Answer:D
Explanation:
It was right o khan academy
#1
As we know that

now plug in all data into this


now from the formula of strain




#2
As we know that
pressure * area = Force
here we know that


now force is given as

#3
As we know that density of water will vary with the height as given below

here we know that


now density is given as


#4
as we know that pressure changes with depth as per following equation

here we know that

now we will have



here we will have

so it is 20.1 m below the surface
#5
Here net buoyancy force due to water and oil will balance the weight of the block
so here we will have




so it is 3.48 cm below the interface
Answer:
5 m/s
Explanation:
Here we can see there is no external force acted on a two masses when we consider the motion. If there is no external forces then momentum is conserved.
Initial momentum = Final momentum
0.5 × 10 = 1 × V
V = 5 m/s
Answer:
.
Explanation:
The efficiency of a machine is the percentage of energy input that was turned into useful energy.
The power rating of this lamp is
(same as
,) meaning that
of energy is supplied to this lamp every second.
The question states that
out of that
of energy input would be turned into heat, which is not useful energy output in this scenario. Assuming that all other forms of energy loss is negligible. The rest of the
of energy supplied to this lamp would be turned into useful energy output.
Thus, every second, this lamp would receive
of energy input and would outputs
of useful work. The efficiency of this lamp would be:
.
<span>i think the average speed was 54.54 (mph)</span>