Answer:
The range is maximum when the angle of projection is 45 degree.
Explanation:
The formula for the horizontal range of the projectile is given by

The range should be maximum if the value of Sin2θ is maximum.
The maximum value of Sin2θ is 1.
It means 2θ = 90
θ = 45
Thus, the range is maximum when the angle of projection is 45 degree.
If the angle of projection is 0 degree
R = 0
It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.
If the angle of projection is 30 degree.

R = 0.088u^2
If the angle of projection is 45 degree.

R = u^2 / g
The force of friction is 19.1 N
Explanation:
According to Newton's second law, the net force acting on the bag is equal to the product between its mass and its acceleration:

where
is the net force
m is the mass
a is the acceleration
The bag is moving at constant speed, so its acceleration is zero:

Therefore the net force is zero as well:

Here we are interested only in the forces acting along the horizontal direction, therefore the net force is given by:

where
is the horizontal component of the applied force, with
F = 22.5 N

is the force of friction
And solving for
, we find

Learn more about friction:
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Explanation:
It is given that,
Mass of the crate, m = 50 kg
Force acting on the crate, F = 10 N
Angle with horizontal, 
Let N is the normal force acting on the crate. Using the free body diagram of the crate. It is clear that,


N = 486.57 N
or
N = 487 N
If a is the acceleration of the crate. The horizontal component of force is balanced by the applied forces as :




or

So, the normal force the crate and the magnitude of the acceleration of the crate is 487 N and
respectively.
Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.
Net flux through the cylindrical surface is given as

here q = enclosed charge in the surface
so here in order to find the value of q

so now we have

so this is the total flux
now by Gauss's law we can find the electric field




<em>by above expression we can find the electric field at required position</em>