Answer:
The average recoil force on the gun during that 0.40 s burst is 45 N.
Explanation:
Mass of each bullet, m = 7.5 g = 0.0075 kg
Speed of the bullet, v = 300 m/s
Time, t = 0.4 s
The change in momentum of an object is equal to impulse delivered. So,
For 8 shot burst, average recoil force on the gun is :
So, the average recoil force on the gun during that 0.40 s burst is 45 N.
Answer:
2.11 seconds
Explanation:
We use the kinematic equation for the velocity in a constantly accelerated motion under the acceleration of gravity (g):
Answer:
Stopping distance = 40m
Explanation:
Given the following :
Initial speed of vehicle before applying brakes = 72km/hr
Converting km/hr to m/s:
72km/hr = [(72 * 1000)m] / (60 * 60)
72km/hr = 72,000m / 3600s
72km/hr = 20m/s
Deceleration after applying brakes (-a) (negative acceleration) = - 5m/s^2
From the 3rd equation of motion:
v^2 = u^2 + 2as
Where v = final Velocity ; u= Initial Velocity ; a = acceleration and s = distance
Final velocity when the car stops will be 0
Therefore ;
v^2 = u^2 + 2as
0 = 20^2 + 2(-5)(s)
0 = 400 - 10s
10s = 400
s = 400/10
s = 40m
Therefore, the stopping distance of the car = 40 meters
Answer:
a) T = 0.5 s
b) v = 1.2π m/s ≈ 3.77 m/s
Explanation:
It makes two revolutions in one second so makes one revolution in ½ second
circumference of the circle is
C = 2πr = 0.6π m
which it traverses in one time period
0.6π m / 0.5 s = 1.2π m/s
15 explanation: look at the graph
