The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is
<em>T</em> - 25 N = (8 kg) <em>a</em>
where <em>T</em> is the tension in the rope and <em>a</em> is the acceleration.
The hanging mass has a net force of
(6 kg) <em>g</em> - <em>T</em> = (6 kg) <em>a</em>
where <em>g</em> = 9.8 m/s².
Adding these equations together eliminates <em>T</em>, and we can solve for <em>a</em> :
(<em>T</em> - 25 N) + ((6 kg) <em>g</em> - <em>T </em>) = (14 kg) <em>a</em>
33.8 N = (14 kg) <em>a</em>
<em>a</em> = (33.8 N) / (14 kg) ≈ 2.4 m/s²
Then the tension in the rope is
<em>T</em> - 25 N = (8 kg) (2.4 m/s²)
<em>T</em> ≈ 25 N + 19.31 N ≈ 44 N
Answer:
Explanation:
Radius of the pollen is given as
Volume of the pollen is given as
mass of the pollen is given as
so weight of the pollen is given as
Now electric force on the pollen is given
now ratio of electric force and weight is given as
Answer:
D
Explanation:
For this kind of problem, forces add. F = F1 + F2
F1 = 6 N
F2 = 10 N
F = 6N + 10N
F = 16N
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
