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dlinn [17]
2 years ago
12

Determine the torque applied to the shaft of a car that transmits 225 hp

Physics
1 answer:
Arisa [49]2 years ago
8 0

Incomplete question.The complete question is here

Determine the torque applied to the shaft of a car that transmits 225 hp and rotates at a rate of 3000 rpm.

Answer:

Torque=0.51 Btu

Explanation:

Given Data

Power=225 hp

Revolutions =3000 rpm

To find

T( torque )=?

Solution

As

T(Torque)=\frac{W(Work)}{2\pi n(Revolutions) }

As force moves an object through a distance, work is done on the object. Likewise, when a torque rotates an object through an angle, work is done.

So

T=\frac{225*42.207}{2\pi 3000}\\ T=0.51 Btu

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I could be wrong, but I'm pretty sure it's 144kg.
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2 years ago
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A cylinder has 500 cm3 of water. After a mass of 100 grams of sand is poured into the cylinder and all air bubbles are removed b
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Answer: SG = 2.67

Specific gravity of the sand is 2.67

Explanation:

Specific gravity = density of material/density of water

Given;

Mass of sand m = 100g

Volume of sand = volume of water displaced

Vs = 537.5cm^3 - 500 cm^3

Vs = 37.5cm^3

Density of sand = m/Vs = 100g/37.5 cm^3

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3 0
2 years ago
A standing wave of the third overtone is induced in a stopped pipe, 2.5 m long. The speed of sound is The frequency of the sound
NemiM [27]

Answer:

f3 = 102 Hz

Explanation:

To find the frequency of the sound produced by the pipe you use the following formula:

f_n=\frac{nv_s}{4L}

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vs: speed of sound = 340 m/s

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You replace the values of n, L and vs in order to calculate the frequency:

f_{3}=\frac{(3)(340m/s)}{4(2.5m)}=102\ Hz

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2 years ago
A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas
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F=qE

F_p=F_e

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by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

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(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

\frac{m_ev_e^2}{2d}=qE

we can equals these expressions for both proton and electron, because the forces qE are the same:

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4 0
2 years ago
_____ discovered that light also showed properties commonly found in waves.
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