The smallest group of particles in a crystal that retains the shape of the crystal is called the

Answer:

50% of unpolarised light passes through the first filter because, on average, 50% of the waves are aligned with the fiter's axis.  Intensity is reduced by a factor 0.5.

The second filter then reduces the intensity by a factor cos²(θ)

Explanation:

Answer:

a) equal to its weight on the surface of the earth.

Explanation:

The mass of an object is a measure of the amount of matter it contains and is typically expressed in kilograms (kg) or grams (g). Weight, on the other hand, is the force of gravity acting on an object's mass. The weight of an object depends on the strength of the gravitational field it's in. On the surface of the Earth, where the gravitational field is relatively constant, an object's weight is directly proportional to its mass, and the two are equal. This relationship is described by the equation:

Weight (W) = Mass (m) x Acceleration Due to Gravity (g)

On the surface of the Earth, the acceleration due to gravity (g) is approximately 9.8 m/s². Therefore, an object's weight is equal to its mass times 9.8 m/s².

So, on the surface of the Earth, the mass of an object is equal to its weight. In deep space, where there is negligible gravitational force, an object's weight would be nearly zero, but its mass remains the same.

An individual center of gravity can usually be found just below the hips. The correct option is D.

Usually, a person's centre of gravity is situated just below the hips. The location where the body's mass is equally distributed in all directions is referred to as the center of gravity.

The centre of gravity in a standard standingposition is below the hips, specifically in the area of the pelvis. This position gives the body stability and balance, enabling effective movement and control.

Thus, the correct option is D.

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Answer:

4.281 kgm/s upward

Explanation:

Impulse:This can be defined product of force and time. The S.I unit of impulse is Ns.

From Newton's second law of motion,

Impulse = Change in momentum.

I = mΔv....................... Equation 1.

Where m = mass of the ball, Δv = change in velocity of the ball  

and Δv = v -u

Where u = velocity of the ball before it hit the floor, v = velocity of the ball after if hit the floor

I = m(v-u) -------------- Equation 2

But

the initial kinetic energy of the ball = potential energy at the initial height (1.2 m above)

1/2mu² = mgh₁

Where h₁ = initial height. or height of the ball before collision

making u the subject of the equation,

u = √(2gh₁)........................ Equation 3

Where h₁ = 1.2 m g = 9.81 m/s²

Substitute into equation 3

u = √(2×1.2×9.81)

u =√(23.544)

u = -4.852 m/s.

Note: u is negative because the ball was moving downward at the first instance.

Similarly,

v = √(2gh₂)............................. Equation 3

h₂ = height of the ball after collision

Given: h₂ = 0.7 m, g = 9.81 m/s²

Substitute into equation

v = √(2×9.81×0.7)

v = √13.734

v = 3.71 m/s.

Also given: m = 0.5 kg,

Substituting into equation 2

I = 0.5(3.71-(4.852)

I = 0.5(8.562)

I = 4.281 kgm/s. Upward.

Thus the impulse = 4.281 kgm/s upward