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Kamila [148]
2 years ago
5

As a 5.0 x 10^2 newton basketball player jumps from the floor up toward the basket, the magnitude of the force of her feet on th

e floor is 1.0 x 10^3 newtons. As she jumps, the magnitude of the force of the floor on her feet is
1) 5.0 x 10^2 N
2) 1.0 x 10^3 N
3) 1.5 x 10^3 N
4) 5.0 x 10^5 N
Physics
2 answers:
Harlamova29_29 [7]2 years ago
5 0

Answer:

2) 1.0 x 10^3 N

Explanation:

As a 5.0 x 10^2 newton basketball player jumps from the floor up toward the basket, the magnitude of the force of her feet on the floor is 1.0 x 10^3 newtons. As she jumps, the magnitude of the force of the floor on her feet is

1) 5.0 x 10^2 N

2) 1.0 x 10^3 N

3) 1.5 x 10^3 N

4) 5.0 x 10^5 N

The answer is B , because from newton's third law of motion which states that action and reaction are equal and opposite.

so the reaction of the floor on her leg will be equal to the the magnitude of her feet on the floor, this will hold if the floor does not collapse.

Yuliya22 [10]2 years ago
4 0
I believe the answer is (4) The reason that is, is because if the exponents are the same like 10^2 and 10^3, you can add them. Then you would get 10^5. You can go ahead though and multiply 5.0 and 1.0. Now remember that with decimals you don't need the zeros behind the decimal point. So that simplifies it with just 5 x 1. Leaving you with 5.0 x 10^5. 
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A person kicks a ball, giving it an initial velocity of 20.0 m/s up a wooden ramp. When the ball reaches the top, it becomes air
Alex Ar [27]

Answer:

(a) Height is 4.47 m

(b) Height is 4.37 m

Solution:

As per the question:

Initial velocity of teh ball, v_{o} = 20.0 m/s

Angle made by the ramp, \theta = 22.0^{\circ}

Distance traveled by the ball on the ramp, d = 5.00 m

Now,

(a) At any point on the projectile before attaining maximum height, the velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH

where

H = dsin22^{\circ} = 5sin22^{\circ}

g = 9.8 m/s^{2}

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ}

v = \sqrt{400 - 19.6\times 5sin22^{\circ}} = 19.06 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(19sin(22^{\circ}))^{2}}{2\times 9.8} = 2.60 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.60 = 4.47m

(b) now, considering the coefficient of friction bhetween ramp and the ball, \mu = 0.150:

velocity can be given by the eqn-3 of motion:

v^{2} = v_{o}^{2} - 2gH - \mu gd

v^{2} = 20^{2} - 2\times 9.8\times 5sin22^{\circ} - 0.150\times 9.8\times 5

v = \sqrt{400 - 19.6\times 5sin22^{\circ} - 0.150\times 9.8\times 5} = 18.7 m/s

Now, maximum height attained is given by:

h = \frac{(vsin\theta)^{2}}{2g}

h = \frac{(18.7sin(22^{\circ}))^{2}}{2\times 9.8} = 2.50 m

Height from the ground = 5sin22^{circ} + 2.86 = 1.87 + 2.50 = 4.37 m

6 0
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USPshnik [31]
Note that
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Answer:
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