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:
Speed at which it will reach the ground is given as
Total time for which it will remain in air is given as
t = 6.3 s
Explanation:
As we know that the object is projected upwards with speed
now when it will reach the ground then we have
so we have
so we have
Now speed of the object when it reaches the ground is given as
Energy cannot be created nor be destroyed
<span>22.5 newtons.
First, let's determine how much energy the stone had at the moment of impact. Kinetic energy is expressed as:
E = 0.5mv^2
where
E = Energy
m = mass
v = velocity
Substituting known values and solving gives:
E = 0.5 3.06 kg (7 m/s)^2
E = 1.53 kg 49 m^2/s^2
E = 74.97 kg*m^2/s^2
Now ignoring air resistance, how much energy should the rock have had?
We have a 3.06 kg moving over a distance of 10.0 m under a force of 9.8 m/s^2. So
3.06 kg * 10.0 m * 9.8 m/s^2 = 299.88 kg*m^2/s^2
So without air friction, we would have had 299.88 Joules of energy, but due to air friction we only have 74.97 Joules. The loss of energy is
299.88 J - 74.97 J = 224.91 J
So we can claim that 224.91 Joules of work was performed over a distance of 10 meters. So let's do the division.
224.91 J / 10 m
= 224.91 kg*m^2/s^2 / 10 m
= 22.491 kg*m/s^2
= 22.491 N
Rounding to 3 significant figures gives an average force of 22.5 newtons.</span>
