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alisha [4.7K]
2 years ago
7

What is the index of refraction of a material in which the red-light wavelength is 450 nm?

Physics
1 answer:
valentinak56 [21]2 years ago
5 0

Answer:

rrrrrrrrrweeaxxavsbdjjdkgitififififjfjfuriririururj

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A grocery cart with a mass of 16 kg is pushed at constant speed along an aisle by a force fp = 12 n which acts at an angle of 17
olga_2 [115]

Solution for this problem is we will be using the following formula:

Work done by applied force) = Force (applied) x s x cos θ

Where:

Force applied = 12 N

s = 13 m long

Angle = 17

So plugging in the values:


=>W (a) = 12 x 13 x cos 17

= 12 x 12.43196182

= 149.1835 J 

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A car starts from rest and speeds up to 30 m/s in a total of 6 s.
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5 m/s
30 divided by 6 is 5
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A car travels at a constant velocity of 20.0 meters/second for 15 seconds. What is the power of the car if the initial force app
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Work done by the force = Force x displacement. Power = work done/time = F.s/t = F.u.t/t = F.u = 95 x 20 = 1900J.                                                                                                                                                                                               {S=ut because acceleration is zero since car is moving at constant velocity}. 
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This is a measure of the quantity of matter.
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The measure of the quantity of matter would be mass. Mass is measured in kilograms. I hope this helped!:)

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the maximum range of a projectile is 2÷√3 times its actual range what is the angle of the projection for the actual range​
Murrr4er [49]

Answer:

The actual angle is 30°

Explanation:

<h2>Equation of projectile:</h2><h2>y axis:</h2>

v_y(t)=vo*sin(A)-g*t

the velocity is Zero when the projectile reach in the maximum altitude:

0=vo-gt\\t=\frac{vo}{g}

When the time is vo/g the projectile are in the middle of the range.

<h2>x axis:</h2>

d_x(t)=vo*cos(A)*t\\

R=Range

R=d_x(t=2*\frac{vo}{g})

R=vo*cos(A)*2\frac{vo}{g} \\\\R=\frac{(vo)^{2}*2* sin(A)cos(A)}{g} \\\\R=\frac{(vo)^{2} sin(2A)}{g}

**sin(2A)=2sin(A)cos(A)

<h2>The maximum range occurs when A=45°(because sin(90°)=1)</h2><h2>The actual range R'=(2/√3)R:</h2>

Let B the actual angle of projectile

\frac{vo^{2} }{g} =(\frac{2}{\sqrt{3} }) \frac{vo^{2} *sin(2B)}{g}\\\\1= \frac{2 }{\sqrt{3}} *sin(2B)\\\\sin(2B)=\frac{\sqrt{3}}{2}\\\\

2B=60°

B=30°

7 0
2 years ago
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