerged under water?
magnitude _________ N
The force required to hold it completely submerged under water is 0.252 N
As a result of the low density (ρ1 = 0.0839 g/cm3 = 83.9 kg/m3)of the ball compared to that of water (ρ2 =1000 kg/m3), the buoyant force that is acting on the ball is greater than its weight.
Therefore, the minimum force required to hold the ball submerged under water can be calculated using the relation
F = Buoyant force - weight of sphere
Radius = 3.77/2 cm = 0.0377/2 m = 0.01885 m
Volume of sphere = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ = 2.805 e-5 m³
Mass of sphere = 4/3 π r³ ρ1 = 4/3 * 3.142 * 0.01885³ * 83.9 = 0.0023 kg
Weight of sphere = 4/3 π r³ ρ1 g = 4/3 * 3.142 * 0.01885³ * 83.9 * 9.8 = 0.023 N
Volume of water displaced = 4/3 π r³ = 2.805 e-5
Buoyant force = weight of water displaced = 4/3 π r³ ρ2 g = 4/3 π r³ = 4/3 * 3.142 * 0.01885³ * 1000 * 9.8 = 0.275 N
F = 0.275 - 0.023 = 0.252 N
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The force required to hold it completely submerged under water is 0.25 N
The density of the ball ([tex]\rho_b[/tex]) = 0.0839 g/cm³ = 83.9 kg/m³
The density of water [tex]\rho_w[/tex] = 1000 kg/m³
Diameter = 3.77 cm = 0.0377 m
radius of ball = 0.0377/2 = 0.01885 m
The volume (V) = [tex]\frac{4}{3} \pi r^3=\frac{4}{3}*\pi*0.01885^3=2.8*10^{-5}\ m^3[/tex]
Let us assume the acceleration due to gravity (g) = 9.8 m/s², Hence:
The force is required to hold it completely submerged under water (F) is:
[tex]F=\rho_w Vg-\rho_b Vg=1000*(2.8*10^{-5})*9.81-83.9*(2.8*10^{-5})*9.81\\\\[/tex]
F = 0.25 N
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we feel cold in winter when we come out from the quilt but the same room becomes warmer after coming back from outside the room
Answer:
Yes
Explanation:
I think this is because when you go out of the room and going to a hotter room you then get the heat from that room. It then becomes warmer in the room you are coming from because your body got the heat from the outside the room. I think it is because of body temperature.
HOPE THIS HELPED
How many seconds did it take (after starting his descent) for the worker to hit the ground? Answer in units of s.
Answer:
This question is incomplete
Explanation:
The question is incomplete. However, to determine the time (in seconds) it took a worker to hit the ground from an elevated point. The speed the worker was coming with to the ground and the distance between the elevated point and the ground will have to be considered. Thus the formula to be used here will be
Speed (in meter per second) = distance (in meters) ÷ time (in seconds)
time (in seconds) = distance (in meters) ÷ speed (in meter per seconds)
The above photo shows Plates A and B at an mid ocean ridge. If you were to draw arrows on the plates showing plate movement, which way would you draw arrows?
A . The arrows on plates A and B would show the plates moving in the same direction
B . The arrows on plates A and B would show the plates moving towards eachother
C . The arrows on plates A and B would show the plates moving in random directions
D . The arrows on plates A and B would show the plated moving away from each other
Answer:
D . The arrows on plates A and B would show the plated moving away from each other
Explanation:
The correct arrow in this instance will show that the plates A and B would be moving away from each other.
At the mid-ocean ridge, two plates are moving apart and pulling away from one another.
In short, at the mid-oceanic ridge, we have a divergent margin. At a divergent margin, the plates are spreading or moving apart.Is Nuclear Energy renewable or nonrenewable? Use in your own words.
How do you determine: how many significant figures should you to round to when doing addition and subtraction?
Assume you are in the car and the car is moving at a certain speed to
school. Are you at rest or in motion with respect to the school? With
respect to the car?
The rotational inertia of a thin rod about one end is 1/3 ML2. What is the rotational inertia of the same rod about a point located 0.40 L from the end
Answer:
The value is [tex]I = 0.0932 ML ^2[/tex]
Explanation:
From the question we are told that
The rotational inertia about one end is [tex]I_R = \frac{1}{3} ML^2[/tex]
The location of the axis of rotation considered is [tex]d = 0.4 L[/tex]
Generally the mass of the portion of the rod from the axis of rotation considered to the end of the rod is [tex]0.4 M[/tex]
Generally the length of the rod from the its beginning to the axis of rotation consider is
[tex]k = 1 - 0.4 L = 0.6L[/tex]
Generally the mass of the portion of the rod from the its beginning to the axis of rotation consider is
[tex]m = 1- 0.4 M = 0.6 M[/tex]
Generally the rotational inertia about the axis of rotation consider for the first portion of the rod is
[tex]I_{R1} = \frac{1}{3} (0.6 M )(0.6L)^2[/tex]
[tex]I_{R1} = \frac{1}{3} (0.6 M )L^2 0.6^2[/tex]
Generally the rotational inertia about the axis of rotation consider for the second portion of the rod is
[tex]I_{R2} = \frac{1}{3} (0.6 M )(0.6L)^2[/tex]
=> [tex]I_{R2} = \frac{1}{3} (0.4 M )(0.4L)^2[/tex]
=> [tex]I_{R2} = \frac{1}{3} (0.4 M )L^2 0.4^2[/tex]
Generally by the principle of superposition that rotational inertia of the rod at the considered axis of rotation is
[tex]I = \frac{1}{3} (0.6 M )L^2 0.6^2 + \frac{1}{3} (0.4 M )L^2 0.4^2[/tex]
=> [tex]I = \frac{1}{3} ML ^2 [0.6 * (0.6)^2 + 0.4 * (0.4)^2 ][/tex]
=> [tex]I = 0.0932 ML ^2[/tex]
an airplane flies at a speed of 100 m/s and starts to accelerate constantly at a rate of 50 m/s2. how fast is the plane flying after traveling a distance of 1 kilometer?
Answer:
331.7m/s
Explanation:
Given parameters:
Initial velocity = 100m/s
Acceleration = 50m/s²
Distance = 1km = 1000m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we have to apply the right motion equation shown below;
v² = u² + 2aS
v is the final velocity
u is the initial velocity
a is the acceleration
S is the distance
Now insert the parameters and solve;
v² = 100² + (2 x 50 x 1000)
v² = 110000
v = √110000 = 331.7m/s
A gas was compressed to 30.0 mL at 1.5 atm from 65
mL. What was the original pressure?
Two identical bowling balls are moving down a bowling alley so that their centers of mass have the same velocity, but one just slides down the alley, while the other rolls down the alley. Which ball has more energy?
a. Sliding ball
b. Rolling ball
c. They both have the same energy
Answer:
Rolling Ball
Explanation
Two identical bowling balls are moving down a bowling alley so that their centers of mass have the same velocity, but one just slides down the alley, while the other rolls down the alley, rolling ball has more energy.
what is velocity ?Velocity can be referred to the rate of change of the position of the object with respect to time which is basically speeding the object in a specific direction.
Velocity is vector quantity, as it is present both magnitude and direction and The SI unit is meter per second (ms-1). The change in magnitude or the direction of velocity of a body, the object will accelerate.
the final velocity of the object is simple but few calculations and basic conceptual knowledge are needed.
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This is my question
Answer:
N = 470 [N]
Explanation:
The normal force is defined as the reaction exerted by the surface where the body is located in the opposite direction to the weight component.
It can be easily calculated by means of the product of mass by gravitational acceleration.
N = m*g
where:
N = normal force [N] (units of Joules)
m = mass = 47 [kg]
g = gravity acceleration = 10 [N]
N = 47*10
N = 470 [N]
Ancient peoples both feared and revered Solar and Lunar Eclipses as they happened so rarely. Explain the following about eclipses:
1. Why are SOLAR eclipses more rare then lunar eclipses.
2. Explain what must occur for a SOLAR eclipse to occur.
3. Explain what must occur for a LUNAR eclipse to occur.
Answer:
Lunar and solar eclipses occur with about equal frequency. Lunar eclipses are more widely visible because Earth casts a much larger shadow on the Moon during a lunar eclipse than the Moon casts on Earth during a solar eclipse. As a result, you are more likely to see a lunar eclipse than a solar eclipse.
For a total eclipse to take place, the sun, moon and Earth must be in a direct line. The second type of solar eclipse is a partial solar eclipse. This happens when the sun, moon and Earth are not exactly lined up. The sun appears to have a dark shadow on only a small part of its surface.
For a lunar eclipse to occur, the Sun, Earth, and Moon must be roughly aligned in a line
