Answer:
≅50°
Explanation:
We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:
Δx=V₀t+at²/2
And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:
Δx=(V₀cosθ)t+at²/2
Now luckily we are given everything we need to solve (or you found the info before posting here):
- Δx=760 m
- V₀=87 m/s
- t=13.6 s
- a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!
With that we can plug the values in to get:
Answer:
mass of ball 1=m1
mass of ball 2=m2
velocity of ball=r1w1
velocity of ball 2=r2w2
Total angular momentum=m1*v1+m2*v2
but
v1=r1*w1
v2=r2*w2
Substitute values in above equation
Total angular momentum of the system=m1*r1*w1+m2*r2*w2
Answer:
A) m = F/a = 91.7/9.81 = 9.35 kg
B) m = F/a = 59.2/9.81 = 6.03 kg
C) m = F/a = 33.4/9.81 = 3.40 kg
D) m = F/a = 9.65/9.81 = 0.984 kg
Explanation:
Answer:
Its acceleration is certainly negative.
Explanation:
- When an object moving in a straight line decelerates then its velocity certainly decreases.
- Its acceleration is certainly negative.
- As we know that acceleration is the rate of change in velocity and the object is decelerating i.e. its velocity is reducing with time.
where:
are final and initial velocities of the object
are final and initial time of observation.
- Its SI unit is
