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velikii [3]
3 years ago
6

Please this is my senior work I’m trying to pass so help please What is Hooke’s Law

Physics
1 answer:
3241004551 [841]3 years ago
7 0

Answer:

It is a law of mechanics and physics discovered by Robert Hooke. This theory of elasticity says the extension of a spring is proportional to the load applied to it. Many materials obey this law as long as the load does not exceed the material's elastic limit.

Explanation:

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A 5kg wheel rolls off a flat roof of a 50 m tall building at 12m/s.
Olegator [25]

Explanation:

a) Given in the y direction (taking down to be positive):

Δy = 50 m

v₀ = 0 m/s

a = 10 m/s²

Find: t

Δy = v₀ t + ½ at²

50 m = (0 m/s) t + ½ (10 m/s²) t²

t = 3.2 s

b) Given in the x direction:

v₀ = 12 m/s

a = 0 m/s²

t = 3.2 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (12 m/s) (3.2 s) + ½ (0 m/s²) (3.2 s)²

Δx = 38 m

6 0
3 years ago
A child pulls a wagon at a constant velocity along a level sidewalk. The child does this by applying a 22 newton force to the wa
Jlenok [28]

Answer:

The answer is 18 N.

Explanation:

A force can be divided into components x and y components. The component along the x-axis is called the horizontal component and along the y-axis is called the vertical component. In this case, as the force is in a horizontal direction and is also known as x-component of force. The x- component of force is  

Fx = Fcosθ

Fx = 22(cos 35°)

Fx = 22 x 0.819

Fx =  18 N

Child's horizontal pull forces are equal to that of frictional resistance force on the wagon.

5 0
3 years ago
An automobile engine delivers 55.0 hp. How much time will it take for the engine to do 6.22 × 105 J of work? One horsepower is e
Gennadij [26K]

Answer:

15.2 s

Explanation:

Convert hp to W:

55.0 hp × 746 W/hp = 41,030 W

Power = energy / time

41030 W = 6.22×10⁵ J / t

t = 15.2 s

8 0
3 years ago
Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
Ede4ka [16]

Answer:

386.2^{\circ}F

Explanation:

We are given that

P_1=200lbf/in^2

P_2=60lbf/in^2

v_1=200ft/s

v_2=1700ft/s

T_1=500^{\circ}F

Q=0

C_p=1BTU/lb^{\circ}F

We have to find the exit temperature.

By steady energy flow equation

h_1+v^2_1+Q=h_2+v^2_2

C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

Substitute the values

1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}

500+1.598=T_2+115.4

T_2=500+1.598-115.4

T_2=386.2^{\circ}F

7 0
4 years ago
Skater 1 has a mass of 105.0 kg and a velocity of 2.0 m/s to the left. He
mart [117]

The final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.

<h3>What is velocity?</h3>

Velocity can be defined as the ratio of the displacement and time of a body.

To calculate the final velocity of Skater 1 we use the formula below.

Formula:

  • mu+MU = mv+MV............ Equation 1

Where:

  • m = mass of the first skater
  • M = mass of the second skater
  • u = initial velocity of the first skater
  • U = initial velocity of the second skater
  • v = final velocity of the first skater
  • V = final velocity of the second skater.

make v the subject of the equation.

  • v = (mu+MU-MV)/m................ Equation 2

Note: Let left direction represent negative and right direction represent positive.

From the question,

Given:

  • m = 105 kg
  • u = -2 m/s
  • M = 71 kg
  • U = 5 m/s
  • V = -3.4 m/s.

Substitute these values into equation 2

  • v = [(105×(-2))+(71×5)-(71×(-3.4))]/105
  • v = (-210+355+241.4)/105
  • v = 386.4/105
  • v = 3.68 m/s
  • v ≈ 3.7 m/s

Hence, the final velocity of skater 1 is 3.7 m/s to the right. The right option is O A. 3.7 m/s to the right.

Learn more about velocity here: brainly.com/question/25749514

7 0
2 years ago
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