Answer:
The answer of this is question is A.
A) To calculate the charge of each coin, we must apply the expression of the Coulomb's Law:
F=K(q1xq2)/r²
F: The magnitud of the force between the charges. (F=2.0 N).
K: Constant of proporcionality of the Coulomb's Law (K=9x10^9 Nxm²/C²).
q1 and q2: Electrical charges.
r: The distance between the charges (r=1.35 m).
We have the values of F, K and r, so we can calculate q1xq2, because both<span> coins have identical charges:
</span>
q1xq2=(r²xF)/K
q1xq2=(1.35 m)²(2.0 N)/9x10^9 Nxm²/C²
q1xq2=3x10^-10 C
q1=q2=(<span>3x10^-10 C)/2
</span>Then, the charge of each coin, is:
<span>
q1=1.5x</span><span>10^-10 C
</span>q2=1.5x10^-10 C
B) <span>Would the force be classified as a force of attraction or repulsion?
</span>
It is a force of repulsion, because both coins have identical charges and both are postive. In others words, when two bodies have identical charges (positive charges or negative charges), the force is of repulsion.
Answer:
The displacement of the air drop after 3 second is 18.27 m.
Explanation:
Mass of the rain drop = m = 
Weight of the rain drop = W
Duration of time = t = 3 seconds
Drag force on rain drop = 
Motion of the rain drop:
Net force on the rain drop , F= W - D
v = 12.18 m/s
Initial velocity of the rain drop = u = 0 (since, it is starting from rest)
v=u+at (First equation of motion)
(second equation of motion)
s = 18.27 m
The displacement of the air drop after 3 second is 18.27 m.
Answer:
a = 64 ft / s²
Explanation:
The force in a spring is given by Hooke's law
F = -k x
Let's use the initial data to calculate the spring constant
k = F / x
Reduscate to the English system
x = 3 in (1foot/12 in) =0.25 foot
k = 0.3 / 0.25
k = 1.2 lb / foot
Now we can use Newton's second law
F = ma
a = F / m
a = -k x / m
m = w / g
m = 0.3 / 32 = 0.009375
x= 6 in (1foot /12 in)= 0.5 foot
a = - 1.2 0.5 / 0.009375
a = 64 ft / s²
