This is a question on conservation of energy. That is,
mgh + KE1 = KE2
mgh +1/2mv1^2 = 1/2mv2^2
gh + 1/2v1^2 = 1/2v2^2
Where, h = 0.2 m, v1 =3.04 m/s
Therefore,
v2 = Sqrt [2(gh+1/2v1^2)] = Sqrt [2(9.81*0.2 + 1/2*3.04^2)] = 7.26 m/s
Now, Volumetric flow rate, V/time, t = Surface area, A*velocity, v
Where,
V = Av = πD^2/4*3.04 = π*(2.51/100)^2*1/4*3.04 = 1.504*10^-3 m^3/s
At 0.2 m below,
V = 1.504*10^-3 m^3/s = A*7.26
A = (1.504*10^-3)/7.26 = 2.072*10^-4 m^2
But, A = πr^2
Then,
r = Sqrt (A/π) = Sqrt (2.072*10^-4/π) = 0.121*10^-3 m
Diameter = 2r = 0.0162 m = 1.62 cm
You're most likely to build up enough static charge to receive a shock by walking around in a carpeted restaurant in the desert. (A)
Walking on carpet is the fastest way to accumulate charge, and the dry desert air prevents the charge from dribbling off of you and away.
When I walked on stones in the Sinai Desert, the dry wind with a little bit of sand or dust in it built up enough static charge on me that I got a shock every time I stood less than a foot away from my partner.
I had the same experience a few years later near Ouarzazate in the interior of Morocco.
When you hear people say "the desert is dry", they mean it's <em>DRY ! </em>
Answer:
Acceleration is -1.2 m/s² and distance covered is 135 m.
Explanation:
A train going at a speed of 18m / s brakes and stops in 15s calculates its acceleration and the distance traveled when braking
Given that,
The initial speed of the train, u = 18 m/s
Final speed, v = 0
Time, t = 15 s
We need to find acceleration and distance traveled when braking. Let a is acceleration and distance traveled.
Acceleration,
Using third equation of motion,
Hence, acceleration is -1.2 m/s² and distance covered is 135 m.
The force on the box is:
F = mgsin∅
If we multiply by this with the distance it traveled, we will know the work done by the box.
W = dmgsin∅
This work will be converted to elastic potential energy in the spring which is:
1/2 kx². Equating these and substituting values:
1/2 * 170 * x² = 4 * 13 * 9.81 * sin(30)
x = 1.73 m
The box's maximum speed will at the point right before contact with the spring, when the compression is 0.
