Answer:
the displacement would be 32 m
Explanation:
there is no formula for displacement, but you can use the velocity formula as you have the velocity and time so it looks like:
![8 = \frac{x}{4}](https://tex.z-dn.net/?f=8%20%3D%20%5Cfrac%7Bx%7D%7B4%7D)
with x being the displacement
Answer: Energy dissipation
As water passes over a spillway and down the chute, potential energy converts into increasing kinetic energy. Failure to dissipate the water's energy can lead to scouring and erosion at the dam's toe (base).
HOPE THIS HELPS
Given that,
Current = ∞
We know that,
Ohm's law :
Ohm's law is defined as,
The voltage of the circuit is directly proportional to the current of the circuit.
![V\propto I](https://tex.z-dn.net/?f=V%5Cpropto%20I)
Or,
The voltage of the circuit is equal to the product of current and resistance.
In mathematically,
![V=IR](https://tex.z-dn.net/?f=V%3DIR)
Where, V = voltage
I = current
R = resistance
According to ohm's law,
The current in the circuit is
![I=\dfrac{V}{R}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BV%7D%7BR%7D)
If the current is very less then the resistance will be infinity.
If the is reach to infinity then the resistance will be very low.
Hence, The resistance becomes very low.
Explanation:
its easy everone can do it .....let me know
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)
= ![\frac{45}{50.075}](https://tex.z-dn.net/?f=%5Cfrac%7B45%7D%7B50.075%7D)
v = 0.898 m/s
Now, equation for energy is as follows.
E = ![\frac{1}{2}mv^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E%7B2%7D)
= ![\frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20%2875%20%5Ctimes%2010%5E%7B-3%7D%20%2B%2050%29%20%5Ctimes%20%28600%29%5E%7B2%7D)
= 13500 J
Now, energy after the impact will be as follows.
E' = ![\frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{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 = ![\frac{\Delta E}{E}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20E%7D%7BE%7D)
= ![\frac{13480}{13500} \times 100](https://tex.z-dn.net/?f=%5Cfrac%7B13480%7D%7B13500%7D%20%5Ctimes%20100)
= 99.85
= 99.9%
Thus, we can conclude that percentage n of the original system energy E is 99.9%.