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vichka [17]
3 years ago
7

Need help on this thanks​

Physics
1 answer:
lisov135 [29]3 years ago
5 0

Answer:

increase

Explanation:

You might be interested in
A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10r, where a and fare in m/s^2
xz_007 [3.2K]

Answer:

Mistake in question

The correct question

A drag car starts from rest and moves down the racetrack with an acceleration defined by a = 50 - 10t , where a and t are in m/s² and seconds, respectively. After reaching a speed of 125 m/s, a parachute is deployed to help slow down the dragster. Knowing that this deceleration is defined by the relationship a = - 0.02v², where v is the velocity in m/s, determine (a) the total time from the beginning of the race until the car slows back down to 10 m/s, (b) the total distance the car travels during this time.

Explanation:

Given the function

a = 50 —10t

The car started from rest u = 0

And it accelerates to a speed of 125m/s

Then, let find the time in this stage

Acceleration can be modeled by

a = dv/dt

Then, dv/dt = 50—10t

Using variable separation to solve the differentiation equation

dv = (50—10t)dt

Integrating both sides

∫ dv = ∫ (50—10t)dt

Note, v ranges from 0 to 125seconds, so we want to know the time when it accelerate to 125m/s. So t ranges from 0 to t'

∫ dv = ∫ (50—10t)dt

v = 50t —10t²/2. Equation 1

[v] 0<v<125 = 50t —10t²/2 0<t<t'

125—0 = 50t — 5t² 0<t<t'

125 = 50t' — 5t'²

Divide through by 5

25 = 10t' — t'²

t'² —10t' + 25 = 0

Solving the quadratic equation

t'² —5t' —5t' + 25 = 0

t'(t' —5) —5(t' + 5) = 0

(t' —5)(t' —5) = 0

Then, (t' —5) = 0 twice

Then, t' = 5 seconds twice

So, the car spent 5 seconds to get to 125m/s.

The second stage when the parachute was deployed

We want to the time parachute reduce the speed from 125m/s to 10m/s,

So the range of the velocity is 125m/s to 10m/s. And time ranges from 0 to t''

The function of deceleration is give as

a = - 0.02v²

We know that, a = dv/dt

Then, dv/dt = - 0.02v²

Using variable separation

(1/0.02v²) dv = - dt

(50/v²) dv = - dt

50v^-2 dv = - dt

Integrate Both sides

∫ 50v^-2 dv = -∫dt

(50v^-2+1) / (-2+1)= -t

50v^-1 / -1 = -t

- 50v^-1 = -t

- 50/v = - t

Divide both sides by -1

50/v = t. Equation 2

Then, v ranges from 125 to 10 and t ranges from 0 to t''

[ 50/10 - 50/125 ] = t''

5 - 0.4 = t''

t'' = 4.6 seconds

Then, the time taken to decelerate from 125s to 10s is 4.6 seconds.

So the total time is

t = t' + t''

t = 5 + 4.6

t = 9.6 seconds

b. Total distanctraveleded.

First case again,

We want to find the distance travelled from t=0 to t = 5seconds

a = 50—10t

We already got v, check equation 1

v = 50t —10t²/2 + C

v = 50t — 5t² + C

We add a constant because it is not a definite integral

Now, at t= 0 v=0

So, 0 = 0 - 0 + C

Then, C=0

So, v = 50t — 5t²

Also, we know that v=dx/dt

Therefore, dx/dt = 50t — 5t²

Using variable separation

dx = (50t —5t²)dt

Integrate both sides.

∫dx = ∫(50t —5t²)dt

x = 50t²/2 — 5 t³/3 from t=0 to t=5

x' = [25t² — 5t³/3 ]. 0<t<5

x' = 25×5² — 5×5³/3 —0

x' = 625 — 208.333

x' = 416.667m

Stage 2

The distance moved from

t=0 to t =4.6seconds

a = -0.002v²

We already derived v(t) from the function above, check equation 2

50/v = t + C.

When, t = 0 v = 125

50/125 = 0 + C

0.4 = C

Then, the function becomes

50/v = t + 0.4

50v^-1 = t + 0.4

Now, v= dx/dt

50(dx/dt)^-1 = t +0.4

50dt/dx = t + 0.4

Using variable separation

50/(t+0.4) dt = dx

Integrate both sides

∫50/(t+0.4) dt = ∫ dx

50 In(t+0.4) = x

t ranges from 0 to 4.6seconds

50In(4.6+0.4)—50In(4.6-0.4) = x''

x'' = 50In(5) —50In(4.2)

x'' = 8.72m

Then, total distance is

x = x' + x''

x = 416.67+8.72

x = 425.39m

The total distance travelled in both cases is 425.39m

5 0
3 years ago
Read 2 more answers
Student pushes a 50 N block across the floor for a distance of 15 m how much work was done to move the block
Talja [164]

Answer:

750 J

Explanation:

We have a student that pushes a 50N block  across the floor for a distance of 15m. The question is asking how much work was done to move the block.

To solve this, we must know that we are looking for a certain thing called joules. And to get the answer, we must follow the formula of W = FS

F being the force and S being the distance.

W = FS

W = (50)(15)

W = 750

Therefore, 750 joules is our answer.

7 0
3 years ago
A motorcyclist changes the velocity of his bike from 20.0 meters/second to 35.0 meters/second under a constant acceleration of 4
shutvik [7]
You can use the equation V=Vo+at since the acceleration is constant. Plugging in the values you know, you will get an answer of 3.75 seconds
8 0
3 years ago
A heat engine operates between two reservoirs at 300K. During each cycle, it absorbs 200 calories from the high temperature rese
yanalaym [24]

(a) Zero

The maximum efficiency (Carnot efficiency) of a heat engine is given by

\eta=1-\frac{T_C}{T_H}

where

T_C is the low-temperature reservoir

T_H is the high-temperature reservoir

For the heat engine in the problem, we have:

T_C = 300K

T_H = 300K

Therefore, the maximum efficiency is

\eta=1-\frac{300}{300}=0

(b) Zero

The efficiency of a heat engine can also be rewritten as

\eta = \frac{W}{Q_H}

where

W is the work performed by the engine

Q_H is the heat absorbed from the high-temperature reservoir

In this problem, we know

\eta=0

Therefore, since the term Q_H cannot be equal to infinity, the numerator of the fraction must be zero as well, which means

W = 0

So the engine cannot perform any work.

5 0
3 years ago
A force of 100 N is applied to a crate at an angle of 60° to the horizontal surface. The force causes the crate to move 2 m acro
MAVERICK [17]

Answer: B 100 J

Explanation: Pic Attached. Hope This Helps.

4 0
2 years ago
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