<span>Stress
Stress is a condition of mental strain or pressure on the body and coming about because of unfavorable or demanding conditions that the person cannot handle at that moment.</span>
Answer:
The work done to get you safely away from the test is 2.47 X 10⁴ J.
Explanation:
Given;
length of the rope, L = 70 ft
mass per unit length of the rope, μ = 2 lb/ft
your mass, W = 120 lbs
mass of the 70 ft rope = 2 lb/ft x 70 ft
= 140 lbs.
Total mass to be pulled to the helicopter, M = 120 lbs + 140 lbs
= 260 lbs
The work done is calculated from work-energy theorem as follows;
W = Mgh
where;
g is acceleration due gravity = 32.17 ft/s²
h is height the total mass is raised = length of the rope = 70 ft
W = 260 Lb x 32.17 ft/s² x 70 ft
W = 585494 lb.ft²/s²
1 lb.ft²/s² = 0.0421 J
W = 585494 lb.ft²/s² = 2.47 X 10⁴ J.
Therefore, the work done to get you safely away from the test is 2.47 X 10⁴ J.
<u>Answer:</u>
<em>The initial distance between the trains is 1450 m.
</em>
<u>Explanation:</u>
In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.
Train B’s acceleration is
and it accelerated for 50 seconds.
<em>
</em>
<em>t=50 s
</em>
<em>initial speed u=72km/h
</em>
<em>we have to convert this speed into m/s </em>
<em>
</em>
<em>Distance covered in accelerating phase
</em>
<em>
</em>
<em>
</em>
If a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.
<em>Here length of train A+length of train
</em>
<em>Hence the initial distance between the trains =
</em>
Answer:
5.95 m
Explanation:
Given that the biggest loop is 40.0 m high. Suppose the speed at the top is 10.8 m/s and the corresponding centripetal acceleration is 2g
For the car to stick to the loop without falling down, at the top of the ride, the centripetal force must be equal to the weight of the car. That is,
(MV^2) / r = mg
V^2/ r = centripetal acceleration which is equal to 2g
2 × 9.8 = 10.8^2 / r
r = 116.64 /19.6
r = 5.95 m
Explanation:
It is given that,
Bandwidth of a laser source, 
(b) Let t is the time separation of sections of sections of the light wave that can still interfere. The time period is given by :



(a) Let h is the coherence length of the source. It is given by :

c is the speed of light

l = 0.0099 m
Hence, this is the required solution.