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Alenkinab [10]
3 years ago
13

A machine uses a force of 200 newtons to do 20,000 joules of work in 20 seconds. Find the distance the object moved and the powe

r of the machine.
Physics
1 answer:
Sindrei [870]3 years ago
7 0
Work=force * distance
20,000 joules= 200 newtons * d 
divide both sides by 200

100= distance

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If a child starts from rest at point A and lands in the water at point B, a horizontal distance L = 2.52 m from the base of the
tamaranim1 [39]

Answer:

The height of the water slide is 0.878 m

Explanation:

Given that,

Distance = 2.52 m

Suppose Children slide down a friction less water slide that ends at a height of 1.80 m above the pool.

We need to calculate the time

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

Put the value in the equation

1.80=0+\dfrac{1}{2}\times9.8\times t^2

t^2=\dfrac{1.80\times2}{9.8}

t=\sqrt{\dfrac{1.80\times2}{9.8}}

t=0.606\ sec

We need to calculate the velocity

Using formula of velocity

v = \dfrac{d}{t}

Put the value into the formula

v=\dfrac{2.52}{0.606}

v=4.15\ m/s

We need to calculate height

Using conservation of energy

\dfrac{1}{2}mv^2=mgh

h=\dfrac{v^2}{2g}

Put the value into the formula

h=\dfrac{4.15^2}{2\times9.8}

h=0.878\ m

Hence, The height of the water slide is 0.878 m.

4 0
3 years ago
. Calculate the kinetic energy of a 100.0-kg meteor approaching the Earth at a speed of 10.0 km/s. Remember that 1 km = 1000 m.
anastassius [24]
Ke = (1/2)mv²

m = 100kg, v = 10 km/s = 10*1000 = 10000m/s

Ke = (1/2)*100*10000 

Ke = 500000 Joules
7 0
3 years ago
Read 2 more answers
Given the 1-m stick shown below, which is held by a thread at its center. Block 1 is 15 N held at the 10 cm mark, while block 2
sattari [20]
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3 years ago
A block of a plastic material floats in water with 42.9% of its volume under water. What is the density of the block in kg/m3?
adell [148]

To solve this problem we will apply the principle of buoyancy of Archimedes and the relationship given between density, mass and volume.

By balancing forces, the force of the weight must be counteracted by the buoyancy force, therefore

\sum F = 0

F_b -W = 0

F_b = W

F_b = mg

Here,

m = mass

g =Gravitational energy

The buoyancy force corresponds to that exerted by water, while the mass given there is that of the object, therefore

\rho_w V_{displaced} g = mg

Remember the expression for which you can determine the relationship between mass, volume and density, in which

\rho = \frac{m}{V} \rightarrow m = V\rho

In this case the density would be that of the object, replacing

\rho_w V_{displaced} g = V\rho g

Since the displaced volume of water is 0.429 we will have to

\rho_w (0.429V) = V \rho

0.429\rho_w= \rho

The density of water under normal conditions is 1000kg / m ^ 3, so

0.429(1000) = \rho

\rho = 429kg/m^3

The density of the object is 429kg / m ^ 3

7 0
3 years ago
I need help on this
ZanzabumX [31]
What is the question asking?
6 0
3 years ago
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