Answer:
1.74 m/s
Explanation:
From the question, we are given that the mass of the an object, m1= 2.7 kilogram(kg) and the mass of the can,m(can) is 0.72 Kilogram (kg). The velocity of the mass of an object(m1) , V1 is 1.1 metre per seconds(m/s) and the velocity of the mass of can[m(can)], V(can) is unknown- this is what we are to find.
Therefore, using the formula below, we can calculate the speed of the can, V(can);
===> Mass of object,m1 × velocity of object, V1 = mass of the can[m(can)] × velocity is of the can[V(can)].----------------------------------------------------(1).
Since the question says the collision was elastic, we use the formula below
Slotting in the given values into the equation (1) above, we have;
1/2×M1×V^2(initial velocity of the first object) + 1/2 ×M(can)×V^2(final velocy of the first object)= 1/2 × M1 × V^2 m( initial velocity of the first object).
Therefore, final velocity of the can= 2M1V1/M1+M2.
==> 2×2.7×1.1/ 2.7 + 0.72.
The velocity of the can after collision = 1.74 m/s
Answer:
a) I₁ = 11.2 Lux
, vertical direction
, b) I₂ = 1.44 Lux
Explanation:
a) A polarized is a system that absorbs light that is not polarized in the direction of its axis, therefore half of the non-polarized light must be absorbed
consequently the above the processed light has half of the incident intensity and the directional of the polarized
I₁ = I₀ / 2
I₁ = 22.4 / 2
I₁ = 11.2 Lux
is polarized in the vertical direction
b) The polarized light falls on a second polarizer, therefore it must comply with the law of Malus
I₂ = I₁ cos² θ
I₂ = 11.2 cos² 69
I₂ = 1.44 Lux
The universal law of gravitation states that:
Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of distance between them.
It means that if the gravitational force is F, then if the distance is decreased by 5 times, then the new gravitation force is:
F/5² = F/25
Answer:
25 to the right
Explanation:
there you go friend your awsome
Answer:
Explanation:
Given
Distance between two loud speakers 
Distance of person from one speaker 
Distance of person from second speaker 
Path difference between the waves is given by

for destructive interference m=0 I.e.




frequency is given by

where 

For next frequency which will cause destructive interference is
i.e.
and 


frequency corresponding to this is

for 


Frequency corresponding to this wavelength
