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ladessa [460]
3 years ago
12

An ocean fishing boat is drifting just above a school of tuna on a foggy day. Without warning, an engine backfire occurs on anot

her boat 1.0 km away. How much time elapses before the backfire is heard by the fish (a)
Physics
1 answer:
otez555 [7]3 years ago
3 0

Answer:

0.641 seconds

Explanation:

Distance (d) = 1.0 km

Using the relation :

Speed = distance / time

Time = distance / speed

Speed of sound in seawater is about 1560 m/s

Distance = 1.0km =

1km = 1000m

Time = 1000 / 1560

Time = 0.6410256

Time = 0.641 s

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A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with
Julli [10]

Explanation:

Given that,

Capacitor C=1.66\ \mu F

Resistor R=80.0\ \Omega

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

f_{c}=\dfrac{1}{2\pi R C}

Where, R = resistor

C = capacitor

Put the value into the formula

f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}

f_{c}=1198.45\ Hz

(B). We need to calculate the V_{R} when f = \dfrac{1}{2f_{c}}

Using formula of  V_{R}

V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})

Put the value into the formula

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=2.280\ Volt

(C). We need to calculate the V_{R} when f = f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=3.606\ Volt

(D). We need to calculate the V_{R} when f = 2f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=4.561\ Volt

Hence, This is the required solution.

8 0
3 years ago
This problem follows up on a discussion from lecture. A wind turbine with an efficiency of 45% for converting wind energy into e
Volgvan

Answer:

4.1 m

Explanation:

10 kW = 10000 W

20mi/h = 20*1.6 km/mi = 32 km/h = 32 * 1000 (m/km) *(1/3600) hr/s = 8.89 m/s

The power yielded by the wind turbine can be calculated using the following formula

P = \frac{1}{2} \rho v^3 A C_p

where \rho = 1.2 kg/m^3 is the air density, v = 8.89 m/s is the wind speed, A is the swept area and C_p = 0.45 is the efficiency

10000 = 0.5 * 1.2 * 8.89^3 * A * 0.45

10000 = 190A

A = 10000 / 190 = 52.7 m^2

The swept area is a circle with radius r being the blade length

\pi r^2 = A = 52.7

r^2 = 52.7 / \pi = 16.79

r = \sqrt{16.79} = 4.1 m

4 0
3 years ago
What is the speed of a wave that has a frequency of 200 Hz and a
Luba_88 [7]
The wave speed to this question is 400 meters
8 0
3 years ago
A uniformly charged thin ring has radius 15.0 cm and total charge 20.0 nC . An electron is placed on the ring's axis a distance
Ivanshal [37]

Answer:

b) 1.67×10^7 m/s

Explanation:

The solution is attached in the attachment section

8 0
3 years ago
What are some ways houses along the coastlines can protect themselves from storm surges?
BARSIC [14]
Build walls around the coast
8 0
3 years ago
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