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3 years ago

A crate with a mass of 1000 kg is being pulled along greased tracks by a winch. The winch is exerting a force of 2000 N

Physics

1 answer:

Vesnalui [34]3 years ago

3 0

<h3><u>We are given:</u></h3>

Coefficient of Kinetic Friction = 0.2

Mass of the crate = 1000 kg

Force applied = 2000 N

<h2 /><h2><u>Net force Acting on the crate in the Horizontal Direction:</u></h2><h3><u>Normal Force applied on the Crate:</u></h3>

Normal force is the force applied on the object by the surface. It is equal and opposite to the force of gravity

So, we can say that Normal force = | Gravitational Force |

Normal Force = | mg |

Normal Force = 1000 * 9.8

Normal Force = 9800 N

<h3><u>Finding the Frictional Force:</u></h3>

We know that:

coefficient of Kinetic friction = Friction force / Normal force

<em>replacing the known values</em>

0.2 = Friction force / 9800

Friction Force= 0.2 * 9800

Friction Force = 1960 N

<h3><u>Net force acting on the Crate:</u></h3>

We know that a force of 2000 N is being applied on the crate in the Horizontal direction

Frictional force is always opposite to the horizontal force. So, we can say that:

Applied force - Friction Force = Net Force

<em>replacing the variables</em>

2000 - 1960 = Net force

Net Force = 40N

Therefore, a net force of 40N is being applied on the Crate

<h2></h2><h2><u>Acceleration of the Crate:</u></h2>

From newton's second law of motion:

F = ma

<em>replacing the variables</em>

40 = 1000 * a

a = 40/1000

a = 0.04 m/s²

Hence, the crate will have an acceleration of 0.04 m/s²

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Answer:

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Explanation:

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$y_1 = -120sin70^o = -112.763 km$

After the 2nd flight its coordinate would be:

$x_2 = x_1 - 75sin60^o = -41.04 - 64.95 = -106km$

$y_2 = y_1 + 75cos60^o = -112.763 + 37.5 = -75.263 km$

So in order to fly back to its HQ it must fly a distance and direction of

$s = \sqrt{y_2^2 + x_2^2} = \sqrt{75.263^2 + 106^2} = \sqrt{5664.519169 + 11236} = \sqrt{16900.519169} = 130 km$

$tan\theta = \frac{y_2}{x_2} = \frac{75.263}{106} = 0.71$

$\theta = tan^{-1}0.71 = 0.62 rad \approx 35.38^o$ north of east

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A perfectly flexible cable has length L, and initially it is at rest with a length Xo of it hanging over the table edge. Neglect

zaharov [31]

Answer:

$X=X_o+\dfrac{1}{2}gt^2$

Explanation:

Given that

Length = L

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Lets take the length =X after time t

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Now by energy conservation

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$V=\sqrt{2g(X-X_o)}$

We know that

$\dfrac{dX}{dt}=V$

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$\sqrt{2g}\ dt=(X-X_o)^{-\frac{1}{2}}dX$

At t= 0 ,X=Xo

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$X=X_o+\dfrac{1}{2}gt^2$

So the length of cable after time t

$X=X_o+\dfrac{1}{2}gt^2$

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3 years ago

An atom of gadolinium has an atomic number of 64 and a mass number of 154. how many electrons, protons, and neutrons does it con

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