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PIT_PIT [208]
3 years ago
10

In a basketball game, a player shoots a jump shot. What force actually causes the player to jump into the air? Question 3 option

s: the push of the ball on the floor the floor pushing up on the player the player pushing down on the floor the push of the floor on the ball.
Physics
1 answer:
Julli [10]3 years ago
6 0

Option B is the correct answer.

In a basketball game, a player shoots a jump shot then the floor pushes up on the player.

<h3>Newton's Third Law</h3>

Newton's third law states that when two bodies interact with each other, they apply forces to one another that are equal in magnitude and opposite in direction. The third law is also known as the law of action and reaction.

In the given situation, a player shoots a jump shot. It means that the player pushes the floor downward direction.

Newton's third law is applicable in this situation where the player pushes the floor downward direction, at the same time the floor pushes the player upward. The amount of force applied to the floor by the player is equal in magnitude and opposite in direction as compared to the force applied to the player by the floor.

Hence the option B is the correct answer.

To know more about Newton's third law, follow the link given below.

brainly.com/question/974124.

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How long must a 0.70-mm-diameter aluminum wire be to have a 0.40 a current when connected to the terminals of a 1.5 v flashlight
Natalka [10]
By using Ohm's law, we can find what should be the resistance of the wire, R:
R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega

Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
r=35 mm=0.35 \cdot 10^{-3} m
So the area is
A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2

And by using the resistivity  of the Aluminum, \rho=2.65 \cdot 10^{-8} \Omega m, we can use the relationship between resistance R and resistivity:
R= \frac{\rho L}{A}
to find L, the length of the wire:
L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m
4 0
3 years ago
A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou
kupik [55]

Answer:

The acceleration of the satellite is 0.87 m/s^{2}

Explanation:

The acceleration in a circular motion is defined as:

a = \frac{v^{2}}{r}  (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

v = \frac{2 \pi r}{T} (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite (1.50x10^{7} m) and the Earth radius (6.38x10^{6} m) :

r = 1.50x10^{7} m+6.38x10^{6}m

r = 21.38x10^{6}m

Then, equation 2 can be used:

T = 8.65 hrs \cdot \frac{3600 s}{1hrs} ⇒ 31140 s

v = \frac{2 \pi (21.38x10^{6}m)}{31140s}

v = 4313 m/s

Finally equation 1 can be used:

a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}    

a = 0.87 m/s^{2}

Hence, the acceleration of the satellite is 0.87 m/s^{2}

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So, the statements and the choices are related as:

               (1)-(1)

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               (3)-(4)

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Answer:c

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