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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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"My distance to the center of the earth is about 4000 miles when I am on the surface. If I go to a height of 8000 miles above th

lord [1]

Given,

Distance from the surface to the center of the earth, d=4000 miles

Distance from the center to you at a height of 8000 miles, a= 8000+4000=12000 miles

The gravitational force acting on a person at the surface is equal to his weight.

From Newton's Universal Law of Gravitation, the gravitational force is

$F=\frac{G\times M\times m}{r^2}$

Where G is the gravitational constant, M is the mass of the earth, m is the mass of the object/person, r is the distance between the center of the earth and the object/person

At the surface, this force is equal to the weight of the person, W=mg

i.e.

$F_s=\frac{G\times M\times m}{d^2}=W$

On substituting the of d,

$W=\frac{\text{GMm}}{4000^2}$

At a height of 8000 miles from the surface, the gravitational force is equal to,

$F_a=\frac{GMm}{12000^2}$

On dividing the above two equations,

$\frac{F_a}{W}=\frac{4000^2^{}}{12000^2}=\frac{1}{9}$

Therefore,

$F_a=\frac{1}{9}W$

Therefore at a height of 8000 miles above the surface of the earth, the force of gravity becomes 1/9 time your weight.

5 0

1 year ago

Please help Need good grade

ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the range

Using formula of range

$R=\dfrac{v^2\sin(2\theta)}{g}$ ...(I)

The range for the other angle is

$R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$ ....(II)

Here, distance and speed are same

On comparing both range

$\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}$

$\sin(2\theta)=\sin(2\times(\alpha-\theta))$

$\sin120=\sin2(\alpha-60)$

$120=2\alpha-120$

$\alpha=\dfrac{120+120}{2}$

$\alpha=120^{\circ}$

Hence, The other angle is 120°

6 0

3 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

3 years ago

What energy is directly dependent upon velocity and mass?

erastova [34]

Answer:

Kinetic energy

Explanation:

3 0

3 years ago

An ac generator with Em = 223 V and operating at 399 Hz causes oscillations in a series RLC circuit having R = 222 Ω, L = 147 mH

Doss [256]

Answer:

Xc= 17.267 Ω, Z= 415.5 Ω, I= 0.537 A

Explanation:

Em = 223 V

f= 300 Hz, R = 222 Ω, L = 147 mH, C = 23.1 μF

Capacitive reactance = Xc=?

Xc= $\frac{1}{2\pi fC}$

Xc=1/2pi *399*23.1*10^-6

Xc= 17.267 Ω

b).

Z= $\sqrt{ R^2 + (Xl - Xc)^2}$

Xl= 2π * f * L

Xl= 2π * 399 * 147 * $10^{-3}$

Xl= 368.5 Ω

Z= $\sqrt{ R^2 + (Xl - Xc)^2}$ = $\sqrt{222^2 + (368.5 - 17.267)^2}$

Z= 415.5 Ω

c).

Current:

I= V / Z= Em / Z

I= 223/415.5

I= 0.537 A

3 0

3 years ago