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
Get a direct answer of what???
Answer:
19.3m/s
Explanation:
Use third equation of motion
where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer
![v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s](https://tex.z-dn.net/?f=v%5E2-0%5E2%3D2%289.81m%2Fs%5E2%29%2838%2F2%29%5C%5Cv%5E2%3D9.81m%2Fs%5E2%20%2A38%5C%5Cv%5E2%3D372.78%5C%5Cv%3D%5Csqrt%5B%5D%7B372.78%7D%20%5C%5Cv%3D19.3m%2Fs)
Answer:
576 joules
Explanation:
From the question we are given the following:
weight = 810 N
radius (r) = 1.6 m
horizontal force (F) = 55 N
time (t) = 4 s
acceleration due to gravity (g) = 9.8 m/s^{2}
K.E = 0.5 x MI x ω^{2}
where MI is the moment of inertia and ω is the angular velocity
MI = 0.5 x m x r^2
mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg
MI = 0.5 x 82.65 x 1.6^{2}
MI = 105.8 kg.m^{2}
angular velocity (ω) = a x t
angular acceleration (a) = torque ÷ MI
where torque = F x r = 55 x 1.6 = 88 N.m
a= 88 ÷ 105.8 = 0.83 rad /s^{2}
therefore
angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s
K.E = 0.5 x MI x ω^{2}
K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules
