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Eddi Din [679]
3 years ago
10

If you move for 4 seconds, and travel a distance of 360 meters, how fast were you going?

Physics
2 answers:
Mariulka [41]3 years ago
4 0

Answer:

90 m/s

Explanation:

Time taken = 4 sec

Distance travelled = 360 m

Speed = Distance / Time = 360/4 = 90 m/s

lord [1]3 years ago
3 0
<h3><u>Answer</u> :</h3>

Distance travelled = 360m

Time taken = 4s

We have to find speed.

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

◈ Speed is defined as the ratio of distance travelled to the timr taken.

SI unit : <u>m/s</u>

➝ Speed = Distance / Time

➝ Speed = 360/4

➝ Speed = 90m/s

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Answer:

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Explanation:

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now we will have

EMF = Acos30\frac{dB}{dt}

now we have

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5 0
3 years ago
A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ=7.20 g/m\mu=7.20~\t
vladimir2022 [97]

Answer:

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

\mu =7.20 g/m = 0.0072 kg/m is the mass linear density

Solving the equation,

f=\frac{1}{2(2.00 m)}\sqrt{\frac{160.00 N}{0.0072 kg/m}}=37.3 Hz

The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.

The n=1 mode (fundamental frequency) of an open-open tube is given by

f=\frac{v}{2L}

where

v = 343 m/s is the speed of sound

Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:

L=\frac{v}{2f}=\frac{343 m/s}{2(37.3 Hz)}=4.6 m

4 0
4 years ago
Chuyển động thẳng đều là gì
suter [353]

Question:

What is uniform rectilinear motion?

Answer:

Uniform rectilinear motion is when an object travels at a constant speed with zero acceleration.

8 0
3 years ago
A test charge is introduced into the electric field of a charge. It feels a force of 2F where the electric field is 2E. What wou
dezoksy [38]

Answer:8F

Explanation:

8 0
4 years ago
Help with 5 and the question below. BRAINLIEST FOR THE CORRECT ANSWER! Urgent Physics help!
krok68 [10]

Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

L = 10 Log\frac{I}{I_0}

20 = 10 Log{I}{10^{-12}}

I_1 = 10^{-10}W/m^2

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

L = 10 Log\frac{I}{I_0}

50 = 10Log\frac{I}{10^{-12}}

I_2 = 10^{-7}W/m^2

now we know that

\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}

\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}

r_2 = 47 cm

so now the distance from friend must be 47 cm

8 0
3 years ago
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