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sergeinik [125]
3 years ago
7

Which shows the correct angle of reflection given the incident ray shown? ​

Physics
1 answer:
kipiarov [429]3 years ago
5 0

\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the Concept of the Rays of light.

Since we know that, The light ray on a shiny surface makes a angle with the normal, hence The Angle of incidence is equal to angle of reflection made with Normal Axis in the middle.

Thus Option c.) is correct.

C.) is the Answer.

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Can someone help meeeeee... show how to solve it plzzzzzzzz
liubo4ka [24]
<h2>Right answer: 64 units</h2><h2></h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

F=G\frac{m_{1}m_{2}}{r^2}

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

In this case we have a gravitation force F_{1}=16units, given by the formula written at the beginning. Let’s rename the distance r as d:

F_{1}=G\frac{m_{1}m_{2}}{d^2}     (1)

And we are asked to find the gravitation force F_{2} with a given distance of \frac{d}{2}:

F_{2}=G\frac{m_{1}m_{2}}{({\frac{d}{2})}^{2}}      

F_{2}=G\frac{m_{1}m_{2}}{{\frac{d^{2}}{4}}}     (2)

The gravity constant is the same for both equations, and we are assuming both masses are constants, as well. So, let’s isolate G m_{1}m_{2} in both equations:

From (1):

Gm_{1}m_{2}=F_{1}{d}^{2}     (3)

From (2):

Gm_{1}m_{2}=F_{2}\frac{{d}^{2}}{4}     (4)

If (3)=(4):

F_{1}{d}^{2}=F_{2}\frac{{d}^{2}}{4}     (5)

Now we have to find F_{2}:

F_{2}=F_{1}{d}^{2}\frac{4}{{d}^{2}}      

F_{2}=4F_{1}     (6)

If F_{1}=16 units:

F_{2}=(4)(16 units)        

F_{2}=64 units>>>>This is the new force of attraction     

3 0
3 years ago
When a sound wave moves through a medium such as air, the motion of the molecules of the medium is in what direction (with respe
Alchen [17]

<span>A sound wave is a pressure wave that results from the vibration of the particles o the medium from the source. The motion of the particles in the medium is parallel to the direction of the energy transport. The type of wave formed by a sound wave is the longitudinal wave. </span>A longitudinal wave is characterized by rarefactions. A longitudinal wave is a wave motion wherein the particles in the wave medium are displaced parallel to transport. When motion is detected from the source, the particle next to it vibrates from its rest position and a progressive change in phase vibration is observed at each particle within that wave. The result is that the energy is transported from one region to the other. These combined motions result in the movement of alternating regions of rarefaction in the direction of propagation.      

7 0
3 years ago
Argon in the amount of 1.5 kg fills a 0.04-m3 piston cylinder device at 550 kPa. The piston is now moved by changing the weights
Arlecino [84]

Answer:

               275 kPa

Explanation:

             mass of the gas=m=1.5 kg

             initial volume if the gas=V₁=0.04 m³

             initial pressure of the gas= P₁=550 kPa

as the condition is given final volume is double the initial volume

             V₂=final volume

             V₂=2 V₁

As the temperature is constant

             T₁=T₂=T

\frac{P1V1}{T1}=\frac{P2 V2}{T2}

putting the values in the equation.

\frac{P1V1}{T1}=\frac{P2 *2V1}{T2}

P₂=\frac{P1}{2}

P₂=\frac{550}{2}

P₂=275 kPa

So the final pressure of the gas is 275 kPa.

           

3 0
4 years ago
Scientists want to place a 3400.0 kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of 2480.0
BaLLatris [955]

Answer:

Part a)

r = 6.96 \times 10^6 m

Part b)

F_g = 3.004 \times 10^3 N

Part c)

a = 0.88 m/s^2

Part d)

v = \frac{2480}{2} = 1240 m/s

Explanation:

Part a)

As we know that the orbital speed of the satellite is given as

v = 2480 m/s

now we will have

v = \sqrt{\frac{GM_{mars}}{r}}

now we have

M_{mars} = 6.4191 \times 10^{23} kg

R_{mars} = 3.397 \times 10^6 m

now we have

2480 = \sqrt{\frac{(6.67 \times 10^{-11})(6.4191 \times 10^{23})}{r}}

r = 6.96 \times 10^6 m

Part b)

Here force between mars and satellite is the gravitational attraction force which is given as

F_g = \frac{GM_{mars} m}{r^2}

F_g = \frac{(6.67\times 10^{-11})(6.4191 \times 10^{23})(3400)}{(6.96\times 10^6)^2}

F_g = 3.004 \times 10^3 N

Part c)

Acceleration of satellite is the ratio of gravitational force and mass of the satellite

a = \frac{F_g}{m}

a = \frac{3004}{3400}

a = 0.88 m/s^2

Part d)

As we know by III law of kepler

\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}

here we know that T2 = 8 T1

(\frac{1}{8})^2= \frac{r_1^3}{r_2^3}

\frac{r_1}{r_2} = (\frac{1}{2})^2

so we have

r_2 = 4r_1

as we know that speed is given as

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

so we can say since radius is orbit becomes 4 times so the orbital speed must be half

v = \frac{2480}{2} = 1240 m/s

7 0
4 years ago
A car rounds a curve while maintaining a constant speed. Is there a net force on the car as it rounds the curve?
Nesterboy [21]

Answer:

Yes, Centripetal Force acts on the Car.

Explanation:

When a car undergoes a circular motion, a centripetal force acts onto it.

Centripetal Force = [m*(v^2)]/R , where

  • R is the radius of curve
  • v is the constant speed
  • m is the mass of the car
4 0
4 years ago
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