The equation to be used is the derived formulas for rectilinear motion at a constant acceleration. The formula for acceleration is
a = (v - v₀)/t
where
v and v₀ are the initial and final velocities, respectively
t is the time
a is the acceleration
Since it started from rest, v₀ = 0. Using the formula:
0.15 m/s² = (v - 0)/[2 minutes*(60 s/1 min)]
Solving for v,
v = 18 m/s
Answer: Impulse = 4 kgm/s
Explanation:
From the question, you're given the following parameters:
Momentum P1 = 12 kgm/s
Momentum P2 = 16 kgm/s
Time t = 0.2 s
According to second law of motion,
Force F = change in momentum ÷ time
That is
F = (P2 - P1)/t
Cross multiply
Ft = P2 - P1
Where Ft = impulse
Substitute P1 and P2 into the formula
Impulse = 16 - 12 = 4 kgm/s
The magnitude of the impulse is therefore 4 kgm/s.
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Answer: a. F doubled
b. F reduced by one-quarter i.e
1/4*(F)
c. 1/9*(F)
d. F increased by a factor of 4 i.e 4*F
e. F reduces 3/4*(F)
Explanation: Coulombs law states the force F of attraction/repulsion experience by two charges qA and qB is directly proportional to thier product and inversely proportional to the square of distance d between them. That is
F = k*(qA*qB)/d²
a. If qA is doubled therefore the force is doubled since they are directly proportional.
b. If qA and qB are half, that means thier new product would be qA/2)*qB/2 =qA*qB/4
Which means the product of charge is divided by 4 so the force would be divided by 4 too since they are directly proportional.
c. If d is tripped that is multiplied by 3. From the formula new d would be (3*d)²=9d² but force is inversely proportional to d² so instead of multiplying by 9 the force will be divided by 9
d. If d is cut into half that is divided by 2. The new d would be (d/2)²=d²/4. So d² is divided by 4 so the force would be multiplied by 4
e. If qA is tripled that is multiplied by 3. F would be multiplied by 3 also, if at the same time d is doubled (2*d)²= 4*d² . Force would be divided by 4 at same time. So we have,
3/4*F
I think answer is A. The Brightness of a Star
