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kodGreya [7K]
3 years ago
14

A ray of light approaches a jar of honey at an angle of 30.0. If the angle of refraction is 19.5, what is the refractive index o

f honey?
Physics
1 answer:
Stels [109]3 years ago
7 0

Answer:

n = 1.49

Explanation:

Given that,

Angle of incidence = 30 degrees

Angle of refraction = 19.5 degrees

We need to find the refractive index of the honey. Using Snell's law, we can find it as follows :

n=\dfrac{\text{sin i}}{\text{sin r}}\\\\=\dfrac{\text{sin 30}}{\text{sin 19.5}}\\\\n=1.49

So, the refractive index of honey is 1.49.

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3 years ago
A scaffold of mass 57 kg and length 6.5 m is supported in a horizontal position by a vertical cable at each end. A window washer
Mrac [35]

Explanation:

The given data is as follows.

     m_{1} = 57 kg,        m_{2} = 79 kg

      l_{1} = 6.5 m,         l_{2} = (6.5 - 1.9) m = 4.6 m

(a)  The sum of torque ends about far end is as follows.

    m_{1}g \frac{l_{1}}{2} + m_{2}g \times l_{2} - T \times l_{1} = 0

     57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (6.5 - 1.9) - T \times 6.5 = 0

                     T = 828 N

Therefore, 828 N is the tension in the cable closer to the painter.

(b)  Now, we will calculate the sum about close ends as follows.

m_{1}g \frac{l_{1}}{2} + m_{2}g \times (1.9) - T \times l_{1}    

  57 \times 9.81 \times \frac{6.5}{2} + 79 \times 9.81 \times (1.9) - T \times 6.5 = 0

                            T= 506 N

Therefore, 506 N is the tension in the cable further from the painter.  

8 0
3 years ago
The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo
ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

(V_{f})^{2}=(V_{o})^{2}+2aS\\  So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}

As we find acceleration.Now we need to find time

So

V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s

Now for distance

So

Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m

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The isotope of an atom containing 40 protons and 51 neutrons suddenly has 2 neutrons added to it. What isotope is created?
timama [110]
<span>The isotope of an atom containing 40 protons and 51 neutrons suddenly has 2 neutrons added to it
That is X-93 so it will be
</span><span>Zirconium-93 
</span>hope it helps
7 0
3 years ago
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