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:
168.57 mV
Explanation:
Initial magnetic flux = BA , B magnetic field and A is area of loop
= .35 x 3.14 x .37²
= .15 Weber
Final magnetic flux
= - .2 x 3.14 x .37²
= - .086 Weber
change in flux
.15 + .086
= .236 Weber
rate of change of flux
= .236 / 1.4
= .16857 V
= 168.57 mV
<span>An Object 4 Cm Tall Is Placed 12 Cm From A Divergi... | Chegg.com</span>
Answer:
a)
Y0 = 0 m
Vy0 = 15 m/s
ay = -9.81 m/s^2
b) 7.71 m
c) 3.06 s
Explanation:
The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards
Y(0) = 0 m
Vy(0) = 15 m/s
ay = -9.81 m/s^2 (negative because it points down)
Since acceleration is constant we can use the equation for uniformly accelerated movement:
Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2
To find the highest point we do the first time derivative (this is the speed:
V(t) = Vy0 + a * t
We equate this to zero
0 = Vy0 + a * t
0 = 15 - 9.81 * t
15 = 9.81 * t
t = 0.654 s
At this time it will have a height of:
Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m
The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.
0 = Y0 + Vy0 * t + 1/2 * a * t^2
0 = 0 + 15 * t - 1/2 * 9.81 t^2
0 = 15 * t - 4.9 * t^2
0 = t * (15 - 4.9 * t)
t1 = 0 This is the moment it jumped into the air
0 = 15 - 4.9 * t2
15 = 4.9 * t2
t2 = 3.06 s This is the moment when it falls again.
3.06 - 0 = 3.06 s
