# help with physics..?

Question 10 (4 points)

Use Hooke's law to solve the following problems.

a. A spring with k = 11 N/m is stretched 0.25 m. What force does the spring apply? (1 point)

b. What is the spring constant of a rubber band that exerts a force of 35 N when it is stretched 0.1 m? (1 point)

c. An elastic band has a spring constant of 55 N/m. It will break when the force reaches 2000 N. How far do you need to stretch the band for it to break? (1 point)

d. A 2 kg mass hangs motionless from a spring with spring constant 8 N/m. How far will the spring stretch if the mass is motionless? (1 point)

Question 11 (2 points)

A spring with spring constant 40 N/m is compressed 0.1 m past its natural length. A mass of 0.5 kg is attached to the spring.

a. What is the elastic potential energy stored in the spring? (1 point)

b. The spring is released. What is the speed of the mass as it reaches the natural length of the spring? (1 point)

Question 13 (3 points)

Use the universal law of gravitation to solve the following problems.

a. Two cars of mass 3000 kg are 2 m apart. What is the force of gravity between them? (1 point)

b. The force of gravity between a planet and its moon is 371 N. If the planet has a mass of 4 × 1022 kg and the moon has a mass of 5 × 105 kg, what is the distance between their centers? (1 point)

c. The force of gravity between Tim and John is 1.85 × 10–9 N. If they are 18 m apart and Tim has a mass of 120 kg, what is John's mass? (1 point)

### 2 Answers

- oubaasLv 75 years agoBest Answer
Question 10....Use Hooke's law to solve the following problems.

a. A spring with k = 11 N/m is stretched 0.25 m. What force does the spring apply?

F = k*x = 11*0.25 = 2.75 N

b. What is the spring constant of a rubber band that exerts a force of 35 N when it is stretched 0.1 m?

k = F/x = 35/0.1 = 350 N/m

c. An elastic band has a spring constant of 55 N/m. It will break when the force reaches 2000 N. How far do you need to stretch the band for it to break?

x = F/k = 2000 N/55 N/m = 36.364 m

d. A 2 kg mass hangs motionless from a spring with spring constant 8 N/m. How far will the spring stretch if the mass is motionless? (1 point)

x = F/k = m*g/k = 2*9.8/8 = 2.45 m

Question 11

A spring with spring constant 40 N/m is compressed 0.1 m past its natural length. A mass of 0.5 kg is attached to the spring.

a. What is the elastic potential energy stored in the spring? (1 point)

Esp = 1/2kx^2 = 20*0.1^2 = 0.20 joule

b. The spring is released. What is the speed of the mass as it reaches the natural length of the spring?

V = √2Esp/m = √2*0.20/0.5 = √0.8 = 0.63√2 m/sec

Question 13

Use the universal law of gravitation to solve the following problems.

a. Two cars of mass 3000 kg are 2 m apart. What is the force of gravity between them?

F = m^2*G/d^2 = 9*10^6*6.674*10^-11/4 = 1.50*10^-4 N

b. The force of gravity between a planet and its moon is 371 N. If the planet has a mass of 4 × 1022 kg and the moon has a mass of 5 × 105 kg, what is the distance between their centers? (1 point)

371*d^2 = 4*10*^22*5*10^5*6.674*10^-11

d = (√4*10*^22*5*10^5*6.674*10^-11)/371 = 6.00*10^7 m

c. The force of gravity between Tim and John is 1.85 × 10–9 N. If they are 18 m apart and Tim has a mass of 120 kg, what is John's mass?

1.85*10^–9 = 120*mx*G/18^2

mx = 1.85*10^–9*18^2/(120*6,67*10^-11) = 75 kg

- NCSLv 75 years ago
Q10: "Use Hooke's law to solve the following problems.

a. A spring with k = 11 N/m is stretched 0.25 m. What force does the spring apply?

b. What is the spring constant of a rubber band that exerts a force of 35 N when it is stretched 0.1 m?

c. An elastic band has a spring constant of 55 N/m. It will break when the force reaches 2000 N. How far do you need to stretch the band for it to break?

d. A 2 kg mass hangs motionless from a spring with spring constant 8 N/m. How far will the spring stretch if the mass is motionless?"

A: a. F = kx = 11N/m * 0.25m = 2.8 N

b. k = F / x = 35N / 0.1m = 350 N/m

c. x = F / k = 2000N / 55N/m = 36 m

d. x = F / k = mg / k = 2kg * 9.8m/s² / 8N/m = 2.5 m

Q11: "A spring with spring constant 40 N/m is compressed 0.1 m past its natural length. A mass of 0.5 kg is attached to the spring.

a. What is the elastic potential energy stored in the spring?

b. The spring is released. What is the speed of the mass as it reaches the natural length of the spring?"

A: a. U = ½kx² = ½ * 40N/m * (0.1m)² = 0.20 J

b. KE = 0.20 J = ½mv² = ½ * 0.5kg * v² → v = 0.89 m/s

13: F = GmM / d² where G = 6.674e−11N·m²/kg²

a. Plugging in the given M, m and d, find F = 0.00015 N = 1.5e-4 N

b. d = √(GmM / F) yields d = 6e7 m

c. M = Fd² / Gm yields M = 75 kg

All answers to two significant digits, which seems appropriate for the data.

Hope these help!

thanks..happy new year !!