All we need to do is divide the momentum and weight to find the velocity:
v = m / w
Solve using the values we are given.
v = 952/68 = 14 m/s
Best of Luck!
Answer:
a) x = 1.5 *10⁻⁴cos(524πt) m
b) v = -1.5 *10⁻⁴(524π)sin(524πt) m/s
a = -1.5 *10⁻⁴(524π)²cos(524πt) m/s²
c) x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm
x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm
Explanation:
x = Acos(ωt)
ω = 2πf = 2π(262) = 524π rad/s
x = 1.5 *10⁻⁴cos(524πt)
v = y' = -Aωsin(ωt)
v = -1.5 *10⁻⁴(524π)sin(524πt)
a = v' = -Aω²cos(ωt)
a = -1.5 *10⁻⁴(524π)²cos(524πt)
not sure about the last part as time is generally not given in mm
I will show at 1 second and at 0.001 s to try to cover bases
x(1) = 1.5 *10⁻⁴cos(524π(1))
x(1) = 1.5 *10⁻⁴cos(524π)
x(1) = 1.5 *10⁻⁴(1)
x(1) = 1.5 *10⁻⁴ m = 1.5 *10⁻1 mm
x(0.001) = 1.5 *10⁻⁴cos(524π(0.001))
x(0.001) = 1.5 *10⁻⁴cos(0.524π)
x(0.001) = 1.5 *10⁻⁴(-0.0753268)
x(0.001) = -1.129902...*10⁻⁵ m
x(0.001) = -1.13*10⁻⁵ m = -1.13*10⁻² mm
Answer:

Explanation:
Let the shells speed with respect to ground be 
Let the shells speed with respect to ground be 
mass of shell, 
mass of gun, 
The relative velocity of shell with respect to a free unconstrained gun, 
According to the Newton's third law of motion the direction of the velocity of gun and shell will be in the opposite direction.
So, the relation between the relative velocity and their individual velocity will be:
......................(1)
<u>And according to the conservation of momentum (as the condition is very close to the elastic collision):</u>

substitute the value of
from equation (1)






Answer:
Explanation:
mass of car, m = 1000 kg
initial velocity, u = 20 m/s
final velocity, v = 0 m/s
distance, s = 120 m
Let a be the acceleration of motion
use third equation of motion
v² = u² + 2 as
0 = 20 x 20 + 2 x a x 120
a = - 1.67 m/s²
Let F be the force
Force, F mass x acceleration
F = - 1000 x 1.67
F = - 1666.67 N
The direction of force is towards south and the magnitude of force is 1666.67 N.
The correct answer is: <span>Unscrew one light, if the others remain on it is a parallel circuit.</span>