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aleksandrvk [35]

3 years ago

12

A 60 kg box is lifted by a rope a distance of 10 meters straight up at constant speed. how much power is required to complete th

is task in 5 seconds?

2 answers:

Nadya [2.5K]3 years ago

3 0

Power=Work/Time
The work done is the energy required to lift the box, fighting the force of gravity. So, Work=Potential energy of the box at 10 meters.

W=PE=mgh=(60)(9.8)(10)=5880J
Finally,
P=W/T=(5880)/(5)=1176Watt

So the answer is 1176 Watts

GrogVix [38]3 years ago

3 0

Answer: 1,200Watts

Explanation:

Power is defined as the rate of change in work done.

Mathematically, Power = Workdone/Time

Since work done = Force × Distance

Power = Force×Distance/Time

Given mass = 60kg

Since we are given mass, we will convert it to force using;

F= ma = 60×10 = 600N

Distance = 10meters

Time = 5seconds

Substituting the values in the formula for power, we have;

Power = 600×10/5

Power = 6000/5

Power = 1,200Watts

Therefore, 1,200 watts is required to complete the work

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Instead of moving back and forth, a conical pendulum moves in a circle at constant speed as its string traces out a cone (see fi

tigry1 [53]

Answer:

a

The radial acceleration is $a_c = 0.9574 m/s^2$

b

The horizontal Tension is $T_x = 0.3294 i \ N$

The vertical Tension is $T_y =3.3712 j \ N$

Explanation:

The diagram illustrating this is shown on the first uploaded

From the question we are told that

The length of the string is $L = 10.7 \ cm = 0.107 \ m$

The mass of the bob is $m = 0.344 \ kg$

The angle made by the string is $\theta = 5.58^o$

The centripetal force acting on the bob is mathematically represented as

$F = \frac{mv^2}{r}$

Now From the diagram we see that this force is equivalent to

$F = Tsin \theta$ where T is the tension on the rope and v is the linear velocity

So

$Tsin \theta = \frac{mv^2}{r}$

Now the downward normal force acting on the bob is mathematically represented as

$Tcos \theta = mg$

So

$\frac{Tsin \ttheta }{Tcos \theta } = \frac{\frac{mv^2}{r} }{mg}$

=> $tan \theta = \frac{v^2}{rg}$

=> $g tan \theta = \frac{v^2}{r}$

The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as

$a_c = \frac{v^2}{r}$

=> $a_c = gtan \theta$

substituting values

$a_c = 9.8 * tan (5.58)$

$a_c = 0.9574 m/s^2$

The horizontal component is mathematically represented as

$T_x = Tsin \theta = ma_c$

substituting value

$T_x = 0.344 * 0.9574$

$T_x = 0.3294 \ N$

The vertical component of tension is

$T_y = T \ cos \theta = mg$

substituting value

$T_ y = 0.344 * 9.8$

$T_ y = 3.2712 \ N$

The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is

$T = T_x i + T_y j$

substituting value

$T = [(0.3294) i + (3.3712)j ] \ N$

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3 years ago

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AleksAgata [21]

Answer:

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3 years ago

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White raven [17]

Answer:

refractive index of the unknown material is 1.33.

Explanation:

μ₁ = 1.21

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Let us assume μ be the refractive index of the unknown material

according to snell's law of refraction.

μ₁ sin i = μ₂ sin r

1.21 × sin 41.9° = μ × sin 37.3°

μ = 1.33

hence the refractive index of the unknown material comes out top be 1.33

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