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Rom4ik [11]
4 years ago
8

I need help with question 8 and 9

Physics
1 answer:
Wewaii [24]4 years ago
7 0

8) The height of the shelf is 2.04 m

9) The gravitational potential energy is 100 J

Explanation:

8)

The gravitational potential energy of an object is given by:

PE=mgh

where

m is the mass of the object

g is the acceleration of gravity

h is the heigth of the object, relative to the ground

For the object in this problem, we have

m = 10 kg

g=9.8 m/s^2

And we know that

PE = 200 J

Therefore we can re-arrange the formula to solve for h, the heigth of the object:

h=\frac{PE}{mg}=\frac{200}{(10)(9.8)}=2.04 m

9)

As before, the gravitational potential energy of the object is given by:

PE=mgh

where in this  case, we have:

m = 10 kg is the mass of the object

h = 2.04 m is the height of the object from the ground

While in this case, the acceleration of gravity on this planet is

g=4.9 m/s^2

Substituting,

PE=(10)(4.9)(2.04)=100 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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A turntable, with a mass of 1.5 kg and diameter of 20 cm, rotates at 70 rpm on frictionless bearings. Two 540 g blocks fall from
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Answer:

The turntable's angular speed after the event is 28.687 revolutions per minute.

Explanation:

The system formed by the turntable and the two block are not under the effect of any external force, so we can apply the Principle of Conservation of Angular Momentum, which states that:

I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f} (1)

Where:

I_{T} - Moment of inertia of the turntable, in kilogram-square meters.

r - Distance of the block regarding the center of the turntable, in meters.

m - Mass of the object, in kilograms.

\omega_{o} - Initial angular speed of the turntable, in radians per second.

\omega_{f} - Final angular speed of the turntable-objects system, in radians per second.

In addition, the momentum of inertia of the turntable is determined by following formula:

I_{T} = \frac{1}{2}\cdot M\cdot r^{2} (2)

Where M is the mass of the turntable, in kilograms.

If we know that \omega_{o} \approx 7.330\,\frac{rad}{s}, M = 1.5\,kg, m = 0.54\,kg and r = 0.1\,m, then the angular speed of the turntable after the event is:

I_{T} = \frac{1}{2}\cdot M\cdot r^{2}

I_{T} = 7.5\times 10^{-3}\,kg\cdot m^{2}

I_{T}\cdot \omega_{o} = (2\cdot r^{2}\cdot m +I_{T})\cdot \omega_{f}

\omega_{f} = \frac{I_{T}\cdot \omega_{o}}{2\cdot r^{2}\cdot m +I_{T}}

\omega_{T} = 3.004\,\frac{rad}{s} (28.687\,\frac{rev}{min})

The turntable's angular speed after the event is 28.687 revolutions per minute.

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A 128.0-N carton is pulled up a frictionless baggage ramp inclined at 30.0∘above the horizontal by a rope exerting a 72.0-N pull
Elden [556K]

Answer:

(A) 374.4 J

(B) -332.8 J

(C) 0 J

(D) 41.6 J

(E)  351.8 J

Explanation:

weight of carton (w) = 128 N

angle of inclination (θ) = 30 degrees

force (f) = 72 N

distance (s) = 5.2 m

(A) calculate the work done by the rope

  • work done = force x distance x cos θ
  • since the rope is parallel to the ramp the angle between the rope and

        the ramp θ will be 0

       work done = 72 x 5.2 x cos 0

       work done by the rope = 374.4 J

(B) calculate the work done by gravity

  • the work done by gravity = weight of carton x distance x cos θ
  • The weight of the carton = force exerted by the mass of the carton = m x g
  • the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.

      work done by gravity = 128 x 5.2  x cos 120

      work done by gravity = -332.8 J

(C) find the work done by the normal force acting on the ramp

  • work done by the normal force = force x distance x cos θ
  • the angle between the normal force and the ramp is 90 degrees

       

         work done by the normal force = Fn x distance x cos θ

         work done by the normal force = Fn x 5.2 x cos 90

         work done by the normal force = Fn x 5.2 x 0

         work done by the normal force = 0 J

(D)  what is the net work done ?

  • The net work done is the addition of the work done by the rope,       gravitational force and the normal force

     net work done = 374.4 - 332.8 + 0 =  41.6 J  

(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal

  • work done by the rope= force x distance x cos θ
  • the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp

work done = 72 x 5.2 x cos 20

work done = 351.8 J

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4 years ago
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