longitude and latitude<span />
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Explanation:
As we know that catapult is projected with speed 19.9 m/s
so here we have


similarly we have


Part a)
Horizontal displacement in 1.03 s



Part b)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part c)
Horizontal displacement in 1.71 s



Part d)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Part e)
Horizontal displacement in 5.44 s



Part f)
Vertical direction we have
![y = v_y t - \frac{1]{2}gt^2](https://tex.z-dn.net/?f=y%20%3D%20v_y%20t%20-%20%5Cfrac%7B1%5D%7B2%7Dgt%5E2)


Answer:
270 m/s²
Explanation:
Given:
α = 150 rad/s²
ω = 12.0 rad/s
r = 1.30 m
Find:
a
The acceleration will have two components: a radial component and a tangential component.
The tangential component is:
at = αr
at = (150 rad/s²)(1.30 m)
at = 195 m/s²
The radial component is:
ar = v² / r
ar = ω² r
ar = (12.0 rad/s)² (1.30 m)
ar = 187.2 m/s²
So the magnitude of the total acceleration is:
a² = at² + ar²
a² = (195 m/s²)² + (187.2 m/s²)²
a = 270 m/s²
Answer:
The store energy in the inductor is 0.088 J
Explanation:
Given that,
Inductor = 100 mH
Resistance = 6.0 Ω
Voltage = 12 V
Internal resistance = 3.0 Ω
We need to calculate the current
Using ohm's law


Put the value into the formula


We need to calculate the store energy in the inductor



Hence, The store energy in the inductor is 0.088 J
Answer:
500 N
Explanation:
Since the work done on the spring W = Fx where F = force applied and x = compression length = 0.170 m (since the spring will be compressed its full length when the force is applied)
Since W = 85.0 J and we need to find F,
F = W/x
= 85.0 J/0.170 m
= 500 N
So, the magnitude of force must you apply to hold the platform stationary at the final distance given above is 500 N.