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galben [10]
3 years ago
10

A sled is dragged along a horizontal path at a constant speed of 1.50 m/s by a rope that is inclined at an angle of 30.0° with r

espect to the horizontal. The total weight of the sled is 470 N. The tension in the rope is 230 N. How much work is done by the rope on the sled in a time interval of 10.0 s?
Physics
1 answer:
Allisa [31]3 years ago
4 0

Answer:2987.7 J

Explanation:

Given

sled weight is W=470 N

Tension in rope T=230 N

Time interval t=10 s

velocity of sled v=1.5 m/s

here speed is equal to velocity which is constant therefore acceleration is zero

thus F_{net}=0

Distance traveled in t=10 s

x=v\times t=1.5\times 10=15 m

considering surface to be friction less

thus T\cos 30 will do work on sled

W=F\cdot x

W=T\cos 30\cdot x

W=230\cos 30\times 15

W=2987.78 J\approx 2.987 kJ

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An electric motor with a 1200 hp output drives a machine with an efficiency of 60%. What is the energy output of the machine per
kirill [66]

Answer:

536,904 J/s

Explanation:

The energy output from motor is the input energy in the machine.

We know that efficiency is percentage energy ouput to energy input, and expressed as

n=\frac {E_o}{E_i}

Where n and E represent efficiency and energy respectively, subscripts o and i represent output and input respectively. Since for the machine we have the input energy then the output will be the product of efficiency and input energy

Energy output=0.6*1200 hp=720 hp

Converting hp to J/s we multiply by 745.7

Energy is 720*745.7=536,904 J/s

4 0
3 years ago
Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 wi
Flauer [41]

Answer:

a. Fnet =37.67N

b. The direction = 133.4 from the x axis counter clockwise.

c. Option 2

Explanation:

Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.

Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..

Given that F1 is 71N at 20°, then it is in the first quadrant.

a. Fnet= F1+F2+F3

Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j

Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j

Resolving the vectors into x and y components.

Fnet= -2.04i+37.62j

Magnitude of the vector

Fnet= √((-2.04)^2+(37.62)^2)

Fnet= 37.67N

Fnet is approximately 38N.

b. Direction of the Fnet.

Angle=arctan(y/x)

Angle=arctan(-37.61/2.04)

Angle= -43.37°

The angle is in the negative x axis and positive y axis.

Then the direction becomes 180-43.37

Therefore, the direction of the net force is 133.37°.

c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.

Fnet=ma

Fnet= mv/t

So the velocity is in the direction of the Fnet.

3 0
3 years ago
What advantages does an electromagnet have over a permanent magnet?
KonstantinChe [14]
Electromagnets can be turned off, this makes it easier to release things from the magnetic field. 

Hope this helps :)
4 0
3 years ago
A tank of liquid (SG=0.80) that is 1 ft in diameter and 1.0 ft high is rigidly fixed to a rotating arm having a 2 ft radius. The
IgorLugansk [536]

Answer:

the pressure at B is 527psf

Explanation:

Angular velocity, ω = v / r

ω = 20 /1.5

= 13.333 rad/s

Flow equation from point A to B

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the pressure at B is 527psf

6 0
3 years ago
How long does it take for a train to increase its velocity from 10m/s to 40m/s if it accelerates at 3 m/s
DIA [1.3K]

Answer:

Explanation:

Givens

Vi = 10 m/s

Vf = 40 m/s

a = 3 m/s^2

Formula

a = (vf - vi) /t              Substitute the givens into this formuls

Solution

3 = (40 - 10) / t          Multiply both sides by t

3*t = t(40 - 10)/t        Combine. Cancel t's on the right

3*t = 30                     Divide by 3

3t/3 = 30 / 3

Answer: t = 10 seconds.

6 0
2 years ago
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