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kondor19780726 [428]
4 years ago
13

A student weighing 200N climbs a flight of stairs 7.0m high in 8s. What power is required to perform this task?

Physics
1 answer:
noname [10]4 years ago
3 0

Answer:

Power required, P = 175 watts

Explanation:

It is given that,

Weight of a student, F = mg = 200 N

The student climbs a flight of stairs of height, h = 7 m

Time taken, t = 8 s

We have to find the power required to perform this task. Work done per unit time is called the power required. Mathematically, it is given by :

P=\dfrac{W}{t}

W = work done

t = time taken

P=\dfrac{mgh}{t}

P=\dfrac{200\ N\times 7\ m}{8\ s}

P = 175 watts

Hence, the power required to complete this task is 175 watts.

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This is the Conservation of energy. We say energy is conserved because none is lost in the process.
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4 years ago
What is the speed of a transverse wave in a rope of length 3.1 m and mass 86 g under a tension of 380 n?
stellarik [79]

The speed of a transverse wave( v) = 117.03 m/s

The formula we can use in this case would be:

v = sqrt (T / (m / l))

Where,

v = is the velocity of the transverse wave = unknown (?)

T = is the tension on the rope = 380 N

m = is the mass of the rope = 86.0 g = 0.086 kg

l = is the length of the rope = 3.1 m

Substituting the given values into the equation to search for the speed v:

v = sqrt (380 N/(0.086 kg /3.1 m))

v = sqrt (380 * 3.1/ 0.086)

v = sqrt (13,697.67)

v = 117.03 m/s

speed of a transverse wave( v) = 117.03 m/s

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4 0
1 year ago
According to newtons third law of motion action nd reaction forces are equal in magnitudes nd opposite in direction but they do
snow_lady [41]

Because the action and the reaction forces act on different objects

Explanation:

Newton's third law of motion states that:

<em>"When an object A exerts a force on an object B (action), then object B exerts an equal and opposite force (reaction) on object A"</em>

From the statement above, we clearly see that the two forces mentioned in the law act on different objects. In fact, the action is exerted on object B, while the reaction is exerted on object A.

When we are considering the free-body diagram of an object, we just represent all the forces acting on that object, but not the forces exerted on other objects: this means that the action and the reaction never appear in the same free-body diagram of the same object, so they do not cancel out, simply because they are applied to different objects.

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6 0
3 years ago
2.5 years into minutes?<br><br>​
Mnenie [13.5K]
Is it 1314000 Or something
8 0
3 years ago
You tie the loose end of a 0.1 kg yo-yo string to your finger and then release the yo-yo so that it spins down toward the ground
Nikitich [7]

Answer:

The answer is "5.06 \times 10^{-6} \ kg \ m^2"

Explanation:

\to E_1=0..............(i)\\\\\to E_2= \frac{mV^2}{2} +\frac{Iw^2}{2} - mgh.............(ii)\\\\ \Delta E=0\\\\\to mgh= \frac{mV^2}{2} +\frac{Iw^2}{2} \\\\ \to 2 \ mgh=  mV^2 +Iw^2\\\\ \to 2 \ mgh- mV^2 =Iw^2\\\\ \to  m(2gh- V^2) =Iw^2\\\\ \to I= \frac{m(2gh- V^2)}{w^2}

       = 5.06 \times 10^{-6} \ kg \ m^2

3 0
3 years ago
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