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4 years ago

What is the mechanical advantage of the wheel and axle shown below?

Physics

2 answers:

Salsk061 [2.6K]4 years ago

7 0

Answer:

Mechanical advantage of the wheel and axle is 32.5

Explanation:

Given that,

The radius of the axle, $r_a=0.8\ cm$

The radius of wheel , $r_w=26\ cm$

We need to find the mechanical advantage of the wheel and axle. It can be calculated as :

$MA=\dfrac{r_w}{r_a}$

$MA=\dfrac{26}{0.8}$

MA = 32.5

So, the mechanical advantage of the wheel and axle shown below is 32.5 . Hence, this is the required solution.

BigorU [14]4 years ago

6 0

The correct answer is A. 32.5

Mechanical advantage is the ratio of force that is input into a machine to the force output.

Mechanical advantage of a wheel and axle is calculated by dividing the radius of the wheel by that of the axle.

MA=R/r where R is the radius of the wheel and r is the radius of the axle.

Substituting for the values in the question gives:

MA=26cm/0.8cm

=32.5

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Answer:

Conduction is the transfer of heat between substances that are in direct contact with each other. The better the conductor, the more rapidly heat will be transferred. Metal is a good conduction of heat. Conduction occurs when a substance is heated, particles will gain more energy, and vibrate more.

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4 years ago

Please help me . i need to answer it asap thankyou

alex41 [277]

The amount of heat needed to melt the block of lead is $5.4\cdot 10^4 J$

Explanation:

The amount of heat needed to melt a solid substance already at its melting point is given by the equation

$Q=m \lambda_f$

where

m is the mass of the substance

$\lambda_f$ is the specific latent heat of fusion

In this problem, we have a block of lead already at its melting point, $327^{\circ}C$ . Its mass is

m = 200 g = 0.2 kg

and its specific latent heat of fusion is

$\lambda_f = 2.7\cdot 10^5 J/kg$

Therefore, the amount of heat needed for the block to completely melt is

$Q=(0.2)(2.7\cdot 10^5)=5.4\cdot 10^4 J$

Learn more about specific heat:

brainly.com/question/3032746

brainly.com/question/4759369

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3 years ago

A hockey puck slides off the edge of a table with an initial velocity of 23.2 m/s and experiences no air resistance. The height

Dennis_Churaev [7]

Answer:

15.1°

Explanation:

The horizontal velocity of the hockey puck is constant during the motion, since there are no forces acting along this direction:

$v_x = 23.2 m/s$

Instead, the vertical velocity changes, due to the presence of the acceleration due to gravity:

$v_y(t)= v_{y0} -gt$ (1)

where

$v_{y0}=0$ is the initial vertical velocity

g = 9.8 m/s^2 is the gravitational acceleration

t is the time

Since the hockey puck falls from a height of h=2.00 m, the time it needs to reach the ground is given by

$h=\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2.00 m)}{9.8 m/s^2}}=0.64 s$

Substituting t into (1) we find the final vertical velocity

$v_y = -(9.8 m/s^2)(0.64 s)=-6.3 m/s$

where the negative sign means that the velocity is downward.

Now that we have both components of the velocity, we can calculate the angle with respect to the horizontal:

$tan \theta = \frac{|v_y|}{v_x}=\frac{6.3 m/s}{23.2 m/s}=0.272\\\theta = tan^{-1} (0.272)=15.1^{\circ}$

6 0

3 years ago

A box with mass m = 8 kg is pushed x = 10 m across a level floor by a constant applied force F P = 16.27 N. The coefficient of k

vitfil [10]

Answer:

Final speed of the box after it has moved to x = 10 is given as

v = 3.35 m/s

Explanation:

As we know by work energy theorem that work done by all the forces is equal to the change in its kinetic energy

Here work done by external force + work done by friction = change in kinetic energy of the box

so we have

$W_{ex} + W_{fric} = \frac{1}{2}mv^2$

$F x - \mu mg x = \frac{1]{2}mv^2$

$16.27 (10) - (0.15)(8)(9.8) (10) = \frac{1}{2}(8) v^2$

$162.7 - 117.6 = 4 v^2$

$v = 3.35 m/s$

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3 years ago

Find the horizontal and vertical components of a projectile that is fired with a velocity of 50m/s at 30 degrees relative to the

VashaNatasha [74]

Find this answer on safari and if you can’t reply to me and i’ll help

5 0

3 years ago