Answer:
Explanation:
Given that
Mass of rifle = M
Initial velocity ,u= 0
Mass of bullet = m
velocity of bullet = v
Lets take final speed of the rifle is V
There is no any external force ,that is why linear momentum of the system will be conserve.
Initial linear momentum = Final linear momentum
M x 0 + m x 0 = M x V + m v
0 = M x V + m v
Negative sign indicates that ,the recoil velocity will be opposite to the direction of bullet velocity.
Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant
Given:
P = 687 days
A = 1.52 AU
Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³
Answer: 1.3439 x 10⁵ (days²/AU³)
Answer= 8m/s
Because total Momentum before= total momentum after
Momentum before (p=mu)
p=(4)(12)= 48
p=2(0)=0
So total momentum before=48
Momentum after (p=mu)
Masses combined —2+4=6kg
p=6u
Mb=Ma
48=6u
u=8m/s
Answer:
Total impulse = 
= Initial momentum of the car
Explanation:
Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.
The final velocity of the car is 0 m/s as it is brought to rest.
Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.
The magnitude of impulse is the absolute value of the change in momentum.
Momentum of an object is equal to the product of its mass and velocity.
So, the initial momentum of the car is given as:
The final momentum of the car is given as:
Therefore, the impulse is given as:
Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.
Answer:
Same magnitude of the 10 nc charge cause the electric field is external.
Explanation:
To do a better explanation, let's go and suppose we have an electric field of, 1300 N/C with a 10 nC charge.
As the system we are talking about is really big, and the charge is small, we can assume always if the charge is sitting right in the same point where the electric field is, then, the electric field would not suffer any kind of alteration in it's value. Therefore, no matter what value of the charge is sitting here, the electric field is independent of the charge, so it would not feel any alteration. However, the force that the charge is feeling would be stronger than in the first case.
F = qE
If charge is doubled, then the force would be bigger in the second case than in the first case, but electric field remain the same value.
