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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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Answer:

a) v=0.999124c

b) E=7.566*10^{22}

c) E_a=760 times\ larger

Explanation:

From the question we are told that

Distance to Betelgeuse d_b=430ly

Mass of Rocket M_r=20000

Total Time in years traveled T_d=36years

Total energy used by the United States in the year 2000 E_{2000}=1.0*10^20

Generally the equation of speed of rocket v mathematically given by

v=\frac{2d}{\triangle t}

v=860ly/ \triangle t

where

\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}

\triangle t=\sqrt{(860)^2+(36)^2}

\triangle t=860.7532

Therefore

v=\frac{860ly}{ 860.7532}

v=0.999124c

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2

E=7.566*10^{22}

c)Generally the equation of the energy E_a required to accelerate the rocket mathematically given by

E_a=\frac{E}{E_{2000}}

E_a=\frac{7.566*10^{22}}{1.0*10^{20}}

E_a=760 times\ larger

8 0
3 years ago
Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more t
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The force result in stretching the spring 10.0 centimeters is 2.5N.

<h3>What is Hooke's law?</h3>

If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.

F = kx

where k is the proportionality constant called the spring constant or force constant.

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.

Let the spring constant be very low 0.04N/m

The force applied is

F = 10 cm / 0.04

F = 0.1 m  / 0.04

F = 2.5 N

Thus, the force result in stretching the spring 10cm is 2.5 N.

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