Answer:
Explanation:
Given that,
Lightning flashes one mile (1609 m) away from you.
We need to find the time it take the light to travel that distance. Let the time be t. We know that,
speed = distance/time
So, the required time is 
Answer:
The magnitude of the gravitational force is 4.53 * 10 ^-7 N
Explanation:
Given that the magnitude of the gravitational force is F = GMm/r²
mass M = 850 kg
mass m = 2.0 kg
distance d = 1.0 m , r = 0.5 m
F = GMm/r²
Gravitational Constant G = 6.67 × 10^-11 Newtons kg-2 m2.
F = (6.67 × 10^-11 * 850 * 2)/0.5²
F = 0.00000045356 N
F = 4.53 * 10 ^-7 N
The electrostatic force between two charges q1 and q2 is given by
where
is the Coulomb's constant and r is the distance between the two charges.
If we use F=19.2 N and q1=q2=-3.0 C, we can find the value of r, the distance between the two charges by re-arranging the previous formula:
Answer:
Explanation:
Given that,
At one instant,
Center of mass is at 2m
Xcm = 2m
And velocity =5•i m/s
One of the particle is at the origin
M1=? X1 =0
The other has a mass M2=0.1kg
And it is at rest at position X2= 8m
a. Center of mass is given as
Xcm = (M1•X1 + M2•X2) / (M1+M2)
2 = (M1×0 + 0.1×8) /(M1 + 0.1)
2 = (0+ 0.8) /(M1 + 0.1)
Cross multiply
2(M1+0.1) = 0.8
2M1 + 0.2 =0.8
2M1 = 0.8-0.2
2M1 = 0.6
M1 = 0.6/2
M1 = 0.3kg
b. Total momentum, this is an inelastic collision and it momentum after collision is given as
P= (M1+M2)V
P = (0.3+0.1)×5•i
P = 0.4 × 5•i
P = 2 •i kgm/s
c. Velocity of particle at origin
Using conversation of momentum
Momentum before collision is equal to momentum after collision
P(before) = M1 • V1 + M2 • V2
We are told that M2 is initially at rest, then, V2=0
So, P(before) = 0.3V1
We already got P(after) = 2 •i kgm/s in part b of the question
Then,
P(before) = P(after)
0.3V1 = 2 •i
V1 = 2/0.3 •i
V1 = 6 ⅔ •i m/s
V1 = 6.667 •i m/s
