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2 years ago

If the tension in the rope is 160 n, how much work does the rope do on the skier during a forward displacement of 270 m?

2 answers:

8 0

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on the skier is,

$W=T.S$

⇒ $W=270*160*cos\pi \\W=-43200 J$

Hence work done by the rope is - 43200 J.

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Mandarinka [93]2 years ago

6 0

The correct answer is - 43200 J.

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on the skier is,⇒

Hence work done by the rope is - 43200 J.

What is work ?

Work is the energy transferred to or from an object by applying force along a displacement in physics. It is frequently represented as the product of force and displacement in its most basic form. When a force is applied, it is said to do positive work if it has a component in the direction of the point of application's displacement. If a force has a component that is opposite the direction of displacement at the force's application point, it does negative work.

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A child is trying to throw a ball over a fence. She gives the ball an initial speed of 8.0 m/s at an angle of 40° above the hori

EastWind [94]

Answer:

the child is 1.581 m far from the fence

Explanation:

The diagrammatic illustration that give a better view of what the question denote can be seen in the image attached below.

From the image attached below, let assume that the release point is the origin, then equation of the motion (x) is as follows:

$x - x_o = u_xt$

$\mathtt{x = u_xt \ \ \ since (x_o = 0)}$ ---- (1)

the equation of the motion y is :

$\mathtt{y - y_o =u_yt - 0.5 gt^2}$

$\mathtt{y = u_yt-4.9t^2 \ \ \ since (y_o =0)}$

$\mathtt{ 1= (u \ sin 40^0)t -4.9 \ t^2 }$

$\mathtt{1 = 8 sin 40^0 t - 4.9 t^2}$

$\mathtt{1 = 5.14t - 4.9t^2}$

$\mathtt{4.9t^2 - 5.14t +1 = 0}$

By using the quadratic formula, we have;

$\mathtt{ \dfrac{ -b \pm \sqrt{b^2 - 4ac}}{2a}} }$

where;

a = 4.9, b = -5.14 c = 1

$= \mathtt{ \dfrac{ -(-5.14) \pm \sqrt{(-5.14)^2 - 4(4.9)(1)}}{2(4.9)}} }$

$= \mathtt{ \dfrac{ 5.14 \pm \sqrt{26.4196 -19.6}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14 \pm \sqrt{6.8196}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14+ \sqrt{6.8196}}{9.8} \ \ OR \ \ \dfrac{ 5.14- \sqrt{6.8196}}{9.8}} }$

$= \mathtt{ \dfrac{ 5.14+ 2.6114}{9.8} \ \ OR \ \ \dfrac{ 5.14- 2.6114}{9.8}} }$

$= \mathtt{ \dfrac{ 7.7514}{9.8} \ \ OR \ \ \dfrac{ 2.5286}{9.8}} }$

$= \mathbf{ 0.791 \ \ OR \ \ 0.258} }$

In as much as the ball is traveling upward, then we consider t= 0.258sec.

From equation (1)

$\mathtt{x = u_x(0.258)}$

$\mathtt{x = ucos 40^0 (0.258)}$

$\mathtt{x = 8 \ cos 40^0 (0.258)}$

$\mathbf{x = 1.581 \ m}$

Thus, the child is 1.581 m far from the fence

6 0

3 years ago

A 0.300 kg ball, moving with a speed of 2.5 m/s, has a head-on collision with at 0.600 kg ball initially at rest. Assuming a per

FrozenT [24]

Answer:

1.25 m/s

Explanation:

Given,

Mass of first ball=0.3 kg

Its speed before collision=2.5 m/s

Its speed after collision=2 m/s

Mass of second ball=0.6 kg

Momentum of 1st ball=mass of the ball*velocity

=0.3kg*2.5m/s

=0.75 kg m/s

Momentum of 2nd ball=mass of the ball*velocity

=0.6 kg*velocity of 2nd ball

Since the first ball undergoes head on collision with the second ball,

momentum of first ball=momentum of second ball

0.75 kg m/s=0.6 kg*velocity of 2nd ball

Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg

=1.25 m/s

4 0

3 years ago

You are standing at the top of a cliff that has a stairstep configuration. There is a vertical drop of 7 m at your feet, then a

Zolol [24]

Answer:

1. v = 6.67 m/s

2. d = 9.54 m

Explanation:

1. To find the horizontal velocity of the rock we need to use the following equation:

$d = v*t \rightarrow v = \frac{d}{t}$

<u>Where</u>:

d: is the distance traveled by the rock

t: is the time

The time can be calculated as follows:

$t = \sqrt{\frac{2d}{g}}$

<u>Where:</u>

g: is gravity = 9.8 m/s²

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*7 m}{9.8 m/s^{2}}} = 1.20 s$

Now, the horizontal velocity of the rock is:

$v = \frac{d}{t} = \frac{8 m}{1.20 s} = 6.67 m/s$

Hence, the initial velocity required to barely reach the edge of the shell below you is 6.67 m/s.

2. To calculate the distance at which the projectile will land, first, we need to find the time:

$t = \sqrt{\frac{2d}{g}} = \sqrt{\frac{2*(7 m + 3 m)}{9.8 m/s^{2}}} = 1.43 s$

So, the distance is:

$d = v*t = 6.67 m/s*1.43 s = 9.54 m$

Therefore, the projectile will land at 9.54 m of the second cliff.

I hope it helps you!

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3 years ago

3) A 15.0kg cannonball is ascending after being launched. Neglect drag.

WINSTONCH [101]

Answer:

Explanation:

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3 years ago

Ordinary ocean waves are created by the interaction of the ____ and ____.

dmitriy555 [2]

Wind and amplitude creates the waves in an ordinary ocean

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3 years ago