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kondor19780726 [428]
2 years ago
8

A roller coaster's velocity at the bottom of a hill is 35.0 m/s, 5.00 seconds later it reaches the top of the hill with a veloci

ty of 10.0
m/s. What was the deceleration of the coaster?
O 5.00 m/s/s
O 7.00 m/s/s
O 2.00 m/s/s
O 25.0 m/s/s
Physics
1 answer:
anyanavicka [17]2 years ago
4 0

Answer:

5 m /s^2

Explanation:

Change in velocity = 35 - 10 = 25 m/s

change in time = 5 s

decelleration = 25/5 = 5 m/s^2

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kramer

Answer:

Actions

Explanation:

if u mean what are these called then it's actions

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3 years ago
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What will happen to a negatively charged particle placed at rest between points that have electric potentials 0V and 9V? Explain
finlep [7]

Hey there!

a) The electric field is in direction of decreasing value of potential so the electric field here will points towards 0v from 9 v. But the charge on particle is negative so the force will be towards the 9v point. and so the charge will move towards the 9 V point.

b) Electric potential is a location-dependent quantity that expresses the amount of potential energy per unit of charge at a specified location while.the electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy. Electric potential is always a relative quantity because no one can find out absolute potential at a point, so in reality only potential difference exists. We can assume potential at certain point (say at infinity it is zero) only then we are able to define potential at every point.

c) The knowlendge of electric potential helps us in finding the work done on the particle which is used in work energy theorem to find out the chanrge in kinetic energy and other stuff.

Hope this helps!

8 0
3 years ago
A waterbed has a force of 1300N on the floor. It exerts 347 Pa of pressure. What is the area of the waterbed?
Sergio [31]

Answer:

Pressure = Force/Area

347 = 1300/Area

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3 years ago
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The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the sprin
Korolek [52]

Answer:

a. V=11.84 m/s

b.x=0.052m

Explanation:

a).

Given

K=730 N/m,m=0.053kg, h=1.90m.

v_f^2=v_i^2+2*g*h

v_i^2=2*g*h=2*9.8m/s^2*1.9m

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v_i=V*sin(31)

V=\frac{v_i}{sin(31)}=\frac{6.1m/s}{sin(31)}

V=11.84 m/s

b).

K_k=\frac{1}{2}*K*x^2

No friction on the ball so:

x^2=\frac{2*K_k}{K}

x=\sqrt{\frac{2*0.053kg*9.8m/s^2*1.9m}{730N/m}}

x=\sqrt{2.7x10^{-3}m^2}=0.052m

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I did questions 2 and 3 but I don't know if they are right. Someone help!
Tems11 [23]

You got it right my friend

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