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dlinn [17]
3 years ago
12

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

Physics
1 answer:
Arisa [49]3 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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A block attached to a spring with an unknown spring constant oscillates with a period of 2.0 s. What is the period if
Zigmanuir [339]

Answer:

a) If the mass is doubled, then the period is increased by \sqrt{2}. Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

Explanation:

The statement is incomplete. We proceed to present the complete statement: <em>A block attached to a spring with unknown spring constant oscillates with a period of 2.00 s. What is the period if </em><em>a. </em><em>The mass is doubled? </em><em>b.</em><em> The mass is halved? </em><em>c.</em><em> The amplitude is doubled? </em><em>d.</em><em> The spring constant is doubled? </em>

We have a block-spring system, whose angular frequency (\omega) is defined by the following formula:

\omega = \sqrt{\frac{k}{m} } (1)

Where:

k - Spring constant, measured in newtons per meter.

m - Mass, measured in kilograms.

And the period (T), measured in seconds, is determined by the following expression:

T = \frac{2\pi}{\omega} (2)

By applying (1) in (2), we get the following formula:

T = 2\pi\cdot \sqrt{\frac{m}{k} }

a) If the mass is doubled, then the period is increased by \sqrt{2}. Hence, the period of the system is 2.828 seconds.

b) If the mass is halved, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

c) The period of the system does not depend on amplitude. Hence, the period of the system is 2 seconds.

d) If the spring constant is doubled, then the period is reduced by \frac{\sqrt{2}}{2}. Hence, the period of the system is 1.414 seconds.

8 0
3 years ago
In order to identify them, animal fossils and other fossils nearby can be: A. compared B. piled C. seen
tigry1 [53]

Answer:

A.compared

Explanation:

Fossils help figure out the time that organisms lived. If you know one of the fossils, it can be used as a reference for others around.

5 0
3 years ago
How to convert work done in joules into kilojoules?
Sav [38]

1 Kilojoule [kJ] = 737.562 149 277 27 Foot pound force [ftlbf]

3 0
1 year ago
Read 2 more answers
How to do this question
Anna71 [15]

Answer:

(a) 10 m/s

(b) 22.4 m/s

Explanation:

(a) Draw a free body diagram of the car when it is at the top of the loop.  There are two forces: weight force mg pulling down, and normal force N pushing down.

Sum of forces in the centripetal direction (towards the center):

∑F = ma

mg + N = mv²/r

At minimum speed, the normal force is 0.

mg = mv²/r

g = v²/r

v = √(gr)

v = √(10 m/s² × 10.0 m)

v = 10 m/s

(b) Energy is conserved.

Initial kinetic energy + initial potential energy = final kinetic energy

½ mv₀² + mgh = ½ mv²

v₀² + 2gh = v²

(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²

v = 22.4 m/s

4 0
3 years ago
A 5.0 kg rock is dropped from the top of a building. The speed of the rock after it has fallen for 2.2 seconds is?
PSYCHO15rus [73]
5x2.2/3 Because that's how you calculate velocity.

5 0
3 years ago
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