All categories

3 years ago

How does the input distance of a single fixed pulley compare to the out- put distance?

1 answer:

4 0

A pulley is another sort of basic machine in the lever family. We may have utilized a pulley to lift things, for example, a banner on a flagpole.

<u>Explanation:</u>

The point in a fixed pulley resembles the support of a lever. The remainder of the pulley behaves like the fixed arm of a first-class lever, since it rotates around a point. The distance from the fulcrum is the equivalent on the two sides of a fixed pulley. A fixed pulley has a mechanical advantage of one. Hence, a fixed pulley doesn't increase the force.

It essentially alters the direction of the force. A moveable pulley or a mix of pulleys can deliver a mechanical advantage of more than one. Moveable pulleys are appended to the item being moved. Fixed and moveable pulleys can be consolidated into a solitary unit to create a greater mechanical advantage.

You might be interested in

A weight trainer lifts a 90.0-kg barbell from a stand 0.90 m high and raises it to a height of 1.75 m. What is the increase in t

nirvana33 [79]

Answer:

∆PE = 749.7 J

At 0.9 m high, PE = 793.8 J

At 1.75 m high, PE = 1543.5 J

7 0

2 years ago

A vessel with an unknown volume is filled with 10 kg of water at 90oC. Inspection of the vessel at equilibrium shows that 8 kg o

damaskus [11]

Hey there!

In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).

Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:

$x=\frac{m_{steam}}{m_{total}}$

Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:

$x=\frac{2kg}{10kg} =0.20$

Now, at this temperature and pressure, the volume of a saturated vapor is 2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:

$v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg$

This means that the volume of the container will be:

$V=10kg*0.4727m^3/kg\\\\V=4.73m^3$

Regards!

8 0

2 years ago

1-Find the gravitational force

butalik [34]

$g = \frac{GMm}{r^2}$

where $G$ is the gravitational constant and is about $6.67 * 10^{-11}$

$g = \frac{6.67 * 10^{-11} * 25 * 0.55}{35^2}$

$g = \frac{6.67 * 10^{-11} * 25 * 0.55}{1225} = 0.000000000000749 N$ :D

5 0

3 years ago

Suppose the velocity of an object moving along a line is positive. are position, displacement, and distance traveled equal? expl

Pavlova-9 [17]

No, the object's displacement and distance travelled will be equal, but since the initial position is unknown, the object's position might not match up with its displacement and distance travelled.

We cannot assert that the displacement or distance equals the position because the initial position is not provided. We could reach a different conclusion if the starting position had been zero because the distance from zero is equal to the position.

Find more on velocity at : brainly.com/question/11347225

#SPJ4

4 0

2 years ago

An 18.8 kg block is dragged over a rough, horizontal surface by a constant force of 156 N acting at an angle of angle 31.9◦ abov

Sav [38]

Answer:

a) W = 2635.56 J

b) Wf = 423.27 J

c) c) The Sign of the work done by the frictional force (Wf) is negative (-)

d) W=0

Explanation:

Work (W) is defined as the Scalar product of the force (F) by the distance (d) that the body travels due to this force .

The formula for calculate the work is :

W = F*d*cosα

Where:

W : work in Joules (J)

F : force in Newtons (N)

d: displacement in meters (m)

α :angle that form the force (F) and displacement (d)

Known data

m = 18.8 kg : mass of the block

F= 156 N,acting at an angle θ = 31.9◦°: angle above the horizontal

μk= 0.209 : coefficient of kinetic friction between the cart and the surface

g = 9.8 m/s²: acceleration due to gravity

d = 19.9 m : displacement of the block

Forces acting on the block

We define the x-axis in the direction parallel to the movement of the cart on the floor and the y-axis in the direction perpendicular to it.

W: Weight of the cart : In vertical direction downaward

N : Normal force : In vertical direction the upaward

F : Force applied to the block

f : Friction force: In horizontal direction

Calculated of the weight of the block

W= m*g = ( 18.8 kg)*(9.8 m/s²)= 184.24 N

x-y components of the force F

Fx = Fcosθ = 156 N*cos(31.9)° = 132.44 N

Fy = Fsinθ = 156 N*sin(31.9)° = 82.44 n

Calculated of the Normal force

Newton's second law for the block in y direction :

∑Fy = m*ay ay = 0

N-W+Fy= 0

N-184.24+82,44= 0

N = 184.24-82,44

N = 101.8 N

Calculated of the kinetic friction force (fk):

fk = μk*N = (0.209)*( 101.8)

fk = 21.27 N

a) Work done by the F=156N.

W = (Fx) *d *cosα

W = (132.44 )*(19.9)(cos0°) (N*m)

W = 2635.56 J

b) Work done by the force of friction

Wf = (fk) *d *cos(180°)

Wf = (21.27 )*(19.9) (-1) (N*m)

Wf = - 423.27 J

Wf = 423.27 J :magnitude

c) The Sign of the work done by the frictional force is negative (-)

d) Work done by the Normal force

W = (N) *d *cos(90°)

W = (101.8 )*(19.9) (0) (N*m)

W = 0

4 0

3 years ago