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Anni [7]
2 years ago
10

A free-falling golf ball strikes the ground and exerts a force on it. Which sentences are true about this situation? A golf ball

striking the ground is a collision. The ground exerts an equal force on the golf ball. The ground doesn’t exert a force on the golf ball. The force is zero because the golf ball has little mass.
Physics
2 answers:
Harlamova29_29 [7]2 years ago
8 0

Answer:

The ground exerts an equal force on the golf ball

Explanation:

Third's Newton Law states that:

"When an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A".

In this problem, object A is the golf ball while object B is the ground, so we can say that:

- the golf ball exerts a force on the ground

- the ground exerts an equal and opposite force on the golf ball

Serga [27]2 years ago
5 0
THe ground exerts an equal force on the golf ball

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A satellite that orbits Earth with a speed of v0 must be in an orbit of radius 8RE to maintain a circular orbit, where RE is the
NISA [10]

Answer:

1.024 × 10⁸ m

Explanation:

The velocity v₀ of the orbit 8RE is v₀ = 8REω where ω = angular speed.

So, ω =  v₀/8RE

For the orbit with radius R for it to maintain a circular orbit and velocity 2v₀, we have

2v₀ = Rω

substituting ω =  v₀/8RE into the equation, we have

2v₀ = v₀R/8RE

dividing both sides by v₀, we have

2v₀/v₀ = R/8RE

2 = R/8RE

So, R = 2 × 8RE

R = 16RE

substituting RE = 6.4 × 10⁶ m

R = 16RE

= 16 × 6.4 × 10⁶ m

= 102.4 × 10⁶ m

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8 0
2 years ago
In a movie, Tarzan evades his captors by hiding under water for many minutes while breathing through a long, thin reed. Assume t
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Answer: 0.98m

Explanation:

P = -74 mm Hg = 9605 Pa = 9709N/m^2

= 9605 kg m/s^2/m^2

density of water: rho = 1 g/cc = 1 (10^-3 kg)/(10^-2 m)^-3 = 1000 kg/m^3

Pressure equation: P = rho g h

h = P/(rho g)

h = (9605 kg/m/s^2) / (1000 kg/m^3) / (9.8 m/s^2)

h = 0.98 m

0.98m is the maximum depth he could have been.

8 0
2 years ago
A battleship launches a shell horizontally at 100 m/s from the ship’s deck that’s 50 m above the water. The shell is intended to
Annette [7]

Answer:

The shell will land 10.18m away from the buoy.

Explanation:

In order to solve this problem, we must first do a sketch of what the problem looks like (see attached picture).

Now, there are two cases, one with the tailwind and another with the tailwind. In both cases the shell would have the same vertical initial velocity and acceleration, therefore the shell would hit the water in the same amount of time. So we need to first find the time it takes the shell to hit the water.

In order to do so we can use the following equation:

y_{f}=y_{0}+V_{0}t+\frac{1}{2}at^{2}

now, we know that the final height and the initial velocity are to be zero, so we can simplify the equation like this:

0=y_{0}+\frac{1}{2}at^{2}

and solve for t:

t=\sqrt{\frac{-2y_0}{a}}

now we can substitute the values:

t=\sqrt{\frac{-2(50m)}{-9.81m/t^2}}

t=3.19s

Since it takes 3.19s for the shell to hit the water, that's the amount of time it spends flying horizontally.

So we can consider the shell to move at a constant speed if there was no tailwind, so we can find the  distance from the ship to point A to be:

x_{A}=V_{x}t

x_{A}=(100m/s)(3.19)

x_{A}=319m

We can now find the distance between the ship to point B, which is the point the ball falls due to the tailwind. Since the movement will be accelerated in this scenario, we can find the distance by using the following formula:

x_{f}=V_{x0}t+\frac{1}{2}a_{x}t^{2}

So we can substitute the given values:

x_{f}=(100m/s)(3.19s)+\frac{1}{2}(2m/s^{2})(3.19s)^{2}

Which yields:

x_{f}=329.18m

so now we can use the A and B points to find by how far the shell missed the buoy:

Distance=329.18m-319m=10.18m

So the shell missed the buoy by 10.18m.

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