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kotykmax [81]
2 years ago
11

If a 1,000-kg car is traveling at 20.0 m/s when it reaches the bottom of a circular hill of radius 8.00 m, what normal force doe

s the road exert on the car at the bottom of the hill
Physics
1 answer:
scoray [572]2 years ago
6 0

Answer:

i don't know this one.

Explanation:

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A ball of mass M is suspended by a thin string (of negligible mass) from the ceiling of an elevator.uploaded image
lilavasa [31]

Answer:

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.  T > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor. T > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor. T < mg

(d) The elevator is traveling downward at a constant velocity. T = mg

(e) The elevator is traveling downward and its downward velocity is increasing. T < mg

(f) The elevator is stationary and remains at rest. T = mg

Explanation:

To answer this question, consider all the forces acting on the elevator.

The mass of the ball acting downwards due to gravity = mg

The tension on the string depends on upward or downwards force on the ball. T = m(a+g)

where a is acceleration and increase in velocity causes increase in acceleration, and vice versa. (a = v/t)

(a) The elevator is traveling upward and its upward velocity is decreasing as it nears a stop at a higher floor.

If the upward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a+g) > mg

(b) The elevator is traveling upward and its upward velocity is increasing as it begins its journey towards a higher floor.

If the upward velocity is increasing, its acceleration is also increasing.

Then, T = m(a+g) > mg

(c) The elevator is traveling downward and its downward velocity is decreasing as it nears a stop at a lower floor.

If the downward velocity is decreasing, its acceleration is also decreasing, and acceleration is not equal to Zero

T = m(a-g) < mg

(d) The elevator is traveling downward at a constant velocity

At constant velocity, acceleration is zero, because acceleration is the rate of change of velocity.

T = m(0+g) = mg

(e) The elevator is traveling downward and its downward velocity is increasing

If the downward velocity is increasing, its acceleration is also increasing

T = m(a-g) < mg

(f) The elevator is stationary and remains at rest.

if the elevator is at rest, its acceleration is zero

T = m(0+g) = mg

6 0
3 years ago
Help again please !!!
rusak2 [61]
D I think .... don’t be mad if I’m wrong
5 0
3 years ago
Where does the kinetic energy come fromwhen you roll down a hill?
Lapatulllka [165]
From the change of GPE into KE. Conservation of energy tells us this.
8 0
3 years ago
Gravity can be described as..?
Tju [1.3M]

Answer:

D

Explanation:

Gravity is the force of attraction between two objects.

Each object creates a gravitational field in wich every other object is affected by it.

6 0
3 years ago
Which Box in the diagram above, should this picture go into?
lyudmila [28]
What diagram? There isn’t one
7 0
3 years ago
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