Answer:
D) This is the correct answer
Explanation:
In this exercise the two ball loads are suspended by a thread.
To answer this exercise, let us remember that charges of the same sign repel and charges of a different sign attract.
Therefore, for the system to maintain equilibrium, the two charges must be of the same sign.
When examining the different proposals
A) in this case, as a sphere has no charge, there is no electric force and the induced charge is of the opposite sign, so the spheres attract each other
B) in this case there is an electric force, but being of a different sign, the force is attractive so the system is not in equilibrium
C) as the charges are of different magnitude the system does not have equal angles
D) This is the correct answer, since the charges have the same magnitude and are of the same sign, so the force is repulsive and is counteracted by the weight component
F_e = W sin θ
The forces (what causes the ball to accelerate) are gravity, friction, and the normal force. In this case, gravity is a downward force caused by the gigantic mass of the Earth and the mass of the ball. Keep in mind that a force is acceleration. Acceleration is a change in velocity. The ball speeds up. Than it stops speeding up at a certain point where the frictional force (along with air friction) equals the parallel component of gravity.
Newton's Second Law States- The greater mass of an object, the more force it will take to accelerate the object.
To Make I the subject you need to get it by itself. To do this divide both sides by V and t:
I = E/Vt
Answer:
250 m/s
Explanation:
The mass of the bullet, m₁ = 100 g = 0.1 kg
The mass of the gun, m₂ = 5 kg
The backward velocity of the gun, v₂ = -5 m/s
Given that the momentum is conserved, we have;
The total initial momentum = The total final momentum
The gun and the bullet are at rest, therefore, we have;
The initial momentum = 0
The total final momentum = m₁·v₁ + m₂·v₂
Where;
v₁ = The forward velocity of the bullet
Therefore, we get;
m₁·v₁ + m₂·v₂ = 0
0.1 kg × v₁ + 5 kg × (-5 m/s) = 0
0.1 kg × v₁ = 5 kg × 5 m/s
v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s
The forward velocity of the bullet, v₁ = 250 m/s
Answer:
The specific heat is 3.47222 J/kg°C.
Explanation:
Given that,
Temperature = 13°C
Temperature = 37°C
Mass = 60 Kg
Energy = 5000 J
We need to calculate the specific heat
Using formula of energy
Put the value into the formula
Hence, The specific heat is 3.47222 J/kg°C.
