If he moves the piano 4 meters, with 700 newtons of force
in the direction of the motion, then the work done is
Work = (force) x (distance)
= (700 N) x (4 m) = 2,800 joules .
If he does all of that in 2 seconds, then
he did it at the rate of
(work) / (time) = (2,800 J) / (2 sec) = 1,400 watts
(about 1.88 'horsepower' !)
Answer:
Net force = 13.36 N
Explanation:
Recall the formula for the force between two charges q1 ans q2 separated a distance "d" based on Coulomb's Law:
so, we need to calculate the force exerted on the charge 10 micro Coulombs by the 20 micro-Coulombs charge as they are separated 0.8 meters (haf the distance of the original two charges.
Notice that charges q1 and q3 are both positive charges so the force we obtain will be pointing in the direction towards charge q2. The magnitude of this is:
![F_{13}=k\,\frac{20\,\,10^{-6}\,*\,10\,\,10^{-6}}{0.8^2}=9\,\,10^{9}\,\frac{200\,\,10^{-12}}{0.8^2}= 2812.5 \,\,10^{-3} \,N = 2.8125 \,\,N](https://tex.z-dn.net/?f=F_%7B13%7D%3Dk%5C%2C%5Cfrac%7B20%5C%2C%5C%2C10%5E%7B-6%7D%5C%2C%2A%5C%2C10%5C%2C%5C%2C10%5E%7B-6%7D%7D%7B0.8%5E2%7D%3D9%5C%2C%5C%2C10%5E%7B9%7D%5C%2C%5Cfrac%7B200%5C%2C%5C%2C10%5E%7B-12%7D%7D%7B0.8%5E2%7D%3D%202812.5%20%5C%2C%5C%2C10%5E%7B-3%7D%20%5C%2CN%20%3D%202.8125%20%5C%2C%5C%2CN)
We can calculate now the force exerted by charge q2 on charge q3 using the same reasoning, and noticing that since charge q2 is negative, the force on q3 will be of attraction, then also pointing in the direction towards q2:
![F_{23}=k\,\frac{75\,\,10^{-6}\,*\,10\,\,10^{-6}}{0.8^2}=9\,\,10^{9}\,\frac{750\,\,10^{-12}}{0.8^2}= 10546.875 \,\,10^{-3} \,N = 10.547 \,\,N](https://tex.z-dn.net/?f=F_%7B23%7D%3Dk%5C%2C%5Cfrac%7B75%5C%2C%5C%2C10%5E%7B-6%7D%5C%2C%2A%5C%2C10%5C%2C%5C%2C10%5E%7B-6%7D%7D%7B0.8%5E2%7D%3D9%5C%2C%5C%2C10%5E%7B9%7D%5C%2C%5Cfrac%7B750%5C%2C%5C%2C10%5E%7B-12%7D%7D%7B0.8%5E2%7D%3D%2010546.875%20%5C%2C%5C%2C10%5E%7B-3%7D%20%5C%2CN%20%3D%2010.547%20%5C%2C%5C%2CN)
Now, we need to add both forces to get the final force (recall they are both in the same direction):
Net force = 10.547 N + 2.813 N = 13.36 N
<h2>Calculating the Speed given the Distance and Time</h2><h3>
Answer:</h3>
![4.\overline{4} \frac{\mathrm{km}}{\mathrm{min}}\\](https://tex.z-dn.net/?f=4.%5Coverline%7B4%7D%20%5Cfrac%7B%5Cmathrm%7Bkm%7D%7D%7B%5Cmathrm%7Bmin%7D%7D%5C%5C)
<h3>
Explanation:</h3>
The speed,
, is calculated by
where
is the distance traveled and
is the time. We are given the distance traveled of the bus,
, and the time it has traveled,
. So, the speed of the bus,
, is calculated as...
![r =\frac{200 \mathrm{km}}{50 \mathrm{min}} \\ r = \frac{40 \mathrm{km}}{9 \mathrm{min}}](https://tex.z-dn.net/?f=r%20%3D%5Cfrac%7B200%20%5Cmathrm%7Bkm%7D%7D%7B50%20%5Cmathrm%7Bmin%7D%7D%20%5C%5C%20r%20%3D%20%5Cfrac%7B40%20%5Cmathrm%7Bkm%7D%7D%7B9%20%5Cmathrm%7Bmin%7D%7D)
The speed of the bus is
or
.
Answer:
If you double the force, you double the acceleration, but if you double the mass, you cut the acceleration in half.
Explanation:
Answer:
Explanation:
a ) F = (-kx + kx³/a²)
intensity of field
I = F / m
= (-kx + kx³/a²) / m
If U be potential function
- dU / dx = (-kx + kx³/a²) / m
U(x) = ∫ (kx - kx³/a²) / m dx
= k/m ( x²/2 - x⁴/4a²)
b )
For equilibrium points , U is either maximum or minimum .
dU / dx = x - 4x³/4a² = 0
x = ± a.
dU / dx = x - x³/a²
Again differentiating
d²U / dx² = 1 - 3x² / a²
Put the value of x = ± a.
we get
d²U / dx² = -2 ( negative )
So at x = ± a , potential energy U is maximum.
c )
U = k/m ( x²/2 - x⁴/4a²)
When x =0 , U = 0
When x= ± a.
U is maximum
So the shape of the U-x curve is like a bowl centered at x = 0
d ) Maximum potential energy
put x = a or -a in
U(max) = k/m ( x²/2 - x⁴/4a²)
= k/m ( a² / 2 - a⁴/4a²)
= k/m ( a² / 2 - a²/4)
a²k / 4m
This is the maximum total energy where kinetic energy is zero.