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2 years ago

A sound wave has a frequency of 265 hz and a wavelength measured at 1.3

Physics

1 answer:

pochemuha2 years ago

8 0

Answer:

V = f λ speed of wave in terms of frequency and wavelength

t = S / V time for wave to travel a distance S

t = 91.4 m / 344.5 m/s = .265 sec time to travel 91.4 m

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Which is an advantage of storing data digitally?

ra1l [238]

Answer:

Explanation:

Because I said so

5 0

3 years ago

Read 2 more answers

Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

$r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m$

The radius is 1.195 m

Time period would be given by

$T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s$

Time period of the motion is 2.8375 s

Angular speed is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s$

The angular speed of the motion is 2.21433 rad/s

4 0

4 years ago

A 5.75 mm high firefly sits on the axis of, and 11.3 cm in front of, the thin lens A, whose focal length is 5.77 cm . Behind len

weeeeeb [17]

Answer

given,

focal length of lens A = 5.77 cm

focal length of lens B= 27.9 cm

flies distance from mirror = 11.3 m

now,

Using lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q}$

$\dfrac{1}{5.77} = \dfrac{1}{11.3} + \dfrac{1}{q}$

q =11.79 cm

image of lens A is object of lens B

distance of lens = 59.9 - 11.79 = 48.11

now, Again applying lens formula

$\dfrac{1}{f} = \dfrac{1}{p} + \dfrac{1}{q'}$

$\dfrac{1}{27.9} = \dfrac{1}{48.11} + \dfrac{1}{q'}$

q' =66.41 cm

hence, the image distance from the second lens is equal to q' =66.41 cm

6 0

3 years ago

When you go camping, you burn wood. Are you contributing to air pollution?

Jet001 [13]

Yes because of the smoke you are creating in the air

3 0

3 years ago

You have two square metal plates with side lengths of (6.50 C) cm. You want to make a parallel-plate capacitor that will hold a

gtnhenbr [62]

Answer:

The necessary separation between the two parallel plates is 0.104 mm

Explanation:

Given;

length of each side of the square plate, L = 6.5 cm = 0.065 m

charge on each plate, Q = 12.5 nC

potential difference across the plates, V = 34.8 V

Potential difference across parallel plates is given as;

$V = \frac{Qd}{L^2 \epsilon_o} \\\\d = \frac{V L^2 \epsilon_o}{Q}$

Where;

d is the separation or distance between the two parallel plates;

$d = \frac{VL^2 \epsilon_o}{Q} \\\\d = \frac{34.8*(0.065)^2 *8.854*10^{-12}}{12.5*10^{-9}} \\\\d = 0.000104 \ m\\\\d = 0.104 \ mm$

Therefore, the necessary separation between the two parallel plates is 0.104 mm

6 0

4 years ago