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3 years ago

Which statement best compares the accelerations of two objects in free fall?

Physics

1 answer:

scoray [572]3 years ago

4 0

Answer:

increasing the masses of the objects and decreasing the distance between the objects

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What is the value of work done on an object when a

tino4ka555 [31]

W = F * s
Here, F = 50 N
s = 15 m

Substitute their values,
W = 50 * 15
W = 750 J

In short, Your Answer would be 750 Joules

Hope this helps!

7 0

3 years ago

The graph below shows how the speed of a bus change during part of a journey. Use this graph to answer questions 3-5

Minchanka [31]

Answer:

O to A and D to E are accelorating.

Explanation:

You can tell because the graphed points have positive slopes in these locations.

6 0

3 years ago

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A horizontal sheet of negative charge has a uniform electric field E = 3000N/C. Calculate the electric potential at a point 0.7m

vesna_86 [32]

Answer:

Electric potential, E = 2100 volts

Explanation:

Given that,

Electric field, E = 3000 N/C

We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m

The electric potential is given by :

$V=E\times d$

$V=3000\ N/C\times 0.7\ m$

V = 2100 volts

So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.

6 0

3 years ago

When you get into a car with hot black leather in the middle of the summer and it hurts your skin whe you sit this is an example

NeTakaya

C.the heat is conducted to your skin

3 0

3 years ago

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Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

7 0

3 years ago