Mass, m = 4g = 0.004 kg
Velocity, = 50cm/s = 0.5m/s
Distance, 10cm = 0.1m
The wall would have to resist the energy acquired by the bullet.
Kenetic Energy of bullet = Resistance offered by the wall.
1/2 mv² = Resistance Force * Distance
(1/2) * 0.04 * 0.5 * 0.5 = F * 0.1
0.5 * 0.04 * 0.5 * 0.5 = F * 0.1
0.5 * 0.04 * 0.5 * 0.5/0.1 = F
0.05 = F
Therefore, Resistance offered by the wall = 0.05 N
Answer:
size get smaller maybe there is no object remaining
Answer:
Explanation:
The relationship between the refractive index and the critical angle is given as follows:
where,
η = refractive index = 1.67
θc = critical angle =?
Therefore,
<span>Place a test charge in the middle. It is 2cm away from each charge.
The electric field E= F/Q where F is the force at the point and Q is the charge causing the force in this point.
The test charge will have zero net force on it. The left 30uC charge will push it to the right and the right 30uC charge will push it to the left. The left and right force will equal each other and cancel each other out.
THIS IS A TRICK QUESTION.
THe electric field exactly midway between them = 0/Q = 0.
But if the point moves even slightly you need the following formula
F= (1/4Piε)(Q1Q2/D^2)
Assume your test charge is positive and make sure you remember two positive charges repel, two unlike charges attract. Draw the forces on the test charge out as vectors and find the magnetude of the force, then divide by the total charge to to find the electric field strength:)</span>
Answer:
Light's angle of refraction = 37.1° (Approx.)
Explanation:
Given:
Index of refraction = 1.02
Base of refraction = 1
Angle of incidence = 38°
Find:
Light's angle of refraction
Computation:
Using Snell's law;
Sin[Angle of incidence] / Sin[Light's angle of refraction] = Index of refraction / Base of refraction
Sin38 / Light's angle of refraction = 1.02 / 1
Sin[Light's angle of refraction] = Sin 38 / 1.02
Sin[Light's angle of refraction] = [0.6156] / 1.02
Sin[Light's angle of refraction] = 0.6035
Light's angle of refraction = 37.1° (Approx.)
