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3 years ago

What is reflection of light

2 answers:

4 0

Reflection is when light bounces off an object. If the surface is smooth and shiny, like glass, water or polished metal, the light will reflect at the same angle as it hit the surface. ... This is called specular reflection

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balandron [24]3 years ago

4 0

It's called specular reflection. It's when the light bounces off objects like water,glass,metal, and many other objects. Hope that helps

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What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma

Sindrei [870]

Answer:

E = 1.50 × $10^{8}$ V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

$\mu _b = \frac{B}{2\mu _o}$ ...............1

and

energy density due to electric filed is

$\mu _e = \frac{\epsilon _o E^2}{2}$ ...............2

and here $\mu _b = \mu _ e$

so that

E = $\frac{B}{\sqrt{\mu _o \times \epsilon _o}}$ ...................3

put here value and we get

$E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}$

E = 3 × $10^{8}$ × 0.50

E = 1.50 × $10^{8}$ V/m

6 0

3 years ago

A 77.3 g mass is attached to a horizontal spring with a spring constant of 12.5 N/m and released from rest with an amplitude of

DENIUS [597]

Answer:

speed of the mass is 3.546106 m / s

Explanation:

given data

mass = 77.3 g = 77.3 × $10^{-3}$ kg

spring constant k = 12.5 N/m

amplitude A = 38.9 cm = 38.9 × $10^{-2}$ m

to find out

the speed of the mass

solution

we will apply here conservation energy that is

K.E + P.E = Total energy ..................1

so that Total energy = K.E max = P.E max

we know amplitude so we find out first P.E max that is

PE max = K.E + P.E

(1/2)kA² = (1/2)mv² + (1/2)kx²

kA^² = mv²+ kx²

so here v² will be

v² = k(A² - x²) / m

v = √[(k/m)×(A² - x²)] ............2

here x = (1/2)A so from from 2 equation

v = √[(k/m)×(A² - (A/2)²)]

v = √[(k/m)×(3/4×A²)]

now put all value

v = √[(12.5/ 77.3 × $10^{-3}$ )×(3/4×(38.9 × $10^{-2}$ )²)]

v = 3.546106 m / s

speed of the mass is 3.546106 m / s

6 0

3 years ago

The air that flows over the top part of an airplane's wing moves faster than the air that flows across the bottom. This faster m

kaheart [24]

As an airplane moves through the air, its wings cause changes in the speed and pressure of the air moving past them. These changes result in the upward force called lift.

The Bernoulli principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in the pressure exerted by the fluid.

A wing is shaped and tilted so the air moving over it moves faster than the air moving under it. As air speeds up, its pressure goes down. So the faster-moving air above exerts less pressure on the wing than the slower-moving air below. The result is an upward push on the wing—lift!

4 0

3 years ago

A ball is kicked at 30.0 m/s at an angle of 20.0°. Beneath the tilted columns calculate the vertical and horizontal components.

RSB [31]

The velocities at time t are

• Horizontal:

v = (30.0 m/s) cos(20.0º)

• Vertical:

v = (30.0 m/s) sin(20.0º) - g t

(where g = 9.80 m/s² is the magnitude of the acceleration due to gravity)

If you only want the initial velocities, they are

• Horizontal:

v = (30.0 m/s) cos(20.0º) ≈ 28.2 m/s

• Vertical:

v = (30.0 m/s) sin(20.0º) ≈ 10.3 m/s

(just set t = 0)

As far as starting equations go, you can derive everything from the definition for average acceleration:

a = ∆v / ∆t = (final v - initial v) / t

→ v = u + a t

(here, u stands in for "initial v" and v is simply velocity at time t )

There is no acceleration in the horizontal direction, while the ball is essentially in free-fall in the vertical direction.

6 0

3 years ago

A sled drops 50 meters in height on a hill. The mass of the rider and sled is 70 kg and the sled is going 10 m/s at the bottom o

navik [9.2K]

Answer:

Efficiency = 10.2 %

Explanation:

Given the following data;

Mass = 70 kg

Height = 50 m

Velocity = 10 m/s

We know that acceleration due to gravity is equal to 9.8 m/s².

To find the efficiency of energy conversion from potential to kinetic;

First of all, we would determine the potential energy;

P.E = mgh

P.E = 70 * 9.8 * 50

P.E = 34300 J

For the kinetic energy;

K.E = ½mv²

K.E = ½ * 70 * 10²

K.E = 35 * 100

K.E = 3500

Therefore, Input energy, I = 34300 J

Output energy, O = 3500 J

Next, we find the efficiency;

Efficiency = O/I * 100

Substituting into the formula, we have;

Efficiency = 3500/34300 * 100

Efficiency = 0.1020 * 100

Efficiency = 10.2 %

4 0

3 years ago