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lutik1710 [3]
2 years ago
4

Suppose that the spotlight shines so that different parts of the beam reflect off of different two surfaces, one inclined at an

angle α (from the horizontal) and one inclined at an angle β. What would the angular separation Δθ be between the rays reflected from the two surfaces? Assume that the light comes at an angle θa to the vertical. Express your answer in terms of some or all of the angles θa, α, and β.
Physics
1 answer:
Amanda [17]2 years ago
6 0

Answer:

Explanation:

For reflection from first surface ,

angle of incidence  = angle of reflection

θa = angle of reflection

Now reflecting surface turns by angle ( β - α ) , so reflected ray will turn by twice the angle. So angle between two reflected rays will be equal to

2( β - α )

So Δθ = 2( β - α )

This is the required relation.

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A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coas
SVEN [57.7K]

Answer:

the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s

Explanation:

given information

car's mass, m = 1200 kg

h_{A} = 100 m

v_{A} = v_{A}

h_{B} = 150 m

v_{B} = 0

according to conservative energy

the distance from point A to B, h = 150 m - 100 m = 50 m

the initial speed v_{A}

final speed  v_{B} = 0

thus,

v_{B}² = v_{A}² - 2 g h

0 = v_{A}² - 2 g h

v_{A}² = 2 g h

v_{A} = √2 g h

    = √2 (9.8) (50)

    = 31.3 m/s

8 0
2 years ago
1) A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup
azamat

Answer:

\dot{W} = 339.84 W

Explanation:

given data:

flow Q = 9 m^{3}/s

velocity = 8 m/s

density of air = 1.18 kg/m^{3}

minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as

\dot{W} =\dot{m}\frac{V^{2}}{2}

here \dot{m}is mass flow rate and given as

\dot{m} = \rho*Q

\dot{W} =\rho*Q\frac{V^{2}}{2}

Putting all value to get minimum power

\dot{W} =1.18*9*\frac{8^{2}}{2}

\dot{W} = 339.84 W

7 0
2 years ago
D=? V=100mL M=1.5kg=___g
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If you go to a website known as 'Wolfram' and type in the question they should be able to answer it.  It's basically a very smart calculator.  I'd add a link but I don't want to seem like I'm advertising, and it's against Brainly rules.
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2 years ago
An observer O is standing on a platform of length L = 90 m on a station. A rocket train passes at a relative (constant) speed of
Natali [406]

Answer:

Explanation:

Since the front and back of the rocket simultaneously line up with forward and backward end of the platform respectively .

Then length of the platform = length of the train rocket .

A )

Time to cross a particular point on the platform

= length of rocket train / .96 x 3 x 10⁸

= 90 /  .96 x 3 x 10⁸

= 31.25 x 10⁻⁸ s

B)  Rest length of the rocket = length of platform = 90 m

C ) length of platform  as viewed by moving observer =

\frac{90}{\sqrt{1-\frac{v^2}{c^2 } } }

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= 321 m

D )  For the observer on platform time taken = 31.25 x 10⁻⁸ s

for the observer in the rocket , time will be dilated so time recorded by observer in motion ,

31.25\times10^{-8} \times \sqrt{1-\frac{.96^2}{1} }

8.75 x 10⁻⁸ s .

8 0
2 years ago
Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec
AysviL [449]
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
E= \frac{\sigma}{2\epsilon_0}
where
\sigma is the charge density
\epsilon_0 is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
E=30 N/C
7 0
2 years ago
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