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Vikki [24]
3 years ago
8

Does a mirror work because of refraction

Physics
2 answers:
vichka [17]3 years ago
5 0
A mirror works from reflection. Not refraction. Easy stuff c'mon
Semenov [28]3 years ago
3 0

No.  A mirror works because of reflection.

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What is the minimum force required to increase the energy of a car by 84 J over a distance of 38 m? Assume the force is constant
konstantin123 [22]

Answer:

2.210N

Explanation:

Workdone = Force x distance

Distance = 38m , Workdone = 84J

Hence 84J = Force x 38m

Force = 84J / 38m

Force = 2.210N =2.2N

4 0
3 years ago
If a pulley system with an ideal mechanical advantage of 2,000,000 is used in lifting a 2,000 lb. car, how far would the car mov
krek1111 [17]

Answer:

C

Explanation:

4 0
3 years ago
A charge -353e is uniformly distributed along a circular arc of radius 5.30 cm, which subtends an angle of 48°. What is the line
vladimir2022 [97]

Answer:

- 1.3 x 10⁻¹⁵ C/m

Explanation:

Q = Total charge on the circular arc = - 353 e = - 353 (1.6 x 10⁻¹⁹) C = - 564.8 x 10⁻¹⁹ C

r = Radius of the arc = 5.30 cm = 0.053 m

θ = Angle subtended by the arc = 48° deg = 48 x 0.0175 rad = 0.84 rad        (Since 1 deg = 0.0175 rad)

L = length of the arc

length of the arc is given as

L = r θ

L = (0.053) (0.84)

L = 0.045 m

λ = Linear charge density

Linear charge density is given as

\lambda =\frac{Q}{L}

Inserting the values

\lambda =\frac{-564.8\times 10^{-19}}{0.045}

λ = - 1.3 x 10⁻¹⁵ C/m

4 0
3 years ago
What is the wavelength of the wave
sladkih [1.3K]

Explanation:

it is equal to the speed (v) of a wave train in a medium divided by its frequency (f): λ = v/f. Waves of different wavelengths.

8 0
3 years ago
Read 2 more answers
A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre
pantera1 [17]

Answer:

Part a)

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

T = \frac{m}{L}xg + Mg

so the speed of the wave at that position is given as

v = \sqrt{\frac{T}{\mu}}

here we know that

\mu = \frac{m}{L}

now we have

v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}

v = \sqrt{xg + \frac{MLg}{m}}

Part b)

time taken by the wave to reach the top is given as

t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}

t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})

t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})

t = 12 s

4 0
3 years ago
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