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3 years ago

A crane performs 1820 J of work by lifting an object 35 meters. How much force did the crane exert?

Physics

1 answer:

RUDIKE [14]3 years ago

5 0

Answer:

52 N

Explanation:

Given:

Work done= 1820 J, distance= 35 meters Force = ?

Formula for work done

W = F • d

Find F.

F = W / d

Substitute the given into the equation.

F = 1820 J / 35 m

= 52 N

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Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.160. If the patch i

Olegator [25]

Answer:

14.1 m/s

Explanation:

From the question,

μk = a/g...................... Equation 1

Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.

make a the subject of the equation

a = μk(g).................. Equation 2

Given: μk = 0.160, g = 9.8 m/s²

Substitute into equation 2

a = 0.16(9.8)

a = 1.568 m/s²

Using,

F = ma

Where F = force, m = mass.

Make m the subject of the equation

m = F/a................... Equation 3

m = 160/1.568

m = 102.04 kg.

Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.

Assuming the initial velocity of the skier to be zero

Fd+mgμ = 1/2mv²........................Equation 4

Where v = speed of the skier after crossing the patch, d = distance/width of the patch.

v = √2(Fd+mgμ)/m)................ Equation 5

Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16

Substitute these values into equation 5

v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]

v = √197.57

v = 14.1 m/s

v = 9.86 m/s

4 0

3 years ago

A rock, with a density of 3.55 g/cm^3 and a volume of 470 cm^3, is thrown in a lake. a) What is the weight of the rock out of th

SIZIF [17.4K]

Answer:

a) Weight of the rock out of the water = 16.37 N

b) Buoyancy force = 4.61 N

c) Mass of the water displaced = 0.47 kg

d) Weight of rock under water = 11.76 N

Explanation:

a) Mass of the rock out of the water = Volume x Density

Volume = 470 cm³

Density = 3.55 g/cm³

Mass = 470 x 3.55 = 1668.5 g = 1.6685 kg

Weight of the rock out of the water = 1.6685 x 9.81 = 16.37 N

b) Buoyancy force = Volume x Density of liquid x Acceleration due to gravity.

Volume = 470 cm³

Density of liquid = 1 g/cm³

$\texttt{Buoyancy force}= \frac{470\times 1\times 9.81}{1000} = 4.61 N$

c) Mass of the water displaced = Volume of body x Density of liquid

Mass of the water displaced = 470 x 1 = 470 g = 0.47 kg

d) Weight of rock under water = Weight of the rock out of the water - Buoyancy force

Weight of rock under water = 16.37 - 4.61 =11.76 N

3 0

3 years ago

Resistance increases with the length of the wire

Marta_Voda [28]

If the length of the wire increases, then the amount of resistance will also increase.

1. Take a long piece of wire and cut it 10 pieces. Those pieces should all be different sizes, one should be 5___ (units in meter, cm, inches, etc.), and the next should be 5 ___ (units in meter, cm, inches, etc.) more than the one before.
2. Take one piece of wire and measure the resistance using ___ and record the results in the data table.
3. Repeat the previous step with all the pieces of wire.
4. Compare and contrast the results you have found.

I hope this helps a bit :)

4 0

3 years ago

The idea that the Earth sits motionless in the Universe at the center of a revolving globe of stars, with the Moon and planets i

andrew-mc [135]

Earth sits motionless in the universe at the center of a revolving globe of starts , with the moon and planets in orbit around the earth, is the surrounding model of the uninverse

6 0

2 years ago

What is the net charge of a metal ball if there are 21,749 extra electrons in it?

pickupchik [31]

Answer:

$Q=3.47\times 10^{-15}\ C$