100 J of work in 10 seconds

Nick’s swimming goggles are made of blue plastic. When he straps them on, he notices that the goggles change the color of the ob

Liono4ka [1.6K]

A white ring buoy appears blue because the blue plastic absorbs all colors of light except blue. Only the blue light reflected from the ring buoy passes through the blue plastic.

3 0

3 years ago

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Two small charged spheres are located on the y-axis. One is at y = 1.00 m, the other is at y = −1.00 m, and they both have a cha

yuradex [85]

Answer:

(a) 23.946 kV

(b) -0.077 J

Explanation:

(a) The electric potential is given by the following formula:

$V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}$ (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = q2 = 1.60*10^{-6}C

r1 and r2 are the distance from the charges to the point in which electric potential is evaluated.

Firs, you calculate the distance r1 and r2 by taking into account the position of the charges

$r_1=\sqrt{(1.00)^2+(0.670)^2}m=1.20m\\\\r_2=\sqrt{(-1.00)^2+(0.670)^2}m=1.20m$

Next, you replace the values of the parameters to calculate V:

$V=(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}+(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}\\\\V=23946.66\ V=23.946\ kV$

(b) The potential electric energy is given by:

$U_T=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}\\\\r_{1,2}=2.00m\\\\r_{1,3}=1.20m\\\\r_{2,3}=1.20m\\\\U_T=(8.98*10^9)[\frac{(1.6*10^{-6})^2}{2.00m}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}]J\\\\U_T=-0.077J$

5 0

3 years ago

A wind is blowing a small ice fishing shed along the ice to the east with a 800N force. The shed weighs 2,200N. What would the l

e-lub [12.9K]

Answer:

0.36

Explanation:

The maximum force of friction exerted by the surface is given by:

$F_f = \mu N$ (1)

where

$\mu$ is the coefficient of friction

N is the normal reaction

The shed's weight is 2200 N. Since there is no motion along the vertical direction, the normal reaction is equal and opposite to the weight, so

N = 2200 N

The horizontal force that is pushing the shed is

F = 800 N

In order for it to keep moving, the force of friction (which acts horizontally in the opposite direction) must be not greater than this value. So the maximum force of friction must be

$F_f = 800 N$