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lara31 [8.8K]
3 years ago
13

A 50 kg student climbs 3m to the top of a set of stairs. Calculate the change in the student’s gravitational potential energy fr

om the bottom to the top of the stairs.
Physics
1 answer:
erastova [34]3 years ago
7 0
Gpe = mgh
gpe = 50*10*3
gpe = 1500 J
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Anton [14]

Answer:

C

Explanation:

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Two objects, A and B, are in contact with one another. Initially, the temperature of A is 50 °C and the temperature of B is 100
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because kinetic energy is directly proportional to temperature so the hottor the object, the more kinetic energy.

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Can a body have zero velocity and finite acceleration?Explain​
sergejj [24]

Answer:

Kinda? Depends what the question is fully asking

Explanation:

Acceleration is a change in velocity. So I guess if the velocity of something is -2 m/s and its positively accelerating at a value of +1 m/s, then that means every second its velocity changes by +1m/s.

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After another second it'll be going 0 m/s.

After another itll be going +1 m/s and so on.

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5 0
3 years ago
1. A 1.5 kg ball moves in a circle that is 0.5 m in radius at a speed of 5.1 m/s.
kolezko [41]

Answer: a= 52.02 m/s²

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3 0
3 years ago
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

s_x=v_x_0 t

t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s

Now for the vertical distance

vy_o=0

than the equation of motion becomes

s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2

Now using this acceleration the value of electric field is calculated as

E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0
3 years ago
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