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Irina18 [472]
2 years ago
12

A horse does 860 j of work in 420 seconds while pulling a wagon. what is the power output of the horse? round your answer to the

nearest whole number.
Physics
2 answers:
kicyunya [14]2 years ago
6 0

the correct answer is 2w

GenaCL600 [577]2 years ago
6 0

By definition, the power is given by:

P = \frac{W}{t}

Where,

W: work done

t: time spent doing work

Therefore, substituting values we have:

P = \frac{860}{420}\\P = 2.05 watts

Rounding the nearest whole we have:

P = 2 watts

Answer:

The power output of the horse is:

P = 2 watts

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melamori03 [73]
If you subscribe I’ll answer QF Aotrx
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If the Force exerted by the intern is doubled and the distance is halved, does the done by the intern increase, decrease, or rem
Jlenok [28]

Remain the same

Explanation:

If the force exerted by the intern is doubled and the distance is halved, the work done by the intern remains the same.

 Work done is the force applied to move a body through a distance.

Work done = F x d

where F is the applied force

            d is the distance moved

Now;

if:

    f = 2f

    d = \frac{1}{2}d

Input the parameter:

 Work done = fxd = 2f x \frac{1}{2}d  = fd

The work done will still remain the same

learn more:

Work done brainly.com/question/9100769

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3 0
2 years ago
Joel uses a claw hammer to remove a nail from a wall. He applies a force of 40 newtons on the hammer. The hammer applies a force
jarptica [38.1K]

Hi!


Mechanical advantage is defined as the<em> ratio of force produced by an object to the force that is applied to it.</em>

In our case, this would be the ratio of the force applied by the claw hammer on the nail to the force Joel applies to the claw hammer, which is

160:40 or 4:1

So the mechanical advantage of the hammer is four.


Hope this helps!


3 0
2 years ago
Read 2 more answers
A solid ball is released from rest and slides down a hillside that slopes downward at 65.0" from the horizontal
PilotLPTM [1.2K]
Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline 
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
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3 0
2 years ago
What's the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmi
Murrr4er [49]

Complete Question:

The Voyager 1 spacecraft is now beyond the outer reaches of our solar system, but earthbound scientists still receive data from the spacecraft s 20-W radio transmitter. Voyager is expected to continue transmitting until about 2025, when it will be some 25 billion km from Earth.

What s the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmitter on Voyager transmits equally in all directions(isotropically).  In fact, the antenna on Voyager focuses the signal in a beam aimed at the earth, so this problem over-estimates the size of the receiving dish needed.

Answer:

d = 2,236 m.

Explanation:

The received power on Earth, can be calculated as the product of the intensity (or power density) times the area that intercepts the power radiated.

As we assume that  the transmitter antenna is ominidirectional, power is spreading out over a sphere with a radius equal to the distance to the source.

So, we can get the power density as follows:

I = P /A = P / 4*π*r², where P = 20 W, and r= 25 billion km = 25*10¹² m.

⇒ I = 20 W / 4*π* (25*10¹²)² m²

The received power, is just the product of this value times the area of the receiver antenna, which we assumed be a circle of diameter d:

Pr = I. Ar =( 20W / 4*π*(25*10¹²)² m²) * π * (d²/4) = 10⁻²⁰ W

Simplifying common terms, we can solve for d:

d= √(16*(25)²*10⁴/20) = 2,236 m.

3 0
2 years ago
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