A) To calculate the charge of each coin, we must apply the expression of the Coulomb's Law:
F=K(q1xq2)/r²
F: The magnitud of the force between the charges. (F=2.0 N).
K: Constant of proporcionality of the Coulomb's Law (K=9x10^9 Nxm²/C²).
q1 and q2: Electrical charges.
r: The distance between the charges (r=1.35 m).
We have the values of F, K and r, so we can calculate q1xq2, because both<span> coins have identical charges:
</span>
q1xq2=(r²xF)/K
q1xq2=(1.35 m)²(2.0 N)/9x10^9 Nxm²/C²
q1xq2=3x10^-10 C
q1=q2=(<span>3x10^-10 C)/2
</span>Then, the charge of each coin, is:
<span>
q1=1.5x</span><span>10^-10 C
</span>q2=1.5x10^-10 C
B) <span>Would the force be classified as a force of attraction or repulsion?
</span>
It is a force of repulsion, because both coins have identical charges and both are postive. In others words, when two bodies have identical charges (positive charges or negative charges), the force is of repulsion.
Answer: 1100 ft lb/s and 2 H.P
Explanation:
To calculate for the power developed in the elevator motor in ft.lb/s, we multiply the distance and the weight of the elevator and divide the product by the time.
Power = (10 ft)(2200 lb) / 20 s = 1100 ft.lb/s
Next, convert the calculated value to HP.
1100 ft.lb/s x (1 HP/ 550 ft.lb/s) = 2 HP
hope this helps please give brainliest!
Answer:
The pressure of the gas will increase
Explanation:
When gas is put into a container, for example, a balloon, the gas expands to fill the space it can occupy. Since gas is not a solid or a liquid, its particles are all over the place - they are constantly moving and vibrating. As such, when too much gas is blown into a balloon, it will pop. So, when the volume of the container decreases, the pressure of the gas will increase the smaller it gets. Vice versa, the greater the space, the less pressure that will be present in the container.
Answer:
Explanation:
Velocity of Top most point of wheel is twice the Velocity of centre of mass of wheel
Thus angular velocity is given by
![K.E.=\frac{1}{2}m_{stone}V^2+2\left [ \frac{1}{2}m_{cyl}\left [ \frac{V}{2}\right ]^2\right ]+2\left [ \frac{1}{2}I\omega ^2\right ]](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B1%7D%7B2%7Dm_%7Bstone%7DV%5E2%2B2%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7Dm_%7Bcyl%7D%5Cleft%20%5B%20%5Cfrac%7BV%7D%7B2%7D%5Cright%20%5D%5E2%5Cright%20%5D%2B2%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7DI%5Comega%20%5E2%5Cright%20%5D)
![K.E.=\frac{V^2}{2}\left [ m_{stone}+\frac{m_{roller}}{2}\right ]+I\omega ^2](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7BV%5E2%7D%7B2%7D%5Cleft%20%5B%20m_%7Bstone%7D%2B%5Cfrac%7Bm_%7Broller%7D%7D%7B2%7D%5Cright%20%5D%2BI%5Comega%20%5E2)
![K.E.=\frac{0.319^2}{2}\left [ 672+\frac{82}{2}\right ]+\frac{82\times 0.343^2\times 0.465^2}{2}](https://tex.z-dn.net/?f=K.E.%3D%5Cfrac%7B0.319%5E2%7D%7B2%7D%5Cleft%20%5B%20672%2B%5Cfrac%7B82%7D%7B2%7D%5Cright%20%5D%2B%5Cfrac%7B82%5Ctimes%200.343%5E2%5Ctimes%200.465%5E2%7D%7B2%7D)
The two kinds of motion are horizontal and vertical
