Answer :
Answer
given,
mass of the drop, m = 0.0014 g
speed of the drop, u = 8.1 m/s
a) Change in momentum is equal to impulse
final velocity of the drop, v = 0 m/s
J = m ( v - u )
J = 0.0014 x 10⁻³ x ( 0 - 8.1 )
J = -1.134 x 10⁻⁵ kg.m/s
impulse of the roof = - J = 1.134 x 10⁻⁵ kg.m/s
b) time, t = 0.37 m s
impact of force = ?
we know
J = F x t
1.134 x 10⁻⁵ = F x 0.37 x 10⁻³
F = 0.031 N
the magnitude of the force of the impact is equal to F = 0.031 N
The magnitude of the impulse delivered to your roof is 1.134 x 10⁻⁵ kg.m/s.
The magnitude of the force of the impact is 0.0307 N.
The given parameters;
- mass of the raindrop, m = 0.0014 g = 0.0014 x 10⁻³ kg
- speed of the rain, v = 8.1 m/s
- time of motion, t = 0.37 ms
The impulse delivered to the roof is calculated as follows;
J = ΔP = mv
[tex]J = (0.0014\times 10^{-3})(8.1)\\\\J = 1.134 \times 10^{-5} \ kg.m/s[/tex]
The magnitude of the force of impact is calculated by applying Newton's second law of motion as follows;
F = ma
[tex]F = \frac{mv}{t} \\\\F = \frac{(0.0014\times 10^{-3})(8.1)}{0.37\times 10^{-3}} \\\\F = 0.0307 \ N[/tex]
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