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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?
Physics
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

         

3 0
3 years ago
A 10 N force and a 15 N force are acting from a single point in opposite directions. What additional force must be added to prod
AleksAgata [21]

Answer:

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Explanation:

10+5=15

15=15

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White raven [17]

Answer:

refractive index of the unknown material is 1.33.

Explanation:

μ₁ = 1.21

incidence angle (i) = 41.9°

refraction angle (r) = 37.3°

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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