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2 years ago

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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?

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 $\mathbf{\frac{I}{I_{max}}=0.4535}$

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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.

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3 years ago

50 points if brainliest !!!!!!!!!!

Natasha_Volkova [10]

Answer:

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

$force = 700 \: newtons$

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

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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

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3 years ago

Read 2 more answers