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damaskus [11]
2 years ago
8

an object has a mass of 50kg, a final height of 20m and an initial height of 8m. what is the amount of work done

Physics
1 answer:
Andrei [34K]2 years ago
7 0

amount of work done is 5880 J

Given:

mass of object = 50kg

Final height = 20m

initial height = 8m

To Find:

amount of work done

Solution:

work is done when a force acts upon an object to cause a displacement. You can calculate the energy transferred, or work done, by multiplying the force by the distance moved in the direction of the force.

The work done by gravity is given by the formula,

W = mgh

W = 50 x 9.8 x ( 20-8)

= 5880 J

So the work done is 5880 J

Learn more about Work done here:

brainly.com/question/25239010

#SPJ4

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A round object of mass 10 kg and radius 0.5 m rolls without slipping down a hill from a height of 4.5 m. If its speed at the bot
Mice21 [21]

Answer:

moment of inertia is 2.72 kg m²

Explanation:

given data

mass m = 10kg

height h = 4.5 m

radius r  = 0.5 m

speed  v = 6.5 m/s

to find out

moment of inertia

solution

we apply here conservation of energy

that is

mgh = 1/2 ×mv² + 1/2 × Iω²

here I is moment of inertia we find and

we know ω = Velocity / radius = 6.5 / 0.5 = 13

and g = 9.8

so put here all these value

10 (9.8) 4.5 = 1/2 ×(10)(6.5)² + 1/2 × I(13)²

441 = 211.25 + 1/2 × I( 169 )

I = 2.72

so moment of inertia is 2.72 kg m²

7 0
4 years ago
A rock thrown with speed 12.0 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 15.5 m bef
Dimas [21]

Supposing there's no air resistance, horizontal velocity is constant, which makes it very easy to solve for the amount of time that the rock was in the air.


Initial horizontal velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s 

15.5m / 10.3923m/s = 1.49s 

So the rock was in the air for 1.49 seconds. </span>

<span>

Now that we know that, we can use the following kinematics equation: 

d = v i * t + 1/2 * a * t^2 

Where d is the difference in y position, t is the time that the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>

<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s 

So: 

d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2 
d = 8.94 + -10.89</span>

d = -1.95<span>

<span>This means that the initial y position is 1.95 m higher than where the rock lands. </span></span>

5 0
3 years ago
Humans are a part of the Earth system. List at least three examples of how you, in particular, influence one or more of Earth’s
Naily [24]

Answer:

True, The humans have modified their environment the ecological system

Explanation:

  • Human is a part of the earth system and they are included in the ecological system that in particular is concerned with the provision of the feedback mechanism and cyclic flow of the system of energy such as the atmosphere, cryosphere, and the biosphere and the hydrosphere.
  • The human has already and influences the earth's physical system alike that of the atmospheric system in terms of the climate changes and the global warming that is taking place at a massive scale.  
  • They are influencing the soil and land systems in terms of the depletion resources and the disturbing the hydrological cycles on earth that further influence the amounts of rainfall patterns.
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8 0
4 years ago
A wheel rotates about a fixed axis with an initial angular velocity of 13 rad/s. During a 8-s interval the angular velocity incr
____ [38]

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}

Where:

\omega_{0}: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

\omega_{f} = \omega_{0} + \alpha t

Where:

\omega_{f}: is the final angular velocity = 57 rad/s

\alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2}

Hence, the number of revolutions is:

\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev

Therefore, the number of revolutions is 44.6.

       

I hope it helps you!

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