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2 years ago

A jetliner flies at a constant speed covering 467 miles in 3.3 hours. What is the speed of the plane in miles per hour?

Physics

1 answer:

stellarik [79]2 years ago

8 0

Answer:

141.152 miles per hour is the speed of the plane in miles per hour

Explanation:

Speed of plane = Total distance travelled/total time taken -

v = D/t

Substituting the given values in the above equation, we get

v = 467/3.3 miles /hour

v = 141.152 miles per hour

141.152 miles per hour is the speed of the plane in miles per hour

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This is a problem about a child pushing a stack of two blocks along a horizontal floor. The masses of the blocks, and the coeffi

Rufina [12.5K]

Answer:

N = 23.4 N

Explanation:

After reading that long sentence, let's solve the question

The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis

N - w₁ -w₂ =

N = m₁ g + m₂ g

N = g (m₁ + m₂)

let's calculate

N = 9.8 (0.760 + 1.630)

N = 23.4 N

This is the force of the support of the two blocks on the surface.

7 0

2 years ago

Will mark brainliest!

MaRussiya [10]

Answer:

Explanation:

Fa - u*m*g = m*a

Fa = u*m*g + m*a

Fa - m*a = u*m*g

u = $\frac{Fa - m*a}{m*g}$

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3 years ago

What type of waves moves energy forward, but the source moves up and down?

stich3 [128]

I'm pretty sure its transverse waves

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2 years ago

How much does a 59 kg woman weigh on earth?

leonid [27]

9.8........................

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3 years ago

Read 2 more answers

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

2 years ago