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coldgirl [10]
3 years ago
15

There is a 200n boulder stuck in ur yard. You want to use 2 meter long lever to pry it loose from the ground. If you are able to

generate 50n of force to the other end of the lever, where should the fulcrum be placed? How far from the boulder? How far from u?

Physics
1 answer:
Kaylis [27]3 years ago
7 0
40cm from boulder and 160cm from you.

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The time required from the passing of one crest to the next is called the wave's
Strike441 [17]
That's the wave's ' period '.
It's the reciprocal of the wave's frequency.
8 0
3 years ago
A non-reflective coating that has a thickness of 198 nm (n = 1.45) is deposited on top of a substrate of glass (n = 1.50). What
spin [16.1K]

Answer:

The  wavelength is \lambda_ 1 =  574.2 nm

Explanation:

From the question we are told that  

      The  thickness is t =  198 nm  =  198 *10^{-9 }\ m

      The refractive  index of the non-reflective coating is  n_m  =  1.45  

       The  refractive  index of glass is n_g  = 1.50

       

Generally the condition for  destructive  interference is mathematically represented as

            2 *  n_m *  t  *  cos (\theta) =  n  *  \lambda

Where \thata \theta is the angle of refraction which is  0° when the light is strongly transmitted

    and  n is the order maximum interference

        so  

             \lambda = \frac{2 *  n *  t  *  cos (\theta )}{n}

at the point n =  1  

           \lambda _1 = \frac{2 *  1.45  *  198*10^{-9}  *  cos (0 )}{1}

           \lambda_1  = 574.2 *10^{-9}

          \lambda_1  = 574.2 nm

at  n =2  

         \lambda _2  =  \frac{\lambda _1 }{2}

         \lambda _2  =  \frac{574.2*10^{-9} }{2}

         \lambda _2  =  2.87 1 *10^{-9} \ m

         \lambda _2  =  287. 1  nm

Now we know that the wavelength range of visible light is  between

           390 \ nm \to  700 \ nm

   So the wavelength of visible light that is been transmitted is  

          \lambda_ 1 =  574.2 nm

           

6 0
4 years ago
What force does the water exert (in addition to that due to atmospheric pressure) on a submarine window of radius 44.0 cm at a d
Butoxors [25]
Calculate the pressure due to sea water as density*depth.
That is, 
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Atmospheric pressure is  101.3 kPa
Total pressure is  94423 + 101.3 = 94524 kPa (approx)

The area of the window is π(0.44 m)^2 = 0.6082 m^2

The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx
3 0
4 years ago
A ball is thrown vertically upward from the top of a 100 foot tower, with an initial velocity of 10 ft/sec. Its position functio
yan [13]

All you need to know is that velocity is the derivative with respect to time of position.

Therefore: ds/dt = -32t +10

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8 0
3 years ago
g A boat is anchored 2000 ft from shore anddirects its searchlight towards an automobile travelingdown the straight road. At the
Darya [45]

Answer:

Explanation:

The distance of searchlight will act as radius R and velocity of car may be supposed to be tangential velocity v  . We are required to calculate  angular velocity ω .

v = 80 ft /s

R = 3000 ft

ω = v / R

= 80  / 3000 = .027 rad / s

For angular acceleration the formula is

angular acceleration α = a / R

a is linear acceleration = 15 ft / s²

α = 15 / 3000 = .005 rad / s².

3 0
3 years ago
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