450 g =0.9921 lb I hope this helps you out.
I DONT know FiGURE it out YOURSELF
Answer:
1) F = 24 N
2) Distance = 1 m
Explanation:
We are given;
Mass; m = 120 g = 0.12 kg
Initial velocity; u = 20 m/s
Final velocity; v = 0 m/s since it came to rest.
Time; t = 0.1 s
We can calculate acceleration from Newton's first equation of motion;
a = (v - u)/t
a = (0 - 20)/0.1
a = -200 m/s²
1) magnitude of the resistance will be;
F = ma
F = 0.12 × (-200)
F = -24 N
Since, we are dealing with the magnitude, we will take the absolute value. Thus, F = 24 N
2) To find the distance moved by the bullet, we know that;
Distance = Average speed × time
Thus;
Distance = ((v + u)/2) × t
Distance = ((0 + 20)/2) × 0.1
Distance = 1 m
Equations of the vertical launch:
Vf = Vo - gt
y = yo + Vo*t - gt^2 / 2
Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
and I will approximate g to 10m/s^2
=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s
The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.
Answer: 11.25 m/s
Answer:
the relative speed should be 40mph
Explanation:
