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stealth61 [152]
3 years ago
15

A piston in an internal combustion engine applies torque during 150° of each full rotation of the crankshaft to which it is conn

ected. Suppose an 8-cylinder engine (8 such pistons) is running with its crankshaft turning at 2500 rpm and produces 410J of work each second. What is the torque applied to the crank shaft by one of the engine’s eight pistons?
Physics
1 answer:
aalyn [17]3 years ago
4 0

Answer:

T=1.566 N.m

Explanation:

Given that:

rotational speed of the scrank shaft, N = 2500 rpm

power produced by one cylinder, P = 410 W

We know, in case of rotational power:

P=\frac{2\pi.N.T}{60}

where: T= torque

Substituting the respective values in the above eq.

410=\frac{2\pi\times2500\times T}{60}

T= \frac{410\times60}{2\pi\times2500}

T=1.566 N.m is the torque applied by the each piston of the engine.

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<h2>Answer:</h2>

C.

<h2>Explanation:</h2>

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3 years ago
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From a branch 35 m high, a 0.75 kg bird dives into a small fish tank containing
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Answer:

ΔT = 1.22*10^-3 °C

Explanation:

First, you calculate the potential energy of the bird when it is at 35 m high. The potential energy is also the mechanical energy of the bird in this case.

U=mgh

m: mass of the bird = 0.75kg

g: gravitational constant = 9.8m/s^2

h: height = 35m

U=(0.75kg)(9.8m/s^2)(35m)=257.25\ J

All this energy is given to the water. You use the following formula in order to calculate the change in temperature:

Q=mc\Delta T

m: mass of the water = 50kg

c: specific heat of water = 4186 J/kg°C

Q is equal to U (potential energy of the bird) because the bird gives all its energy to water. By doing ΔT the subject of the formula you obtain:

\Delta T=\frac{Q}{mc}=\frac{257.25J}{(50kg)(4186J/kg°C)}=1.22*10^{-3}\ \°C

hence, the maximum rise in temperature is 0.00122 °C

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D. Graphing the force as a function of distance and calculating the area under the curve.

Explanation:

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A basketball with a mass of 0.5 kilograms is accelerated at 2
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Answer: 1N

Explanation: its not 0N.

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3 years ago
1. Differentiate between speed and velocity.
Mazyrski [523]

Answer:

1. Speed and velocity both involve a numeric rate describing the distance traveled by a body in a unit of time. However, speed describes the rate of a body traveling in any direction in a unit of time, while velocity describes the rate of a body traveling in a particular direction in a unit of time.

2. Answers may vary, but should resemble the following:

Average velocity explains the velocity the body traveled overall, not taking into consideration each spot in the trip. If a car moves at 65 km/h on average, it may have slowed down for some parts and sped up for others. Overall though, it would have made a certain distance of travel within a specified unit of time that totals the average velocity of 65 km/h.

Instantaneous velocity explains the velocity of a body at a particular instant of the trip. The instantaneous velocity of a car stopped at a stop sign would be 0 m/s even if it was moving before and will continue to move after this stop. The velocity at that particular instant is the instantaneous velocity.

Uniform velocity is when the distance being covered is changing uniformly with time. For example, if a car moves 20 km every 30 minutes and continues to do so in the same direction, it's traveling with a uniform velocity.

3. a=v2−v1t

a=20 m/s−60 m/s6 s

a=−406

a = –6.7 m/s2

4. v2 = v1 + at

v2 = 14 m/s + (3 m/s2 × 6 s)

v2 = 14 + 18

v2 = 32 m/s

5. v=st

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6. First, convert the minutes to seconds. Since there are 60 seconds in one minute, multiply:

60 × 15 (minutes) = 900 seconds

s = v × t

s = 6 m/s × 900 s

s = 5,400 m

7. t=sv

t=80 km35 km/hr

t = 2.29 hr

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a=50 m/s−15 m/s4 s

a=35 m/s4 s

a = 8.75 m/s2

9. vav=v1+v22

vav=15 m/s+50 m/s2

vav=65 m/s2

vav = 32.5 m/s

10. a=v2−v1t

a=0 m/s−11.5 m/s3.5 s

a = –3.29 m/s2

Explanation:

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