Answer:
= 4.38 × 10³⁴kgm²/s
Explanation:
Given that,
mass of moon m = 9.5 × 10²²kg
Orbital radius r = 4.28 × 10⁵km
Orbital period T = 28.9days
T = 28.9 × 24 × 60 × 60
= 2,496,960s
Angular momentum of the moon about the planet
L = mvr
L = mr²w
You can use the displacement method or the eureka can so basically in the displacement can what you have to do is to put some water into a measuring cylinder and measure its volume before adding the irregular shaped object and then measuring the level of water which had been displaced and then eureka can you can check online
This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.
Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.
<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>
<em />
<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>
Answer:
(a) v = 3..6 m/s
(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.
Explanation:
from the question we have the following:
mass of the car (Mc) = 24,000 kg
initial velocity of the car (u) = 4 m/s
mass of water (Mw) = 3000 kg
final velocity of the car (v) = ?
(a) we can calculate the final momentum of the car by applying the conservation of momentum where
initial momentum = final momentum
Mc x U = (Mc + Mw) x V
24000 x 4 = (24000 + 3000) x v
96,000 = 27000v
v =3.6 m/s
(b) The rain falling downward has been able to affect the horizontal motion of the car by reducing it's velocity from 4 m/s to 3.6 m/s.
