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cestrela7 [59]
2 years ago
8

A 2000 kg car accelerates from 0 to 25 m / s in 21.0 s. How much is the average power delivered by the motor? (1 hp = 746 W)

Physics
2 answers:
mash [69]2 years ago
3 0

Answer: 40.2 hp

Explanation:

The mass of the car is 2000kg.

in 21 seconds, it accelerates from 0m/s to 25m/s, so the average acceleration is:

a = (25m/s)/(21s) = 1.2 m/s^2

We know that force = mass*acceleration, then:

F = 2000kg*1.2m/s^2 = 2400N.

Now, the Power can be written as:

P = W/t where W is work = F*d (force per distance) and t is time.

then we have:

P = (F*d)/t and d/t  is the average velocity, and we know the velocity v = 25m/s, so the average velocity will be (1/2)*25m/s = 12.5m/s

P = F*12.5m/s

and the force is F = 2400N

P = 2400 N*12.5m/s = 30,000 W

and we know that 1hp = 746 W

then, P = (30,000/746) hp = 40.2 hp

DIA [1.3K]2 years ago
3 0

Answer:

P=40hp

Explanation:

Hello,

In this cased, for the given velocities, we can compute the work done by the car as shown below, considering the kinetic energy only:

W=\frac{1}{2}m(v_f^2-v_0^2)

Whereas the mass is 2000 kg, the final velocity is 25 m/s and the initial velocity is 0 m/s, therefore the work is:

W=\frac{1}{2}*2000kg*[(25\frac{m}{s})^2  -(0\frac{m}{s})^2]\\\\W=\frac{1}{2}*2000kg*625\frac{m^2}{s^2} \\\\W=6.25x10^5J

Now, the power:

P=\frac{W}{t}=\frac{6.25x10^5}{21.0s}=29762W

And in horsepower:

P=29762W*\frac{1hp}{746W} \\\\P=40hp

Best regards.

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Answer:

Explanation:

a. The source of centripetal force on the car is  (3) the static friction force.

b. ac = v²/R = (20²)/50 = 8 m/s²

c.  Fc = m(ac) = 1500(8) = 12 kN

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A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

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maksim [4K]

Answer:

The work is -67.76 J

Explanation:

The law of conservation of energy is considered one of one of the fundamental laws of physics and states that the total energy of an isolated system remains constant. except when it is transformed into other types of energy.

This is summed up in the principle that energy can neither be created nor destroyed in the universe, only transformed into other forms of energy.

In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.

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You know:

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Replacing:

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<u><em>The work is -67.76 J</em></u>

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