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2 years ago

How to find average speed in physics

1 answer:

7 0

Answer: you divide total distance by time. To get the time, divide total distance by speed. To get distance, multiply speed times the amount of time.

Explanation:

I hope this helps

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When she rides her bike, she gets to her first classroom building 36 minutes faster than when she walks. Of her average walking

postnew [5]

Answer:

$d=2.4\ miles$

Explanation:

Given:

average walking speed, $v_w=3\ mph$

average biking speed, $v_b=12\ mph$

<u>According to given condition:</u>

$t_w=t_b+\frac{36}{60}$

where:

$t_w=$ time taken to reach the building by walking

$t_b=$ time taken to reach the building by biking

We know that,

$\rm time=\frac{distance}{speed}$

so,

$\frac{d}{v_w} =\frac{d}{v_b} +\frac{36}{60}$

$\frac{d}{3}=\frac{d}{12} +\frac{3}{5}$

$d=2.4\ miles$

7 0

3 years ago

Read 2 more answers

Question in pic.. thanks!

11Alexandr11 [23.1K]

That's false. Displacement would be (r2 - r1) .

3 0

3 years ago

A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b

nekit [7.7K]

<h2>Answer:</h2>

In circuits, the average power is defined as the average of the instantaneous power over one period. The instantaneous power can be found as:

$p(t)=v(t)i(t)$

So the average power is:

$P=\frac{1}{T}\intop_{0}^{T}p(t)dt$

But:

$v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)$

So:

$P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}$

$P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}(\frac{1+cos(2\omega t)}{2} )dt \\\\P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}[\frac{1}{2}+\frac{cos(2\omega t)}{2}]dt \\\\P=\frac{v_{m}i_{m}}{T}[\frac{1}{2}(t)\right|_0^T +\frac{sin(2\omega t)}{4\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2T}[(t)\right|_0^T +\frac{sin(2\omega t)}{2\omega} \right|_0^T] \\ \\ P=\frac{v_{m}i_{m}}{2}$

In terms of RMS values:

$V_{RMS}=V=\frac{v_{m}}{\sqrt{2}} \\ \\ I_{RMS}=I=\frac{i_{m}}{\sqrt{2}} \\ \\ Then: \\ \\ P=VI$

7 0

3 years ago

the distance between two successive troughs of wave is 0.4m. If the frequency of the source is 825Hz, calculate the speed of the

neonofarm [45]

Answer:

speed=330m/s

Explanation:

the speed of wave is given as

speed(meter per second) =frequency(hertz) * wavelength(meters)

so using the above formula we substitute the figures given in the question in the formula we get

speed = 0.4*825

speed =330m/s

note m/s is the si unit for speed which is read as meter per second

therefore speed =330m/s

7 0

3 years ago

A circular loop in the plane of a paper lies in a 0.45 T magnetic field pointing into the paper. The loop's diameter changes fro

Amanda [17]

Answer:

(A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

Given that,

Magnetic field = 0.45 T

The loop's diameter changes from 17.0 cm to 6.0 cm .

Time = 0.53 sec

(A). We need to find the direction of the induced current.

Using Lenz law

If the direction of magnetic field shows into the paper then the direction of the induced current will be clockwise.

(B). We need to calculate the magnetic flux

Using formula of flux

$\phi_{1}=BA\cos\theta$

Put the value into the formula

$\phi_{1}=0.45\times(\pi\times(8.5\times10^{-2})^2)\cos0$

$\phi_{1}=0.01021\ Wb$

We need to calculate the magnetic flux

Using formula of flux

$\phi_{2}=BA\cos\theta$

Put the value into the formula

$\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0$

$\phi_{2}=0.00127\ Wb$

We need to calculate the magnitude of the average induced emf

Using formula of emf

$\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})$

Put the value into t5he formula

$\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})$

$\epsilon=0.016867\ V$

$\epsilon=16.87\ mV$

(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

$I=\dfrac{\epsilon}{R}$

Put the value into the formula

$I=\dfrac{0.016867}{2.5}$

$I=0.00675\ A$

$I=6.75\ mA$

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

5 0

3 years ago