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3 years ago

How much work is done by the force of gravity when a 45 N object falls to the ground from a height of 4.6 m?

Physics

2 answers:

saul85 [17]3 years ago

5 0

Work Done = 45N x 4.6m = 207J

shutvik [7]3 years ago

4 0

Answer:

207Nm

Explanation:

Work on said to be done when an applied force causes an object to move through a particular distance. mathematically,

Work done = Force(f)×distance in the direction of force.

Given force of gravoty = 45N

Height = corresponding Distance= 4.6m

Work done = 45×4.6

Work done= 207Nm

This shows that 207Nm of work is done by the force of gravity when a 45 N object falls to the ground from a height of 4.6 m.

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The conducting path between the right hand and the left hand can be modeled as a 9.0-cm-diameter, 140-cm-long cylinder. The aver

Crank

Answer:

The potential difference is 121.069 V

Solution:

As per the question:

Diameter of the cylinder, d = 9.0 cm = 0.09 m

Length of the cylinder, l = 40 cm = 1.4 m

Average Resistivity, $\rho = 5.5\ \Omega-m$

Current, I = 100 mA = 0.1 A

Now,

To calculate the potential difference between the hands:

Cross- sectional Area of the Cylinder, A = $\pi (\frac{d}{2})^{2} = 6.36\times 10^{- 3}\ m^{2}$

Resistivity is given by:

$\rho = R\frac{A}{l}$

$R = \rho \frac{l}{A}$

$R = 5.5\times \frac{1.4}{6.36\times 10^{- 3}} = 1210.69\ \Omega$

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4 0

3 years ago

Three forces act on an object. Two of the forces are at an angle of 100◦to each other and have magnitude 25N and 12N. The third

seraphim [82]

Answer:

F₄ = 29.819 N

Explanation:

Given

F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N

F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N

F₃ = (0 i + 0 j + 4 k) N

Then we have

F₁ + F₂ + F₃ + F₄ = 0

⇒ F₄ = - (F₁ + F₂ + F₃)

⇒ F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N

The magnitude of the force will be

F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N

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3 years ago

A 25kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in motion. Once the chair is in

SCORPION-xisa [38]

Answer:

$\mu_k=0.51$

Explanation:

Given that

Mass , m = 25 kg

We know that when body is in rest condition then static friction force act on the body and when body is in motion the kinetic friction force act on the body .That is why these two forces are given as follows

Static friction force ,fs= 165 N

Kinetic friction force ,fk = 127 N

If the body is moving with constant velocity ,it means that acceleration of that body is zero and all the forces are balanced.

Lets take coefficient of kinetic friction = μk

The kinetic friction is given as follows

fk = μk m g

Now by putting the values

127 = μk x 25 x 9.81

$\mu_k=\dfrac{127}{25\times 9.81}$

$\mu_k=0.51$

Therefore the value of coefficient of kinetic friction will be 0.51

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Marysya12 [62]

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