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LUCKY_DIMON [66]
3 years ago
8

A 762 kg car experiences a braking force of 9045 N and skids to a stop in 4.3 seconds. What is the speed of the car just before

the braking force was applied?
Physics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0

Answer:

The solution for this problem is:

We will be using the formula for force which is F = ma 

=>10,000 = 2000 * a 

but we need to solve for acceleration so divide both sides by 2000, we will get:

=>a = 5 m/s^2 

Let the initial velocity was u m/s 

=>By v = u - at 

=>0 = u - 5 x 6 

Since acceleration is constant the velocity can be computed by multiplying the acceleration by 6 seconds. 

=>u = 30 m/s

Explanation:

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An airplane has a speed of 135 mi/h in still air. It is flying straight north so that it is always directly above a north-south
vlabodo [156]

Answer:

The answer is below

Explanation:

Let vₐ be the speed of airplane = 135 mph, vₙ be the speed of the wind = 70 mph and vₐₙ be the speed of the airplane relative to the wind.

The distance (d) = 135 miles, Δt = 1 hour, vₐₙ = 135 miles / 1 hour = 135 mph

vₐ = vₙ + vₐₙ

vₐ = vₐₙ

Therefore, vₐ, vₐₙ, vₙ can be represented by an isosceles triangle since vₐ = vₐₙ.

The direction of the wind θ is:

sin(θ / 2) = vₙ / 2vₐ

sin(θ / 2) = 70/ (2*135)

sin(θ / 2) = 0.2593

θ / 2 = sin⁻¹(0.2593) = 15

θ = 30⁰

2α = 180° - 30°

2α = 150°

α = 75°

a) The direction of the wind is 75° in the south east direction while the airplane is heading 30° in the north east direction.

8 0
3 years ago
A 4kg ball moving at 8m\sec
Llana [10]
Conservation of momentum: total momentum before = total momentum after

Momentum = mass x velocity

So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32

Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32

Therefore β = 12.8 m/s
8 0
3 years ago
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
How do animals get the nitrogen they need<br>​
Alenkinab [10]

Answer:

Animals get the nitrogen they need by eating plants or other animals that contain nitrogen.

Explanation:

4 0
3 years ago
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Using language from Newton's laws of motion, describe why the space shuttle seen in the picture is able to take off.
melomori [17]
Beacuse of the gravity
6 0
3 years ago
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