Answer:
The answer is C)The force of gravity from Earth acting on the spacecraft decreased because the distance from Earth increased.
Explanation:
Gravity, a force, is dependent on the mass of the object exerting the gravity and the distance of an outside object from that object. The larger the object, the more gravity it will exert on an outside object. This force decreases as you move away from the object, but it will always still exist and never be equal to 0.
Answer:
0.006075Joules
Explanation:
The final kinetic energy of the system is expressed as;
KE = 1/2(m1+m2)v²
m1 and m2 are the masses of the two bodies
v is the final velocity of the bodies after collision
get the final velocity using the law of conservation of momentum
m1u1 + m2u2 = (m1+m2)v
0.12(0.45) + 0/12(0) = (0.12+0.12)v
0.054 = 0.24v
v = 0.054/0.24
v = 0.225m/s
Get the final kinetic energy;
KE = 1/2(m1+m2)v
KE = 1/2(0.12+0.12)(0.225)²
KE = 1/2(0.24)(0.050625)
KE = 0.12*0.050625
KE = 0.006075Joules
Hence the final kinetic energy of the system is 0.006075Joules
Answer:
The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
Explanation:
Given that
q₁ = 5 μ C
q₂ = - 4 μ C
The distance between charges = 50 cm
d= 50 cm
Lets take at distance x from the charge μ C ,the electrical field is zero.
That is why the distance from the charge - 4 μ C = 50 - x cm
We know that ,electric field is given as
Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.
Static friction is the friction that exists between two or more solids that are not moving with a relative speed. To calculate the static friction coefficient we use the formula Fs=us × n where Fs is the static friction , us is the coefficient of static friction and the n is the normal force.
thus the coefficient of static friction will be 5 N÷ 25 N = 0.2
Hence 0.2 is the coefficient of static friction
