They have positive charge and are located in the nucleus with neutrons
Using the following given values:
Object 1:
Mass = M1 = 2 kg
Velocity before collision = Vb1 = 20 m/s
Velocity after collision = Va1 = -5 m/s
Object 2:
Mass = M2 = 3 kg
Velocity before collision = Vb2 = -10 m/s
Velocity after collision = Va2 = ? m/s<span>
</span>
Obtaining Va2 via law of conservation of momentum:
total momentum after collision = total momentum before collision
M1 * Va1 + M2 * Va2 = M1 * Vb1 + M2 * Vb2
2*-5 + 3Va2 = 2*20 + 3*-10
Va2 = 6.67
Total kinetic energy before collision:
KE1 = (1/2)*M1*Vb1^2 + (1/2)*M2*Vb2^2
<span>KE1 = (1/2)*2*(20)^2 + (1/2)*3*(-10)^2
KE1 = 550 J
</span>Total kinetic energy after collision:
KE2 = (1/2)*M1*Va1^2 + (1/2)*M2*Va2^2
<span>KE2 = (1/2)*2*(-5)^2 + (1/2)*3*(6.67)^2
KE2 = 91.73 J
</span>
Total kinetic energy lost:
Energy lost = KE1 - KE2 = 550 - 91.73 = 458.27 J
<span> In the Universe , all matter and energy contained; no energy or matter </span>
<span>leaves the Universe. </span>
Answer:
35.57*10^21 N.
Explanation:
F = Gmm/r², where G is 6.67*10e-11 & r is distance.
F = 6.67*10^-11 * 6*10^24 *2*10^30 /(1.5*10^11)².
F = 35.57*10^21 newtons.
Answer:
The speed of James is 0.776 m/s
Explanation:
Step 1: Data given
mass of James = 95.0 kg
mass of Ramon = 67.0 kg
We consider James and Ramon and the rope to a single system. This means that the net external forces on the system = 0
.The momentum = 0, so the sum of the momentum of each part must be 0 in total.
Step 2: Calculate the speed of James
m(james) *v(James) = m(Ramon) * v(Ramon)
with m(James) = the mass of James = 95.0 kg
with v(James) = speed of James = TO BE DETERMINED
with m(Ramon) = mass of Ramon = 67.0 kg
with v(Ramon) = speed of Ramon = 1.10 m/s
v(James) = (m(Ramon) * v(Ramon))/ m(james)
v(James) = (67.0 kg* 1.10 m/s) / 95.0 kg
v(James) = 0.776 m/s
The speed of James is 0.776 m/s