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Lapatulllka [165]
3 years ago
7

A 48 kg cart is sitting motionless at the top of the hill. At what height, did the cart start at, to reach 88435 J of energy at

the end of the ride? Group of answer choices 19 meters 456 meters 188 meters 0.19 meters
Physics
1 answer:
svet-max [94.6K]3 years ago
7 0

Answer : The height is 188 meters  

Explanation :   When the cart reached at the end from top of hill then the cart have potential energy .

Given that,

Potential energy = 88435 J

Mass of cart = 48 kg

We know that,

The potential energy is

mgh =88435

h= \dfrac{88435}{48\times9.8}

h = 187.9 = 188\ meters

So, the height of the top is 188 meters.

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How many nanoseconds does it take light to travel 3.50 ft in vacuum?
Fiesta28 [93]
Answer:3.56 nanosecond

In this case, you are asked the time and given the light distance(3.5ft)
To answer this question you would need to know the velocity of light. Speed of light is <span>299792458m/s. Then the calculation would be:

time= distance/speed
time= 3.5 ft / (</span>299792458m/s) x 0.3048 meter/ 1 ft=  3.56 10^{-9} second or 3.56 nanosecond
6 0
3 years ago
Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which
antiseptic1488 [7]

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = K_{f} - K₀

          K_{f}  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        K_{f}  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               K_{f} = (2.7 -2.625) 10¹¹

              K_{f} = 7.5 10⁹ J

5 0
3 years ago
A device consisting of four heavy balls connected by low-mass rods is free to rotate about an axle. It is initially not spinning
zubka84 [21]

The angular speed of the device is 1.03 rad/s.

<h3>What is the conservation of angular momentum?</h3>

A spinning system's ability to conserve angular momentum ensures that its spin will not change until it is subjected to an external torque; to put it another way, the rotation's speed will not change as long as the net torque is zero.

Using the conservation of angular momentum

L_{i}=L_{f}

Here,  = the system's angular momentum before the collision

L_{i} = 0 + mv

= (0.005)(450)(0.752)

= 1.692 kgm²/s

The moment of inertia of the system is given by

I = 2(M₁R₁² + M₂R₂²)+ mR₁²

= 2[(1.2)(0.8)² +(0.5)(0.3)²]+0.005(0.8)²

= 1.6292 kgm²

Here,  = Iω

So,

1.692 = 1.6292(ω)

ω = 1.03 rad/s

To know more about the conservation of angular momentum, visit:

brainly.com/question/1597483

#SPJ1

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1 year ago
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Is friction and pushing similar ????
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Answer:

Yes, they are.

Explanation:

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Assume that the body's muscle mechanism can be approximated by a spring with a uniform continuous mass distribution that follows
tatiyna

Based on Hooke's law, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.

<h3>What is the spring constant?</h3>

The spring constant or stiffness constant of an elastic spring is constant which describes the extent a bit forceapplied to an elastic spring will extend it.

  • Spring constant, K = force/extension

Assuming, a body's muscle mechanism is a spring obeying Hooke's law, the effective mass of the spring with mass m is 1/3 of the mass of the spring = m/3

The potential energy that can be stored = ke^2 / 2

where K is spring constant and e is the extension produced.

Therefore, the spring constant of the the body's muscle mechanism is the ratio of force to extension, the effective mass is m/3 and the potential energy that can be stored is ke^2 / 2.

Learn more about Hooke's law at: brainly.com/question/12253978

5 0
2 years ago
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