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denis-greek [22]
3 years ago
11

Electromagnetic waves have the ability to travel through empty space because they do not require a medium to travel through.

Physics
2 answers:
matrenka [14]3 years ago
7 0

True. Electromagnetic energy does not require a medium.

uranmaximum [27]3 years ago
5 0
True is the answer I believe
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As light travels through a medium, it moves in a straight line path,but when light passes from one medium into another medium, the light path bends,Refraction takes place. 
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Does the charge that flows into the capacitor during the charging go all the way through the capacitor and back to the battery,
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3 years ago
The flywheel is rotating with an angular velocity ω0 = 2.37 rad/s at time t = 0 when a torque is applied to increase its angular
nika2105 [10]

Answer:

ω = 12.023 rad/s

α = 222.61 rad/s²

Explanation:

We are given;

ω0 = 2.37 rad/s, t = 0 sec

ω =?, t = 0.22 sec

α =?

θ = 57°

From formulas,

Tangential acceleration; a_t = rα

Normal acceleration; a_n = rω²

tan θ = a_t/a_n

Thus; tan θ = rα/rω² = α/ω²

tan θ = α/ω²

α = ω²tan θ

Now, α = dω/dt

So; dω/dt = ω²tan θ

Rearranging, we have;

dω/ω² = dt × tan θ

Integrating both sides, we have;

(ω, ω0)∫dω/ω² = (t, 0)∫dt × tan θ

This gives;

-1[(1/ω_o) - (1/ω)] = t(tan θ)

Thus;

ω = ω_o/(1 - (ω_o × t × tan θ))

While;

α = dω/dt = ((ω_o)²×tan θ)/(1 - (ω_o × t × tan θ))²

Thus, plugging in the relevant values;

ω = 2.37/(1 - (2.37 × 0.22 × tan 57))

ω = 12.023 rad/s

Also;

α = (2.37² × tan 57)/(1 - (2.37 × 0.22 × tan 57))²

α = 8.64926751525/0.03885408979 = 222.61 rad/s²

6 0
3 years ago
A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.
Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity = 1.5 ra/s

\alpha = Angular acceleration

\theta = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2

Acceleration while slowing down is -0.0047 rad/s²

t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s

Time taken to slow down is 335.103 seconds

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0

Solving the equation

t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

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2 years ago
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