Answer:
F = - 3.56*10⁵ N
Explanation:
To attempt this question, we use the formula for the relationship between momentum and the amount of movement.
I = F t = Δp
Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say
v = d / t
t = d / v
Given that
m = 26 g = 26 10⁻³ kg
d = 50 mm = 50 10⁻³ m
t = d/v
t = 50 10⁻³ / 600
t = 8.33 10⁻⁵ s
F t = m v - m v₀
This is so, because the bullet bounces the speed sign after the crash is negative
F = m (v-vo) / t
F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵
F = - 3.56*10⁵ N
The negative sign is as a result of the force exerted against the bullet
Lungs vacoules on if those 2
Answer:
k = 104.46 N/m
Explanation:
Here we can use energy conservation
so we will have
initial gravitational potential energy = final total spring potential energy
as we know that she falls a total distance of 31 m
while the unstretched length of the string is 12 m
so the extension in the string is given as
![x = L - L_o](https://tex.z-dn.net/?f=x%20%3D%20L%20-%20L_o)
![x = 31 - 12 = 19 m](https://tex.z-dn.net/?f=x%20%3D%2031%20-%2012%20%3D%2019%20m)
so we have
![mgH = \frac{1}{2}kx^2](https://tex.z-dn.net/?f=mgH%20%3D%20%5Cfrac%7B1%7D%7B2%7Dkx%5E2)
![62(9.81)(31) = \frac{1}{2}k (19^2)](https://tex.z-dn.net/?f=62%289.81%29%2831%29%20%3D%20%5Cfrac%7B1%7D%7B2%7Dk%20%2819%5E2%29)
![k = 104.46 N/m](https://tex.z-dn.net/?f=k%20%3D%20104.46%20N%2Fm)
It moves to 56 km per hours
Tensional forces which is associated with normal faults