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xeze [42]
3 years ago
15

A shot-putter exerts an unbalanced force of 140 N on a shot giving it an

Physics
1 answer:
vekshin13 years ago
8 0

Answer:

7.37 kg

Explanation:

By Newtons Second Law of Unbalanced Forces

Force = Mass × Acceleration

140N = Mass × 19m/s^2

Mass = 140N / 19m/s^2

= 7.37 kg

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A(n) 55.5 g ball is dropped from a height of 53.6 cm above a spring of negligible mass. The ball compresses the spring to a maxi
Serggg [28]

Answer:

The spring force constant is  k=243\ \frac{N}{m} .

Explanation:

We are told the mass of the ball is m=0.0555\ kg, the height above the spring where the ball is dropped is h=0.536\ m,  the length the ball compresses the spring is d=0.04897\ m and the acceleration of gravity is 9.8\ \frac{m}{s^{2}} .

We will consider the initial moment to be when the ball is dropped and the final moment to be when the ball stops, compressing the spring. We supose that there is no friction so the initial mechanical energy E_{mi} is equal to the final mechanical energy E_{mf} :

                                                    E_{mf}=E_{mi}

Initially there is only gravitational potential energy because the force of the spring isn't present and the speed is zero. In the final moment there is only elastic potential energy because the height is zero and the ball has stopped. So we have that:

                                                   \frac{1}{2}kd^{2}=mgh

If we manipulate the equation we have that:

                                                    k=\frac{2mgh}{d^{2} }

                                         k=\frac{2\ 0.0555\ kg\ 9.8\frac{m}{s^{2}}\ 0.536\ m}{(0.04897)^{2}m^{2}}

                                              k=\frac{0.58306\ \frac{kgm^{2}}{s^{2}}}{2.398x10^{-3}m^{2}}

                                                     k=243\ \frac{N}{m}

                                                   

                             

5 0
3 years ago
If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose
Volgvan

Answer:

a) A=0.125 m

b) T = 1.72 s

c) f= 0.58 Hz

Explanation:

a) As we are told that the maximum displacement from the equilibrium position was 0.125 m (from which it was released at zero initial speed), this is the amplitude of the resultant SHM, so, A=0.125 m

b) In order to find the period, we must get the total time needed to complete a full cycle (which means that the block must pass twice through the equilibrium point). We are told that at t=0.860 sec, the block has reached to the other end of the trajectory, and it  has passed through the equilibrium point only once.

This means that the period must be exactly the double of this time:

T = 2*0. 860 sec = 1.72 sec.

c) In a SHM, the frequency is defined just as the inverse of the period (like in a uniform circular movement), so we can get the frequency  f as follows:

f = 1/T = 1/ 1.72 s= 0.58 Hz

8 0
3 years ago
Jupiter's gravity is about 2.3 times stronger than Earth's gravity. Calculate the
AlexFokin [52]

Answer:

Explanation:

There is no change in mass. Therefore, mass =  70 k g

Weight on Jupiter  ≈  1578  N

4 0
2 years ago
According to Coulomb's law, the force of attraction or repulsion between two charges is
Maru [420]

Answer:

B is the answer

Explanation:

thank you I hope it helps you

6 0
2 years ago
Read 2 more answers
How much force is needed to stop a 100 kg football player if he decelerates at 20 m/s-2
Brut [27]
We know, f = ma
Here, m = 100 kg
a = 20 m/s²

Substitute their values, 
f = 100 * 20
f = 2000 N

In short, Your Answer would be 2000 N

Hope this helps!
7 0
2 years ago
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