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horsena [70]
4 years ago
6

A racing car is moving around the circular track of radius 300 meters shown above. at the instant when the car's velocity is dir

ected due east, its acceleration is directed due south and has a magnitude of 3 meters per second squared. when viewed from above, the car is moving clockwise at 30 m/s clockwise at 10 m/s counterclockwise at 30 m/s counterclockwise at 10 m/s
Physics
1 answer:
kaheart [24]4 years ago
4 0

Answer:

the car is moving clockwise at 30 m/s

Explanation:

For an object moving in circular motion:

- The velocity is always tangential to the circle

- The acceleration (centripetal acceleration) is towards the centre of the circle, so perpendicular to the velocity)

In this case, the velocity is directed due east, while the acceleration is directed due south: so the car is "curved towards south while moving", this means that the car is travelling clockwise.

Moreover, the centripetal acceleration is given by:

a=\frac{v^2}{r}

where

a = 3 m/s^2 is the acceleration

v is the speed

r = 300 m is the radius of the track

Solving for v,

v=\sqrt{ar}=\sqrt{(3 m/s^2)(300 m)}=30 m/s

So the answer is

the car is moving clockwise at 30 m/s

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The correct answer is C) towards the center of the circle.

Although the object is moving at a constant speed it is constantly accelerating due to the constant change in direction as it describes the circular path. This causes a constant change in velocity as velocity is a vector quantity.

For the object to maintain the circular path there has to be centripetal force acting on the object and this centripetal force is directed towards the center of the circle.

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3 years ago
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3 years ago
If a 0.15 kg ball falls and has a KE of 20 J just before striking the ground, from what height did it fall. A. 1.36m B. 3m C. 13
RUDIKE [14]
According to the conservation of mechanical energy, the kinetic energy just before the ball strikes the ground is equal to the potential energy just before it fell. 

Therefore, we can say KE = PE
We know that PE = m·g·h

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We can solve for h:

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The correct answer is: the ball has fallen from a height of 13.6m.

5 0
3 years ago
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po
gregori [183]

Answer:

E = 0    r <R₁

Explanation:

If we use Gauss's law

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in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

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6 0
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