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Flura [38]
2 years ago
5

Daniel is 50.0 meters away from a building. He observes that his line-of-sight to the tip of the building makes an angle of 63.0

° with the horizontal. What is the height of the building?
Physics
1 answer:
valentina_108 [34]2 years ago
8 0

Answer:

willie tower

Explanation:

mataas kasi

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Two narrow, parallel slits separated by 0.850 mm are illuminated by 600-nm light, and the viewing screen is 2.80 m away from the
CaHeK987 [17]

Answer:

a) 7.947 radians

b) \mathbf{\frac{I}{I_{max}}=0.4535}

Explanation:

y = Distance from central bright fringe = 2.5 mm

λ = Wavelength = 600 nm

L = Distance between screen and source = 2.8 m

d = Slit distance = 0.85 mm

tan\theta =\frac{y}{L}\\\Rightarrow tan\theta =\frac{2.5}{2800}=0.000892\\\Rightarrow \theta=tan^{-1}0.000892=0.05115^{\circ}

\Delta r= dsin\theta\Rightarrow \Delta r=dsin0.05115=7.589\times10^{-7}

a) Phase difference

\phi=\frac{2\pi}{\lambda}\Delta r\\\Rightarrow \phi=\frac{2\pi}{600\times 10^{-9}}7.589\times10^{-7}=7.947\ rad

∴ Phase difference between the two interfering waves on a screen at a point 2.50 mm from the central bright fringe is 7.947 radians

b) \frac{I}{I_{max}}=cos^2\frac{\phi}{2}\\\Rightarrow \frac{I}{I_{max}}=cos^2\frac{7.947}{2}\\\Rightarrow \mathbf{\frac{I}{I_{max}}=0.4535}

∴ Ratio of the intensity at this point to the intensity at the center of a bright fringe \mathbf{\frac{I}{I_{max}}=0.4535}

8 0
3 years ago
How do think animals grow from babies to adults
Sladkaya [172]

Answer:

As animals get older, their bodies begin to change as well as their instincts and priorities. They get bigger and might adapt or develop.

7 0
3 years ago
50 points if brainliest !!!!!!!!!!
Natasha_Volkova [10]

Answer:

weight = mg \\  = 70 \times 10 \\  = 700 \: newtons

force = 700 \: newtons

force = mass \times acceleration \\ 700 = 70 \times a \\ a = 10 \:  {ms}^{ - 2}

4 0
3 years ago
A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit
elena55 [62]

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

                                            = 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

                             T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }  second

Squaring the terms and solving it for 'g'

                             g = 4 π² \frac{(R+h)^{3} }{R^{2}T^{2}  }   m/s²

Substituting the values in the above equation

                   g = 4 π² \frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}  

                                    g = 4.84 m/s²      

Therefore, the weight

                                     w = m x g   newton

                                         = 5832 Kg x 4.84 m/s²

                                         = 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N                

8 0
3 years ago
The position of a particle moving along the x axis varies in time according to the expression x = 4t 2, where x is in meters and
Pie

Answer:

a) X = 17.64 m

b) X = 17.64 + 4∆t^2 + 16.8∆t

c) Velocity = lim(∆t→0)⁡〖∆X/∆t〗 = 16.8 m/s

Explanation:

a) The position at t = 2.10s is:  

                 X = 4t^2

                 X = 4(2.10)^2

                 X = 17.64 m

b) The position at t = 2.10 + ∆t s  will be:

                 X = 4(2.10 + ∆t)^2

                 X = 17.64 + 4∆t^2 + 16.8∆t  m

c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,

                 ∆X= 4∆t^2 + 16.8∆t

Divide by ∆t on both sides:  

                ∆X/∆t =  4∆t + 16.8

 Taking the limit as ∆t approaches to zero we get:  

              Velocity =lim(∆t→0)⁡〖∆X/∆t〗 = 4(0) + 16.8

              Velocity = 16.8 m/s

 

8 0
3 years ago
Read 2 more answers
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