Answer:
B is the answers for the question
Answer:
2.77 * 10^5 m/s
Explanation:
Let us recall that kinetic energy is given by 1/2 mv^2
Where;
m = mass of the body
v = velocity of the body
In this case,
m = 3.38 * 10^31 kg
KE= 1.30 * 10^42 J
KE = 1/2 mv^2
v = √2KE/m
v = √2 * 1.30 * 10^42/3.38 * 10^31
v = √7.69 * 10^10
v = 2.77 * 10^5 m/s
Answer: C and D
The equipment would have stayed in the same exact location indefinitely until the very moment the astronaut applied force to it.
The equipment will continue moving in the same direction indefinitely unless another force is applied to stop it.
Explanation: According to Newton's first law of motion which state that; A body at rest will continue to be at rest, or in linear motion will continue to move in a straight line, unless an external force act on it.
The equipment would have stayed in the same exact location indefinitely until the very moment the astronaut applied force to it.
immediately the astronaut apply force to the object by pushing in, Newton's first law will be manifested in which the equipment will continue moving in the same direction indefinitely unless another force is applied to stop it.
Given Information:
Voltage of circuit A = Va = 208 Volts
Current of circuit A = Ia = 40 Amps
Voltage of circuit B = Vb = 120 Volts
Current of circuit B = Ib = 20 Amps
Required Information:
Ratio of power = Pa/Pb = ?
Answer:
Ratio of power = Pa/Pb = 52/15
Explanation:
Power can be calculated using Ohm's law
P = VI
Where V is the voltage and I is the current flowing in the circuit.
The power delivered by circuit A is
Pa = Va*Ia
Pa = 208*40
Pa = 8320 Watts
The power delivered by circuit B is
Pb = Vb*Ib
Pb = 120*20
Pb = 2400 Watts
Therefore, the ratio of the maximum power delivered by circuit A to that delivered by circuit B is
Pa/Pb = 8320/2400
Pa/Pb = 52/15
Answer: 1.5×10^10 N/C
Explanation:
E= F/q
Where E= magnitude of the electric field
F= force of attraction
q= charge of the given body
Given F= 6.5×10^-8 N
q= 4.3× 10^-18 C
Therefore, E = 6.5×10 ^-8/ 4.3×10^-18
E = 1.5×10^10 N/C
