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solong [7]
3 years ago
14

What is the acceleration of a car that goes from zero to 60m/s in 15s?

Physics
1 answer:
bonufazy [111]3 years ago
3 0

Answer:What is the acceleration of a car that moves from rest to 15.0 m/s in 10.0 s? Vi=0, vf= 15.0 m/s,t=10.0s, a=? a= vf =vi/tA=15.0m/s-0m/s/10.0s = 15.0s/10.0s m/s*1/s =1.50 m/s^2 11.

Explanation:

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A sprinf has a potential energy of 84.08 J and a constant of 342.25 N/m. How far it been stretched? Use potential energy elastic
valina [46]

The spring has been stretched 0.701 m

Explanation:

The elastic potential energy of a spring is the potential energy stored in the spring due to its compression/stretching. It is calculated as

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the elongation of the spring with respect to its equilibrium position

For the spring in this problem, we have:

E = 84.08 J (potential energy)

k = 342.25 N/m (spring constant)

Therefore, its elongation is:

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(84.08)}{342.25}}=0.701 m

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

8 0
3 years ago
Can someone pls help, thank you in advance!
zzz [600]

Answer:

  an object sliding down hill

Explanation:

On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.

7 0
3 years ago
Read this quotation:
Brums [2.3K]

Answer:

the Princess in "St. George and the Dragon"

4 0
3 years ago
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IgorLugansk [536]
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4 0
2 years ago
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The rectangular region of the xy plane shown has nonuniform surface charge density σ = σ0 (1+ y b), where σ0 is a constant. Find
sergiy2304 [10]

Answer:

This is net charge on the surface  is  Q = σ₀ x (y + 2by²)

Explanation:

The surface charge density is defined as the amount of charge Q per unit area A

       σ = dq / dA

       dq = σ dA

Since the surface is a rectangular region we use an xy coordinate system so the area difference  

      dA = dxdy

      dq = σ dx dy

 We replace, evaluate the integral

        ∫ dq = ∫ σ₀ (1 + yb) dxdy

realizamos laintegral de dx

        Q -0 =σ₀ ∫ (1 + yb) (x-0)   dy

Where we evaluate We must recognize that the charge Q must be zero by the time X = 0 and Y = 0. At the starting point Q = 0 for x = 0

 

We perform the other integral (dy)

        Q = σ₀ x (y + 2y² b)

Evaluated between Y = 0 and Y = y

      Q = σ₀ x (y + 2by²)

This is net charge on the surface

8 0
3 years ago
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