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Greeley [361]
3 years ago
10

How does a mirror affect the path of light?

Physics
1 answer:
yulyashka [42]3 years ago
8 0

Light rays change direction when they hit a mirror. The phenomenon is known as reflection. Light rays travels in a straight light. They strike the surface of the mirror at a particular angle called incident angle. It is the angle between the ray and normal at the point of contact. The rays leaves the surface making the same angle with the normal called reflection angle but in different direction.

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According to newtons first law which characteristics of a moving object would remain constant if there were no other forces acti
mars1129 [50]

Answer:

Speed ** is the answer

Explanation:

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A cd has a mass of 17 g and a radius of 6.0 cm. When inserted into a player, the cd starts from rest and accelerates to an angul
GuDViN [60]

Answer:

the net toque is τ=8.03* 10⁻⁴ N*m

Explanation:

Assuming the disk has constant density ρ, the moment of inertia I of is

I = ∫r² dm

since m = ρ*V = ρπR² h , then dm= 2ρπh r dr

thus

I = ∫r²dm = ∫r²2ρπh r dr =2ρπh ∫r³ dr = 2ρπh (R⁴/4- 0⁴/4)= ρπhR⁴ /2= mR²/2

replacing values

I = mR²/2= 0.017 kg * (0.06 m)²/2 = 3.06 *10⁻⁵ kg*m²

from Newton's second law applied to rotational motion

τ= Iα , where τ=net torque and α= angular acceleration

since the angular velocity ω is related with the angular acceleration through

ω= ωo + α*t → α =(ω-ωo)/t =  (21 rad/s-0)/0.8 s = 26.25 rad/s²

therefore

τ= Iα= 3.06 *10⁻⁵ kg*m²*26.25 rad/s² = 8.03* 10⁻⁴ N*m

3 0
3 years ago
A NASA satellite has just observed an asteroid that is on a collision course with the Earth. The asteroid has an estimated mass,
Citrus2011 [14]

Answer:

v = 7934.2 m/s

Explanation:

Here the total energy of the Asteroid and the Earth system will remains conserved

So we will have

-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2

now we know that

v_0 = 660 m/s

M = 5.98 \times 10^{24} kg

m = 5 \times 10^9 kg

r = 4 \times 10^9 m

R = 6.37 \times 10^6 m

now from above formula

GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2

now we have

2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2

now plug in all data

2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2

v = 7934.2 m/s

5 0
3 years ago
This is one of the four planets Mercury,Venus,Earth,and Mars, whose orbits lie nearest the Sun
galben [10]

Answer:

Mercury

Explanation:

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6 0
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A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

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3 years ago
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