Answer:
Power = Energy/time
Energy = Power xtime.
Time= 20hrs
Power = 100Watt =0.1Kw
Energy = 0.1 x 20 = 2Kwhr.
This Answer is in Kilowatt-hour ...
If the one given to you is in Joules
You'd have to Change your time to seconds
Then Multiply it by the power of 100Watts.
Answer:
15.7 m
Explanation:
m = mass of the sled = 125 kg
v₀ = initial speed of the sled = 8.1 m/s
v = final speed of sled = 0 m/s
F = force applied by the brakes in opposite direction of motion = 261
d = stopping distance for the sled
Using work-change in kinetic energy theorem
- F d = (0.5) m (v² - v₀²)
- (261) d = (0.5) (125) (0² - 8.1²)
d = 15.7 m
Answer:
Explanation:
q1 = q2 = e = 1.6 x 10^-19 C
d = 2.7 x 10^-15 m
1. Work done is equal to the potential energy stored between the two charges.
The formula for the potential energy is given by
![U=\frac{Kq_{1}q_{2}}{d}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7BKq_%7B1%7Dq_%7B2%7D%7D%7Bd%7D)
By substituting the values
![U=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{2.7\times 10^{-15}}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B9%5Ctimes%2010%5E%7B9%7D%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B2.7%5Ctimes%2010%5E%7B-15%7D%7D)
U = 8.53 x 10^-14 J
So, the work done is 8.53 x 10^-14 J.
(2) d = 2 x 2.7 x 10^-15 m = 5.4 x 10^-15 m
So, the potential energy is
![U=\frac{Kq_{1}q_{2}}{d}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7BKq_%7B1%7Dq_%7B2%7D%7D%7Bd%7D)
By substituting the values
![U=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{5.4\times 10^{-15}}](https://tex.z-dn.net/?f=U%3D%5Cfrac%7B9%5Ctimes%2010%5E%7B9%7D%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%201.6%5Ctimes%2010%5E%7B-19%7D%7D%7B5.4%5Ctimes%2010%5E%7B-15%7D%7D)
U = 4.27 x 10^-14 J
(3) mass, m = 1 u = 1.67 x 10^-27 kg
So, the kinetic energy is equal to the potential energy.
Let v be the velocity.
![2 \times \frac{1}{2}mv^{2}=8.53\times 10^-14](https://tex.z-dn.net/?f=2%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7Dmv%5E%7B2%7D%3D8.53%5Ctimes%2010%5E-14)
![2 \times \frac{1}{2}\times 1.67\times 10^{-27}\times v^{2}=8.53\times 10^{-14}](https://tex.z-dn.net/?f=2%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5Ctimes%201.67%5Ctimes%2010%5E%7B-27%7D%5Ctimes%20v%5E%7B2%7D%3D8.53%5Ctimes%2010%5E%7B-14%7D)
v = 7.14 x 10^6 m/s
Answer:
See explanation.
Explanation:
If both stars explode in simultaneously in the <em>your </em>frame of reference then obviously you will see the two flashes simultaneously, and therefore, the time difference between the events would be zero.
If however, the stars exploded simultaneously in their frame of reference, then you would not observe the flashes simultaneously. Then the time difference between the events will not be zero, rather, you will observe star B exploding first and star A after.
Taking the distance of Tarzan from the ground before and after he makes the swing:
Ho (initial height) = L(1 - cos45) = 20 (1 - 0.707) = 5.86 meters
Hf (final height) = L(1 - cos30) = 20 (1 - 0.866<span>) = 2.68 meters
</span>
Difference in height = 5.86 - 2.68 = 3.18 meters
PE = KE
mgh = (1/2)mv^2
Solving for v:
v = sqrt (2*g*h)
v = sqrt (2*9.8*3.18)
v = 7.89 m/s
With Tarzan going that fast, it is likely that he will knock Jane off.