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ZanzabumX [31]
3 years ago
13

An older-model car accelerates from 0 to speed v in 9.0 s. A newer, more powerful sports car of the same mass accelerates from 0

to 3v in the same time period. Assuming the energy coming from the engine appears only as kinetic energy of the cars, compare the power of the two cars. (Take Pnew and Pold as the powers of the new car and old car, respectively.)
Physics
1 answer:
Sergio [31]3 years ago
6 0

Answer:The new car has 9 times the power of the old car

Explanation:

Given data

old car

Mass =m

Velocity =v

Time to attain the velocity t= 9secs

New car

Mass =m (same as old car)

Velocity =3v

Time to attain the velocity t= 9secs

We know that the kinetic energy expression is given as

KE=1/2mv²

KE for old car =1/2mv²J

KE for new car =1/2m(3v)²

=1/2m9v²J

Now the expression for power is

P= energy *time

Pold = 1/2mv²*9

4.5(mv²)Hp

Pnew = 1/2m*9v²*9

1/2m*81*v²

40.5(m*v²)Hp

Now comparing the powers of both cars

Pnew/Pold =40.5mv²/4.5mv²

Pnew/Pold = 9

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Explanation:

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Be-36 what hull type is best for use on ponds, small lakes and calm rivers?
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Read 2 more answers
A cylinder with rotational inertia I1 = 3.0 kg · m2 rotates clockwise about a vertical axis through its center with angular spee
erastova [34]

Answer:

ω = 1.83 rad/s clockwise

Explanation:

We are given:

I1 = 3.0kg.m2

ω1 = -5.4rad/s (clockwise being negative)

I2 = 1.3kg.m2

ω2 = 6.4rad/s  (counterclockwise being positive)

By conservation of the momentum:

I1 * ω1 + I2 * ω2 = (I1 + I2) * ω

Solving for ω:

\omega = \frac{I1 * \omega1 + I2*\omega2}{I1+I2}=-1.83rad/s

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8 0
3 years ago
A 2-kg box is pushed by a force of 4 N for 2 seconds. It has an initial velocity vo = 2 m/s to the right. NOTE: Since this probl
IRISSAK [1]

Answer:

Kf= 36 J

W(net) = 32 J

Explanation:

Given that

m = 2 kg

F= 4 N

t= 2 s

Initial velocity ,u= 2 m/s

We know that rate of change of linear momentum is called force.

F= dP/dt

F.t = ΔP

ΔP = Pf - Pi

ΔP = m v  - m u

v= Final velocity

By putting the values

4 x 2 = 2 ( v - 2)

8 =  2 ( v - 2)

4 = v - 2

v= 6 m/s

The final kinetic energy Kf

Kf= 1/2 m v²

Kf= 0.5 x 2 x 6²

Kf= 36 J

Initial kinetic energy Ki

Ki = 1/2 m u²

Ki= 0.5 x 2 x 2²

Ki = 4 J

We know that net work is equal to the change in kinetic energy

W(net) = Kf - Ki

W(net) = 36 - 4

W(net) = 32 J

7 0
3 years ago
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