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3 years ago

PLEASE REAL ANSWERS IM SUPER BEHIND

Physics

1 answer:

natima [27]3 years ago

8 0

Answer:

25000W

Explanation:

Kinetic Energy, EK=(1/2)mv2. Energy is conserved, so the work done by the brakes, W, must be equal to the change in kinetic energy of the car, ∆EK. Therefore: W = ∆EK = (1/2)1000(152-52) = 0.51000200 = 100000 = 1*105J. Since power P = W/T, the average power output of the brakes is equal to the total work done by the brakes over the time that the brakes are applied for, therefore: P = 100000/4 = 25000W.

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Leokris [45]

Because of the power of the light it is very strong so that is why light reaches all the way to earth

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3 years ago

A car travels 20 km South, then turns and travels 30 km East? What is the total displacement?

Ne4ueva [31]

Answer:

Explanation:36.05 km

Given

First car travels $r_1=20\ km$ South

then turns and travels $r_2=30\ km$ east

Suppose south as negative y axis and east as positive x axis

So, $r_1=-20\hat{j}$

$r_2=30\hat{i}$

Displacement is the shortest between initial and final point

Dispalcement $=r=r_1+r_2$

Displacement $=-20\hat{j}+30\hat{i}$

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4 0

3 years ago

What is 6,210 bucks converted to kilobucks?

Vinvika [58]

When 6210 bucks is converted to kilobucks, the result obtained is 6.21 kilobucks

<h3>Conversion scale </h3>

1000 bucks = 1 Kilobuck

Using the above convesion scale, we can express 6210 bucks in kilobucks

<h3>How to express bucks in kilobucks</h3>

1000 bucks = 1 Kilobuck

Therefore,

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Learn more about conversion:

brainly.com/question/2139943

4 0

2 years ago

Read 2 more answers

A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the

olga2289 [7]

Answer:

a) F = 30 N, b) I = 12 N s , c) I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

p₀ = m v

The speed can be found by kinematics

v = v₀ - g t

v = 0 - 10 0.4

v = -4.0 m / s

Final after division

pf = m₁ v₁f + m₂ v₂f

p₀ = pf

M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

F = mg

F = 3 10 = 30 N

b) The expression for momentum is

I = Ft

Before the explosion the only force that acts is the weight

I = mg t

I = 3 10 0.4

I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

M v = m₁ v₁f + m₂ v₂f

v₂f = (M v - m₁ v₁f) / m₂

v₂f = (3 (-4) - 1 6) / 2

v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

I = F t = Dp

F t = pf –po)

F t= [m₁ v₁f + m₂ v₂f]

I = [1 6 + 2 (-9) -0]

I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

ΔI =If – I₀

ΔI = - 12 – (-12)

ΔI = -0 N s

3 0

3 years ago

A race car starts from rest and travels east along a straight and level track. For the first 5.0 s of the car’s motion, the east

Marrrta [24]

Answer:

ax = 6.43m/s²

Explanation:

The acceleration is the time derivative of the velocity function ax = dvx(t)/dt

We have been given the velocity function v(t) and also the velocity v = 12.0m/s and we are requested to calculate the acceleration at this time which we don't know.

So the first step is to calculate the time at which the velocity =12.0m/s and with this time calculate the acceleration. Detailed solution can be found in the attachment below.

7 0

3 years ago