Ask question

Ask question

All categories

Lapatulllka [165]

2 years ago

9

In a single movable pulley, a load of 500 N is lifted by applying 300 N effort. Calculate MA, VR and efficiency.

1 answer:

galben [10]2 years ago

6 0

Answer:

in pulley there are different kinds.but the most common one are fixed,moveable and compound pulley. in this question we asked about movable pulley.

Explanation:

Given request solutions

L=500N a,M.A=? a,M.A=L/E =5/3

E=300N b,V.R=? b,V.R=2

c,efficiency =? c,£=M.A/V.R=5/6

£=IS NOT THE REAL SYMBOLS OF EFFICIENCY BUT I IS LOOK LIKE THIS

I THINK I HELPED

You might be interested in

Which activity is a method of habitat restoration?

irina1246 [14]

A. cleaning up after environmental disasters

7 0

3 years ago

Read 2 more answers

All known matter is made up of Question 1 options:

Tatiana [17]

the answer is (a) molecules

4 0

3 years ago

Read 2 more answers

How much work must be done on a 20-kg go-cart to increase its speed from 5 m/s to 10 m/s?

allochka39001 [22]

b 250 j

hope this helps

6 0

3 years ago

Read 2 more answers

What would happen if the gravity on earth decreases by a small about?

yan [13]

If the gravity decreases on the earth, we will be weightless and pressure will be less exerting on us and we will be able fly

6 0

3 years ago

Read 2 more answers

1. On the planet Microid, trains that run on air tracks much like the air track you used in lab are the principle means of trans

Sedaia [141]

Answer:

a) the mass of the second car = 28000 kg

b) the amount of kinetic energy that was lost in the collision = $246.9*10^3 \ J$

Explanation:

Given that:

mass of the train car $m_1$ = 6300 kg

speed of the train car $v_1$ = 12.0 m/s

mass of the second moving car $m_2$ = ???

speed of the second moving car $v_2$ = 2.2 m/s

After strike;

they both move with a speed $v_f$ = 4.00 m/s

a)

Using the conservation of momentum :

$m_1v_1+m_2v_2 = (m_1 + m_2)v_f$

$(6300*12)+m_2(2.2) = (6300 + m_2)4$

$(75600)+m_2(2.2) = 25200 + 4m_2$

$75600 - 25200 = 4m_2 -2.2m_2$

$50400 = 1.8m_2$

$m_2 = \frac{50400}{1.8}$

$m_2 = 28000 \ kg$

b)

To determine the amount of kinetic energy that was lost in the collision;

we will need to find the difference between the kinetic energy before the collision and after the collision;

i.e

$K.E _{lost} = K.E_{i} - K.E _{f}$

$K.E_{i} = \frac{1}{2} m_1v_1^2 + \frac{1}{2}m_2v_2^2$

$K.E_{i} = \frac{1}{2} (6300)(12)^2 + \frac{1}{2}(28000)(2.2)^2$

$K.E_{i} = 521360 \ J$

$K.E_{f} = \frac{1}{2}(m_1 + m_2 ) v_f ^2$

$K.E_{f} = \frac{1}{2}(6300 + 28000 ) (4)^2$

$K.E_{f} =274400 \ J$

Now; the kinetic energy that was lost in the collision is calculated as follows:

$K.E _{lost} = K.E_{i} - K.E _{f}$

$K.E_{lost} = (521360-274400) \ J$

$K.E_{lost} =246960 \ J$

$K.E_{lost} =246.9 * 10 ^3 \ J$

7 0

3 years ago