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alexdok [17]
2 years ago
8

boy uses a simple pulley to pull a block up a ramp. The height of the ramp is 4 m and the length of the ramp is 10 m. Calculate

the mechanical advantage of the ramp and does the pulley provide any additional advantage?
Physics
1 answer:
Levart [38]2 years ago
8 0

(a) The mechanical advantage of the ramp is determined as 2.5.

(b) The pulley provides additional advantage since it assisted the boy in pulling up the load.

<h3>What is mechanical advantage?</h3>

The mechanical advantage of an object is the ratio of output force to input force, hence it is called force ratio.

<h3>Mechanical advantage of the ramp</h3>

M.A = Lp/h

where;

  • Lp is the length of the inclined plane
  • h is height of the inclined plane

M.A = 10/4

M.A = 2.5

Learn more about mechanical advantage here: brainly.com/question/25984831

#SPJ1

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In a displacement versus time graph, what does the slope of a line at any point indicate?
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In a displacement versus time graph, the slope of the line at any point on the graph indicates the <em>magnitude of velocity</em>.  

(It can't indicate velocity completely, because the graph shows nothing about the direction of the motion.)

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3 years ago
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A 1.50-kg iron horseshoe initially at 550°C is dropped into a bucket containing 25.0 kg of water at 20.0°C. What is the final te
Ber [7]

Answer:

Te =  23.4 °C

Explanation:

Given:-

- The mass of iron horseshoe, m = 1.50 kg

- The initial temperature of horseshoe, Ti_h = 550°C

- The specific heat capacity of iron, ci = 448 J/kgC

- The mass of water, M = 25 kg

- The initial temperature of water, Ti_w = 20°C

- The specific heat capacity of water, cw = 4186 J/kgC

Find:-

What is the final temperature of the water–horseshoe system?

Solution:-

- The interaction of horseshoe and water at their respective initial temperatures will obey the Zeroth and First Law of thermodynamics. The horseshoe at higher temperature comes in thermal equilibrium with the water at lower temperature. We denote the equilibrium temperature as (Te) and apply the First Law of thermodynamics on the system:

                             m*ci*( Ti_h - Te) = M*cw*( Te - Ti_w )

- Solve for (Te):

                             m*ci*( Ti_h ) + M*cw*( Ti_w ) = Te* (m*ci + M*cw )

                             Te = [ m*ci*( Ti_h ) + M*cw*( Ti_w ) ] / [ m*ci + M*cw ]

- Plug in the values and evaluate (Te):

                             Te = [1.5*448*550 + 25*4186*20 ] / [ 1.5*448 + 25*4186 ]

                             Te = 2462600 / 105322

                             Te =  23.4 °C    

7 0
3 years ago
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If you have a yellow sheet of paper and shine a cyan light what color is the paper
Nuetrik [128]
Green would be the colour of the paper
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4 years ago
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A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2
Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel I=25kgm^2

Initial angular velocity of the wheel \omega _0=10rad/sec

Angular acceleration \alpha =15rad/sec^2

(a) We know that \omega =\omega _0+\alpha t

We have given t = 2 sec

So \omega =10+15\times  2=40rad/sec

Now \Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad

(b) After 3 sec \omega =10+15\times 3=55rad/sec

We know that kinetic energy is given by Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J

7 0
3 years ago
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Answer:

Explanation:

Given

mass of boy m_b=70.9\ kg

mass of girl m_g=43.2\ kg

speed of girl after push v_g=4.64\ m/s

Suppose speed of boy after push is v_b

initially momentum of system is zero so final momentum is also zero because momentum is conserved

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0=m_b\cdot v_b+m_g\cdot v_g

v_b=-\frac{m_g}{m_b}\times v_g

v_b=-\frac{43.2}{70.9}\times 4.64  

v_b=-2.82\ m/s

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8 0
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