<span>The component most affected by the collisions is vertical. The ball's vertical will either decrease or increase due to the collision. If the velocity is high during the collsion the ball's vertical will likely be higher and if the ball's velocity is low the vertical will be as well.</span>
Before solving this question, first we have to understand the special theory of relative.
As per classical mechanics, the velocity of light will be different in different frame of reference. The light moves in the ether medium which exists every where in the entire universe.
Let us consider a body which moves with a velocity v. Let light is coming along the direction of the body. As per classical mechanics,the velocity of light with respect to the body will be [ c-v].
Let us consider that light is coming from opposite direction. Hence, the velocity of light with respect to the observer will be c+v.
From above we see that velocity of light is different in both the cases which is wrong.
As per Einstein's special theory of relativity, the velocity of light will be same in every frame of reference i.e c=300000 km/s.
As per the question ,the space craft is moving with a velocity 0.1 c.
We are asked to calculate the velocity of the light with respect to an observer present in Mars.
Considering Einstein's theory of relativity, the velocity of light will be c [300000 km/s] with respect to the person in Mars.
Answer:
what language is that
Explanation:
i don't understand the languge u used please can you change it
Answer:
I = 1.06886 N s
Explanation:
The expression for momentum is
I = F t = Δp
therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor
Let's find the components of the initial velocity
sin 28.2 = v_y / v
cos 28.2= vₓ / v
v_y = v sin 282
vₓ = v cos 28.2
v_y = 42.8 sin 28.2 = 20.225 m / s
vₓ = 42.8 cos 28.2 = 37.72 m / s
since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making
θ = -28.2
v_y = -20.55 m / s
v_x = 37.72 m / s
X axis
Iₓ = Δpₓ =
since the ball moves in the x-axis without changing the velocity, the change in moment must be zero
Δpₓ = m - m v₀ₓ = 0
v_{fx} = v₀ₓ
therefore
Iₓ = 0
Y axis
I_y = Δp_y = p_{fy} -p_{oy}
when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards
v_{fy} = - v_{oy}
Δp_y = 2 m v_{oy}
Δp_y = 2 0.0260 (20.55)
= 1.0686 N s
the total impulse is
I = Iₓ i ^ + I_y j ^
I = 1.06886 j^ N s
Answer:
The angle of refraction is 37°.
Explanation:
let n1 be the refractive index of glass and n2 = 1.0 be the refraction index of air, θ be the angle of incidence , ∅ be the angle of refraction.
then, according to Snell's law:
n1×sin(∅) = n2×sin(θ)
sin(∅) = n2×sin(θ)/n1
= (1.0)(sin(71°))/(1.56)
= 0.606101651
∅ = 37°
Therefore, the angle of refraction is 37°.