To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as
Here
k = Coulomb's Constant
Charge of each object
d = Distance
Our values are given as,
d = 1 m
a) The electric force on charge 
is
Force is positive i.e. repulsive
b) As the force exerted on will be equal to that act on 
,
Force is positive i.e. repulsive
c) If 
, a negative sign will be introduced into the expression above i.e.
Force is negative i.e. attractive
Answer:
5.66 × 10⁻²³ m/s
Explanation:
If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.
Since initial momentum = final momentum,
mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.
My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?
So mv₁ + M × 0 = m × 0 + MV₂
mv₁ = MV₂
V₂ = mv₁/M = 54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s
Answer:
Net displacement = 0
Distance traveled = 2PQ <_up and down
Explanation:
When the body touches the ground two types of Forces will be generated. The Force product of the weight and the Normal Force. This is basically explained in Newton's third law in which we have that for every action there must also be a reaction. If the Force of the weight is pointing towards the earth, the reaction Force of the block will be opposite, that is, upwards and will be equivalent to its weight:
F = mg
Where,
m = mass
g = Gravitational acceleration
F = 5*9.8
F = 49N
Therefore the correct answer is E.
Answer:
Accuracy is how close a measured value is to an accepted value. <u>Precision is how close measurements are to one another.</u> To make measurements, you have to evaluate both the accuracy and the precision to get a correct value.
