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2 years ago

Calculate the power in kilowatts required to lift a 500 kg mass 12 meters in one second

1 answer:

8 0

The weight of the load is (mass) x (gravity).

The work done to lift it is (weight) x (distance) = (mass) x (gravity) x (distance)

The power is (work)/(time) =

   (mass) x (gravity) x (distance) / (time) =

   (500 kg) x (9.8 m/s²) x (12 m) / (1 sec) =

(500 x 9.8 x 12 / 1) (kg-m / sec²) (m) / (sec) =

   (500 x 9.8 x 12 / 1) newton - meter / sec =

(500 x 9.8 x 12 / 1) joule    / sec =

   58,800 watts

   = <em>58.8 kW</em>

   =    about 78.8 horsepower

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Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water wit

melamori03 [73]

Answer:

Explanation:

Given that,

Height of the bridge is 20m

Initial before he throws the rock

The height is hi = 20 m

Then, final height hitting the water

hf = 0 m

Initial speed the rock is throw

Vi = 15m/s

The final speed at which the rock hits the water

Vf = 24.8 m/s

Using conservation of energy given by the question hint

Ki + Ui = Kf + Uf

Where

Ki is initial kinetic energy

Ui is initial potential energy

Kf is final kinetic energy

Uf is final potential energy

Then,

Ki + Ui = Kf + Uf

Where

Ei = Ki + Ui

Where Ei is initial energy

Ei = ½mVi² + m•g•hi

Ei = ½m × 15² + m × 9.8 × 20

Ei = 112.5m + 196m

Ei = 308.5m J

Now,

Ef = Kf + Uf

Ef = ½mVf² + m•g•hf

Ef = ½m × 24.8² + m × 9.8 × 0

Ef = 307.52m + 0

Ef = 307.52m J

Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw

7 0

2 years ago

The weight of Earth's atmosphere exerts an average pressure of 1.01 ✕ 105 Pa on the ground at sea level. Use the definition of p

zloy xaker [14]

Answer:

The weight of Earth's atmosphere exert is $516.6\times10^{17}\ N$

Explanation:

Given that,

Average pressure $P=1.01\times10^{5}\ Pa$

Radius of earth $R_{E}=6.38\times10^{6}\ m$

Pressure :

Pressure is equal to the force upon area.

We need to calculate the weight of earth's atmosphere

Using formula of pressure

$P=\dfrac{F}{A}$

$F=PA$

$F=P\times 4\pi\times R_{E}^2$

Where, P = pressure

A = area

Put the value into the formula

$F=1.01\times10^{5}\times4\times\pi\times(6.38\times10^{6})^2$

$F=516.6\times10^{17}\ N$

Hence, The weight of Earth's atmosphere exert is $516.6\times10^{17}\ N$

8 0

2 years ago

Read 2 more answers

1. What is the momentum of a golf ball with a mass of 62 g moving at 73 m/s?

Anit [1.1K]

Answer:

<h3>The answer is 4.53 kgm/s</h3>

Explanation:

The momentum of an object can be found by using the formula

<h3>momentum = mass × velocity</h3>

From the question

mass = 62 g = 0.062 kg

velocity = 73 m/s

We have

momentum = 0.062 × 73 = 4.526

We have the final answer as

Hope this helps you

4 0

2 years ago

The different in size of each of the rope's pullers, correspond to a difference in the magnitude of the applied force, such that

olga55 [171]

Answer:

F = - 50 N

Hence, the magnitude of resultant force is 50 N and its direction is leftwards.

Explanation:

The magnitude of the resultant force is always equal to the sum of all forces. While, the direction of resultant force will be equal to the direction of the force with greater magnitude:

$Resultant\ Force = F = F_{1} - F_{2}$