All categories

3 years ago

What is the force and distance relationship for a fixed pulley?

Physics

1 answer:

leva [86]3 years ago

6 0

Answer:

Force is a vector quantity which when applied to an object can cause change in its velocity. When Force is applied to an object, the object if stationary moves to a certain distance depending upon the amount of force applied.

A pulley is a wheel with a groove that holds a rope, cable or belt, and is used to help lift an object.

The relationship of Force and Distance is the same for a fixed pulley as well. Except for one thing:

Normally the distance covered by an object is in the direction of the Force.

Using a pulley, an object can be moved with that same Force in a different direction. Using a pulley, a person can apply force in a direction suitable for him while moving the object.

<em> Force = mass of an object * Distance Covered</em>

You might be interested in

A woman is 1.6 m tall and has a mass of 50 kg. She moves past an observer with the direction of the motion parallel to her heigh

eduard

To solve this problem it is necessary to apply the concepts related to linear momentum, velocity and relative distance.

By definition we know that the relative velocity of an object with reference to the Light, is defined by

$V_0 = \frac{V}{\sqrt{1-\frac{V^2}{c^2}}}$

Where,

V = Speed from relative point

c = Speed of light

On the other hand we have that the linear momentum is defined as

P = mv

Replacing the relative velocity equation here we have to

$P = \frac{mV}{\sqrt{1-\frac{V^2}{c^2}}}$

$P^2 = \frac{m^2V^2}{1-\frac{V^2}{c^2}}$

$P^2 = \frac{P^2V^2}{c^2}+m^2V^2$

$P^2 = V^2 (\frac{P^2}{c^2}+m^2)$

$V^2 = \frac{P^2}{\frac{P^2}{c^2}+m^2}$

$V^2 = \frac{(2.1*10^10)^2}{\frac{(2.1*10^10)^2}{(3.8*10^8)^2}+50^2}$

$V = 2.81784*10^8m/s$

Therefore the height with respect the observer is

$l = l_0*\sqrt{1-\frac{V^2}{c^2}}$

$l = 1.6*\sqrt{1-\frac{(2.81*10^8)^2}{(3*10^8)^2}}$

$l = 0.56m$

Therefore the height which the observerd measure for her is 0.56m

8 0

3 years ago

9. A ball is thrown straight up with an initial speed of 30 m/s. How long will it take to reach the top of its trajectory, and h

ziro4ka [17]

Answer:

1.) Time t = 3.1 seconds

2.) Height h = 46 metres

Explanation:

given that the initial velocity U = 30 m/s

At the top of the trajectory, the final velocity V = 0

Using first equation of motion

V = U - gt

g is negative 9.81m/^2 as the object is going against the gravity.

Substitute all the parameters into the formula

0 = 30 - 9.81t

9.81t = 30

Make t the subject of formula

t = 30/9.81

t = 3.058 seconds

t = 3.1 seconds approximately

Therefore, it will take 3.1 seconds to reach to reach the top of its trajectory.

2.) The height it will go can be calculated by using second equation of motion

h = ut - 1/2gt^2

Substitutes U, g and t into the formula

h = 30(3.1) - 1/2 × 9.8 × 3.1^2

h = 93 - 47.089

h = 45.911 m

It will go 46 metres approximately high.

6 0

3 years ago

An inelastic collision is one in which:

Anna35 [415]

Answer:

The last one

Explanation:

the last one. KE before is greater than after. because

1. Velocity before the collision is greater than after

2. Mass before is smaller than after

3 0

3 years ago

A cat chase a mouse across a 1.0 m high table. The mouse steps out of the way, and the car slides off the table and strikes the

Tema [17]

The cat fell 1.0 m from the ground.

Using the formula

$h = v_{oy}t - \frac{gt^2}{2}$

Here, $v_{oy} = 0, \; h = 1.0 \; m.$

Solving for t, the time it spent in the air is

$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.0 \; m)}{9.8 \; m/s^2}} = 0.451753951 \; s$

The cat does not accelerate along the horizontal, so it has constant horizontal velocity. Since it strikes the floor 2.2 m from the table, then

$v_x = \Delta x/t = \frac{2.2\; m}{0.451753951 s} = 4.869907597 \; m/s \Rightarrow 4.9 \; m/s$

5 0

3 years ago

A object is 200 miles from the center of a circle and makes a circuit every 2 hours. How fast is it moving?

Tanya [424]

Answer: 628.31 miles/h

Explanation:

The equation to calculate the average speed $s$ of an object moving in a circular path is:

$s=\frac{2 \pi r}{T}$ (1)

Where:

$r=200 miles$ is the radius of the circumference

$T=\frac{1}{f}$ is the period, which has an inverse relation with the frequency $f=\frac{1 circuit}{2 h}=0.5 h^{-1}$ . Then $T=\frac{1}{0.5 h^{-1}}=2 h$

Solving the equation:

$s=\frac{2 \pi (200 miles)}{2 h}$ (2)

$s=628.31 miles/h$ (3)

4 0

3 years ago