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Sindrei [870]
2 years ago
14

"Suppose a horizontal laser beam is reflected off a plane mirror that is perfectly smooth and flat. At first, the mirror is angl

ed upward at 4 degrees, measured from the vertical. Then, the angle of the mirror is changed to 6 degrees upward. How would the angle formed by the incoming laser beam and the reflected laser beam change
Physics
1 answer:
Rashid [163]2 years ago
3 0

Answer:

The angle in between changed from 8 to 12 degrees

Explanation:

Solution:-

Angle between the incoming and reflected beam is always twice that of the inclination of the mirror.

Initially, Inclination angle θ = 4 degrees

             Angle between incident and reflected beam is twice of the inclination of mirror. Hence,

            Φ = 2*θ = 2*4 = 8 degrees.

Change, Inclination angle θ = 6 degrees

             Angle between incident and reflected beam is twice of the inclination of mirror. Hence,

            Φ = 2*θ = 2*6 = 12 degrees.

- The angle formed by the incoming laser beam and the reflected laser beam will change from 8 degrees to 12 degrees.

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Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l
alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

a)

The magnitude of the gravitational force between two objects is given by the formula

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

m_1=M_p\\m_2 = M_s\\r=R

So the force is

F_A=G\frac{M_p M_s}{R^2}=F_0

For the system planet B - Star B, we have:

m_1 = 4 M_p\\m_2 = M_s\\r=R

So the force is

F=G\frac{4M_p M_s}{R^2}=4F_0

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

m_1 = M_p\\m_2 = 4M_s\\r=R

So the force is

F=G\frac{M_p (4M_s)}{R^2}=4F_0

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

b)

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

G\frac{mM}{r^2}=m\frac{v^2}{r}

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

v=\sqrt{\frac{GM}{r}}

For System A,

M=M_s\\r=R

So the tangential speed is

v_A=\sqrt{\frac{GM_s}{R}}

For system B,

M=M_s\\r=R

So the tangential speed is

v_B=\sqrt{\frac{GM_s}{R}}=v_A

So, the speed of planet B is the same as planet A.

For system C,

M=4M_s\\r=R

So the tangential speed is

v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A

So, the speed of planet C is twice the speed of planet A.

3 0
2 years ago
. Emergency rations are to be dropped from a plane to some stranded hikers. The search and rescue plane is flying at an altitude
almond37 [142]

Answer:

35 m/s down

Explanation:

The horizontal speed of the package is 70 m/s.  So the time needed to reach the hikers is:

1000 m / (70 m/s) = 14.28 s

Taking down to be positive, the initial velocity needed is:

Δy = v₀ t + ½ at²

1500 m = v₀ (14.28 s) + ½ (9.8 m/s²) (14.28 s)²

v₀ = 35 m/s

The package must be launched down with an initial velocity of 35 m/s.

3 0
2 years ago
Read 2 more answers
Three balls are in water. Ball 1 floats, with half of it exposed above the water level. Ball 2, with a density less than the den
Ymorist [56]

Answer:

The magnitude of buoyancy force is equal to that of ball's weight.

Explanation:

Ball 1 is floating on water. Weight of ball 1 is Fg=m1g  is acting vertically downward

Force of buoyancy FB = ρVdisg is acting vertically upward.

Net force acting on the ball is zero, FB=Fg

Answer

The magnitude of buoyancy force is equal to that of ball's weight.

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Which planet might best be described as a large dirty ice ball?
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This planet would be known as pluto
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