As we know that

here we know that


now from above equation we have


so image will form on left side of lens at a distance of 15 cm
This image will be magnified and virtual image
Ray diagram is attached below here
As we know that range of the projectile motion is given by

here we know that range will be same for two different angles
so here we can say the two angle must be complementary angles
so the two angles must be

so it is given that one of the projection angle is 75 degree
so other angle for same range must be 90 - 75 = 15 degree
so other projection angle must be 15 degree
Answer:
The angular speed after 6s is
.
Explanation:
The equation

relates the moment of inertia
of a rigid body, and its angular acceleration
, with the force applied
at a distance
from the axis of rotation.
In our case, the force applied is
, at a distance
, to a ring with the moment of inertia of
; therefore, the angular acceleration is



Therefore, the angular speed
which is

after 6 seconds is


Answer:
It took the projectile 120 s to reach the maximum height.
Explanation:
Given;
maximum height of the projectile, s = 180 km = 180,000 m
initial speed of the projectile, u = 3 km/s = 3000 m/s
final velocity at maximum height, v = 0
Apply the following kinematic equation for average velocity of the projectile;

Therefore, it took the projectile 120 s to reach the maximum height.
Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight is given below.
Explanation:
Measure unstretched length of spring, L. E.g. L = 0.60m.
Set mass to a convenient value (e.g. m = 0.5kg).
Hang mass.
Measure new spring length, L'. E.g. L' = 0.70m.
Calculate extension: e = L' - L = 0.70 – 0.60 = 0.10m
Use mg = ke (in equilibrium weight = tension)
k = mg/e
Don't know what value you are using for example. Suppose it is 10N/kg (same thing as 10m/s²).
k = 0.5*10/0.10 = 50 N/m
Repeat for a few different masses. (L always stays the same.)
Take the average of your k values.