What 2 factors determine the amount of gravitational force an object experiences

The atmosphere is full of nitrogen and oxygen. These are

Alex73 [517]

Types of Gas? I'm not exactly sure what you're asking

3 0

3 years ago

Define Velocity with an example

kolbaska11 [484]

Velocity stands for Displacement w.r.t time

$\\ \bull\tt\longmapsto Velocity=\dfrac{Displacement}{Time}$

Or

$\\ \bull\tt\longmapsto v=\dfrac{ds}{dt}$

It has units m/s.

8 0

2 years ago

Read 2 more answers

A Tennis ball falls from a height 40m above the ground the ball rebounds

worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity v at time t before it hits the ground is

v = -g t

where g = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height y is

y = 40 m - 1/2 g t²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 g t²

==> t = √(80/g) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

v = -g (√(80/g) s)

==> v = -√(80g) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

y = (14 m/s) t - 1/2 g t²

Solve y = 0 for t to see how long it's airborne during this bounce:

0 = (14 m/s) t - 1/2 g t²

0 = t (14 m/s - 1/2 g t)

==> t = 28/g s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

y = (14 m/s) (5 s - 2.86 s) - 1/2 g (5 s - 2.86 s)²

==> (i) y ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But y will converge to 0 as t gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

y = (3.5 m/s) t - 1/2 g t²

Solve y = 0 for t :

0 = (3.5 m/s) t - 1/2 g t²

0 = t (3.5 m/s - 1/2 g t)

==> (iii) t = 7/g m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time t after 5.72 s (but before reaching the ground again) would be

v = 7 m/s - g t

At 6 s, the ball has velocity

(iv) v = 7 m/s - g (6 s - 5.72 s) ≈ 4.26 m/s

4 0

3 years ago

It takes a bus driver 30 minutes to pick up students from four stops. The last stop is at the corner of Green Street and Route 7

deff fn [24]

Answer:

No, the distance from the last stop to the school and the time it takes to travel that distance are required.

7 0

3 years ago

Under the influence of its drive force, a snowmobile is moving at a constant velocity along a horizontal patch of snow. When the

balandron [24]

Answer:

a) Δx = 11.6 m

b) t = 3.9 s

Explanation:

a)

Since the snowmobile is moving at constant speed, and the drive force is 195 N, this means that thereis another force equal and opposite acting on it, according to Newton's 2nd Law, due to there is no acceleration present in the horizontal direction .
This force is just the force of kinetic friction, and is equal to -195 N (assuming the positive direction as the direction of the movement).
Once the drive force is shut off, the only force acting on the snowmobile remains the friction force.
According Newton's 2nd Law, this force is causing a negative acceleration (actually slowing down the snowmobile) that can be found as follows:

$a = \frac{F_{fr} }{m} = \frac{-195N}{128kg} = -1.5 m/s2 (1)$

Assuming the friction force keeps constant, we can use the following kinematic equation in order to find the distance traveled under this acceleration before coming to an stop, as follows:

$v_{f} ^{2} -v_{o} ^{2} = 2* a* \Delta x (2)$

Taking into account that vf=0, replacing by the given (v₀) and a from (1), we can solve for Δx, as follows:

$\Delta x =- \frac{v_{o}^{2}}{2*a} =- \frac{(5.90m/s)^{2}}{2*(-1.5m/s2)} = 11.6 m (3)$

b)

We can find the time needed to come to an stop, applying the definition of acceleration, as follows:

$v_{f} = v_{o} + a*\Delta t (4)$

Since we have already said that the snowmobile comes to an stop, this means that vf = 0.
Replacing a and v₀ as we did in (3), we can solve for Δt as follows:

$\Delta t = \frac{-v_{o} }{a} = \frac{-5.9m/s}{-1.5m/s2} = 3.9 s (5)$

7 0

2 years ago