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4 years ago

A light ray strikes a flat, smooth, reflecting surface at an angle of 80° to the normal. What angle does

Physics

2 answers:

charle [14.2K]4 years ago

7 0

Answer:

80 angle of incidence=angle of reflection

Explanation:

viktelen [127]4 years ago

3 0

Answer:

80 angle of reflection!

Explanation:

Because it is a flat, smooth surface it will reflect like that of a normal plane mirror, so the angle of reflection will be equal to that of the angle of incidence!

Hope this helps you understands!

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if the distance of separation begin two objects is doubled, is the gravitational force between the objects increased or decrease

SashulF [63]

Answer:

force is decreased by a factor of 4.

Explanation:

According to the Newton's law of gravitation, the force of gravitation between the two object is inversely proportional to the square of distance between them. Now the distance is doubled, so the force between the two objects becomes one forth.

Force is decreased by a factor or 4.

7 0

3 years ago

A 282 kg bumper car moving right at 3.50 m/s collides with a 155 kg bumper car moving 1.8 m/s left. Afterwards, the 282 kg car m

Leto [7]

Answer: momentum afterwards = 1040.4kgm/s

Explanation:

8 0

3 years ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

$r_2 = 11.2 nm$

now potential is given as

$V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}$

$V = -0.576 - 0.257 = -0.833 V$

now final potential energy is given as

$U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J$

Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

8 0

3 years ago

Suppose a 1 Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to Ear

vagabundo [1.1K]

Answer:

The time required to send the data from Earth to Moon will be 1.28s while for a two way communication, to send it back to the earth, it will take double time i.e. RTT = 2.56s

Explanation:

Distance between Earth and Moon = 385,000 km = 3.85 x 10⁸m

Speed of data travel = speed of light ≈ 3 x 10⁸m/s

As, v=d/t

t=d/v

$t=\frac{3.85*10^{8} }{3*10^{8}}$

t=1.28s

RTT = Double of single way time taken = 2x1.28

RTT=2.56s

7 0

3 years ago

You are snowboarding down a mountain face at a

Gelneren [198K]

I think it would be 10 m/s

6 0

3 years ago