The efficiency of the scissor is 200%.
<u>Explanation:</u>
Efficiency is defined as the ratio of output of any instrument or device or machine to the input supplied to it. So the greater the output the greater will be the efficiency of the device.
As here the work done by us on the system is said to be 10 J so this will be equal to the input work done on the system. And the work done by the system i.e., the scissor is 20 J, so this will be the output work.
So, the efficiency is the ratio of output to input as shown below.
Efficiency =
= 200
So, the efficiency of the scissor is 200%.
Answer:
umm section 2 and 4
Explanation:
because at section 2 it starts and at section 4 it moves again and stops at 3.
Answer:
5 ohms
Explanation:
Given:
EMF of the ideal battery (E) = 60 V
Voltage across the terminals of the battery (V) = 40 V
Current across the terminals (I) = 4 A
Let the internal resistance be 'r'.
Now, we know that, the voltage drop in the battery is given as:
Therefore, the voltage across the terminals of the battery is given as:

Now, rewriting in terms of 'r', we get:

Plug in the given values and solve for 'r'. This gives,

Therefore, the internal resistance of the battery is 5 ohms.