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otez555 [7]
3 years ago
8

Water waves in a small tank are .06 m long. They pass a given point at a rate of 14.8 waves every three seconds. What is the spe

ed of the water waves? What is the period of these water waves?
Physics
1 answer:
snow_lady [41]3 years ago
4 0

Answer:

Speed = 0.296m/2

Period = 0.203 s

Explanation:

If by 'long' you mean the wavelength of the waves, then the wavelength \lambda=0.06m.

The frequency f of the waves is 14.8 waves every 3 seconds or

f=14.8/3 =4.33Hz.

Now the relationship between wavelength \lambda, frequency f and speed v of the waves is:

v=\lambda f

We put in the values \lambda=0.06m and f=4.933Hz and get:

\boxed{v=0.06*4.922=0.296m/s}

Now the period T is just the inverse of the frequency, or

T=\frac{1}{f}

\boxed{T=\frac{1}{4.933}=0.203\:seconds }

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A 650 × 10–4 F capacitor stores 24 × 10–3 of charge.
alexgriva [62]

C. 0.37V. A capacitor of 650x10⁻⁴F that stores 24x10⁻³C has a potential difference of 0.37V between its plates.

The key to solve this problem is using the capacitance equation C = Q/Vᵃᵇ, where C is the capacitance, Q the charge stored in the plates, and Vᵃᵇ the potential difference between the plates.

A 650x10⁻⁴F capacitor stores 24x10⁻³C, clear Vᵃᵇ for the equation:

C = Q/Vᵃᵇ -----------> Vᵃᵇ = Q/C

Solving

Vᵃᵇ = 24x10⁻³C/650x10⁻⁴F = 0.37V

3 0
3 years ago
An astronaut on a small planet wishes to measure the local value of g by timing pulses traveling down a wire which has a large o
Alekssandra [29.7K]

Answer:

0.53m/s^2

Explanation:

We are given that

Mass of wire=m=3.6 g=3.6\times 10^{-3} kg

1 kg=1000 g

Length of wire=l=1.6 m

Mass of object=m'=3 kg

Time,t=60.1 ms=60.1\times 10^{-3} s

1 ms=10^{-3} s

Speed,v=\frac{distance}{time}=\frac{1.6}{60.1\times 10^{-3}}=26.62 m/s

g=\frac{v^2m}{m'l}

Using the formula

g=\frac{(26.62)^2\times 3.6\times 10^{-3}}{3\times 1.6}=0.53m/s^2

8 0
3 years ago
A jogger runs at a constant rate of 10.0 m every 2.0 seconds. The jogger starts at the origin and runs in the positive direction
Elis [28]

Answer:

(a) 25 m

(b) 75 m

Explanation:

Given that the jogger runs at a constant rate of 10.0 m every 2.0 seconds.

So, the speed of the jogger,

v=\frac{10}{2}=5m/s\;\cdots(i)

Let d be the distance covered by him in time, t s.

As distance=(speed) x (time)

So, d=vt

From equation (i)

\Rightarrow d=5t\;\cdots(ii)

As the jogger starts from origin, so, the distance, d, also represents the position of the jogger at the time t s.

The position-time graph has been shown.

(a) From equation (ii), for t=5.0 s

d=5\times 5=25 m

So, the jogger is at a distance of 25 m from the origin.

(b) Similarly, for t=15.0 s

d=5\times 15=75 m

So, the jogger is at a distance of 75 m from the origin.

8 0
2 years ago
in a lever, a load of 600N is lifted by using 400 effort. If the load is at the distance of 20cm and the effort at the distance
koban [17]

Explanation:

Load (l) = 680N

Effort (E) = 500N

Length slope (l) = 12m

Height slope (h) = 8 m

Output = load * height

680 *8 = 5.44 *103 J

The Input = effort * length = 500 *12 = 6000J

the Mechanical advantage (M.A) = load effort= 600500=1.36

the Velocity ratio (V.R) =lh=128 = 1.5

the Efficiency =M.A100%V.R= 90.6%

8 0
2 years ago
PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)
Cerrena [4.2K]

Answer:

B false

Explanation:

dont have any explanation

8 0
2 years ago
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