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Daniel [21]
2 years ago
5

A surveyor measures the distance across a straight river by the following method: Starting directly across from a tree on the op

posite bank, he walks x = 112 m along the riverbank to establish a baseline. Then he sights across to the tree. The angle from his baseline to the tree is = 37.8°.
How wide is the river?
y = _____________ m
Physics
1 answer:
Zina [86]2 years ago
6 0

y =  11.3380m

<h3>What is Young's double-slit experiment?</h3>

The double-slit experiment  is an experiment, that shows that light has both a wave nature or characteristic and a particle nature or characteristic, and that these natures are inseparable.

So, light is said to have wave–particle duality rather than be only a wave or only a particle. The same is true for electrons and other quantum particles.

According to the question,

The relative to angle θ, its adjacent side has length x and its opposite side is equal to width of the river, y;

tanθ = \frac{y}{d} = y = dtanθ

y =( 112m) tan (37.8° )

y ≈  11.3380m

The width of river is 11.3380m

Learn more about Young's double-slit experiment here:

brainly.com/question/17167388

#SPJ1

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Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω ,
Degger [83]

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41  ,  RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V ,  Rc = 473 Ω

a.

Using law of Ohm

V = I * R

I = Vc / Rc =  364 V / 473 Ω

I = 0.76 A

b.

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

c.

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -)  √ 150.74² 77.0²

RL = 473 (+ / -)  (129.58)

RL₁ = 34.41  ,  RL₂ = 602.58

d.

The higher value have the less angular frequency  

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

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3 years ago
A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force
salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

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Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

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Frictional force f = 0.031 N

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Energy gained by spring

KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J

Energy lost due to friction

W=Fd=0.031\times 0.158=0.0048J

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

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3 years ago
A 0.750kg block is attached to a spring with spring constant 13.5N/m . While the block is sitting at rest, a student hits it wit
vazorg [7]

Answer:

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Given that

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