Answer:
F' = 64 F
Explanation:
The electric force between charges is given by :
Where
q₁ and q₂ are charges
r is the distance between charges
When each charge is doubled and the distance between them is 1/4 its original magnitude such that,
q₁' = 2q₁, q₂' = 2q₂ and r' = (r/4)
New force,
Apply new values,
So, the new force becomes 64 times the initial force.
Answer:
2.5m/s²
Explanation:
a = v/t
Where;
V = velocity (m/s)
a = acceleration (m/s²)
t = time (s).
According to the information provided in this question,
a = ?
v = 10m/s
t = 4
a = 10/4
a = 2.5m/s²
Answer: 1100 ft lb/s and 2 H.P
Explanation:
To calculate for the power developed in the elevator motor in ft.lb/s, we multiply the distance and the weight of the elevator and divide the product by the time.
Power = (10 ft)(2200 lb) / 20 s = 1100 ft.lb/s
Next, convert the calculated value to HP.
1100 ft.lb/s x (1 HP/ 550 ft.lb/s) = 2 HP
hope this helps please give brainliest!
Answer:
The objects become oppositely charged and have equal amounts of charge.
Explanation:
There are three methods for charging objects:
- Conduction: a charged object is brought in contact with a neutral object. Electrons are transferred from the charged object to the neutral one, which also becomes charged
- Induction: a charged object is brought close (but not in contact) to a neutral object. The charges inside the neutral object redistribute, such that those of opposite sign to the charge in the charged object migrate on the side closer to the charged object, while the charges of same sign migrate towards the opposite side. If the neutral object is then grounded, the charges on the opposite side flow to the ground, leaving the neutral object charged as well
- Friction: two objects initially neutral are rubbed against each other. Electrons move from one object to the other one: therefore, one object becomes positively charged while the other one becomes negatively charged. Since the charge gained by one object is equal to charge lost by the other object, it follows that the two objects have same magnitude of charge, but with opposite sign.
