Explanation:
The given data is as follows.
Mass, m = 75 g
Velocity, v = 600 m/s
As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.
where, 
= mass of the projectile
= mass of block
v = velocity after the impact
Now, putting the given values into the above formula as follows.
![75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v](https://tex.z-dn.net/?f=75%2810%5E%7B-3%7D%29%20%5Ctimes%20600%20%3D%20%5B%2875%20%5Ctimes%2010%5E%7B-3%7D%29%20%2B%2050%5D%20%5Ctimes%20v)
= 
v = 0.898 m/s
Now, equation for energy is as follows.
E = 
= 
= 13500 J
Now, energy after the impact will be as follows.
E' = ^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5B75%20%5Ctimes%2010%5E%7B-3%7D%20%2B%2050%5D%280.9%29%5E%7B2%7D)
= 20.19 J
Therefore, energy lost will be calculated as follows.
= E E'
= (13500 - 20) J
= 13480 J
And, n = 
= 
= 99.85
= 99.9%
Thus, we can conclude that percentage n of the original system energy E is 99.9%.
Answer:
<u>The magnitude of the friction force is 8197.60 N</u>
Explanation:
Using the definition of the centripetal force we have:
Where:
- m is the mass of the car
- v is the speed
- R is the radius of the curvature
Now, the force acting in the motion is just the friction force, so we have:
<u>Therefore the magnitude of the friction force is 8197.60 N</u>
I hope it helps you!
Answer: 33 mm
Explanation:
Given
Diameter of the tank, d = 9 m, so that, radius = d/2 = 9/2 = 4.5 m
Internal pressure of gas, P(i) = 1.5 MPa
Yield strength of steel, P(y) = 340 MPa
Factor of safety = 0.3
Allowable stress = 340 * 0.3 = 102 MPa
σ = pr / 2t, where
σ = allowable stress
p = internal pressure
r = radius of the tank
t = minimum wall thickness
t = pr / 2σ
t = 1.5*10^6 * 4.5 / 2 * 102*10^6
t = 0.033 m
t = 33 mm
The minimum thickness of the wall required is therefore, 33 mm
Answer:
the motion that repeat itself in equal interval of time is called periodic motion and it is equal to harmonic motion. for example pendulum
Light travels in waves AND in bundles called "photons".
It's hard to imagine something that's a wave and also a bundle.
But it turns out that light behaves like both waves and bundles.
If you design an experiment to detect waves, then it responds to light.
And if you design an experiment to detect 'bundles' or particles, then
that one also responds to light.
