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2 years ago

An electromagnetic wave is a kind of a wave which,

Physics

2 answers:

Harman [31]2 years ago

7 0

Explanation:

Electromagnetic waves are waves that propagate through space while an electric field and a magnetic field interact with each other.

please mark as brainliest plzz

Anton [14]2 years ago

3 0

Travels in presence of medium or even in a vacuum

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An alternating source drives a series RLC circuit with an emf amplitude of 6.04 V, at a phase angle of +30.3°. When the potentia

Vinvika [58]

Answer:

-8.56V

Explanation:

Our values are given by,

e = 6.04 V

Φ = 30.3

VC = 5.32

We can calculate the voltage across the circuit with the emf formula, that is,

$e(t) = e* sin(wt)$

$e(t) = 6.04 * sin(Φ + π)$

$e(t) = 6.04 * sin(32.5 + 180)$

$e(t) = -3.245 V$

Now, Using Kirchoff Voltage Law,

$e(t) - VR- VL - VC = 0$

$-3.24 - 0 - VL - 5.32 = 0$

Finally we have the potential difference across the inductor.

$VL = - 8.56 v$

5 0

2 years ago

A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas

Nookie1986 [14]

The maximum amount of work performed is

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Explanation:

The efficiency of a real heat engine is given by the equation:

$\eta = 1-\frac{T_C}{T_H}$ (1)

where

$T_C$ is the temperature of the cold reservoir

$T_H$ is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

$\eta = \frac{W_{max}}{Q_H}$

where

$W_{max}$ is the maximum work done

$Q_H$ is the heat absorbed from the hot reservoir

$Q_H$ can be written as

$Q_H=W_{max}+Q_C$

where

$Q_C$ is the heat released to the cold reservoir

So the previous equation can be also written as

$\eta=\frac{W_{max}}{W_{max}+Q_C}$ (2)

By combining eq.(1) and (2) we get

$1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}$

And re-arranging the equation and solving for $W_{max}$ , we find

$W_{max}=\frac{T_H-T_C}{T_C}Q_C$

Learn more about work and heat:

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brainly.com/question/3063912

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8 0

3 years ago

A cricket can travel approximately 8 m/s. How many meters could a cricket travel in 75<br> s?

nika2105 [10]

Answer:

Your answer will be 600meters

7 0

2 years ago

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

irinina [24]

288.51 N is the magnitude of the force that the beam exerts on the hi.nge.

Given

Mass 0f beam = 40 Kg

The horizontal component of the force exerted by the hi_nge on the beam is 86.62 N

Angle between the beam and cable is = 90°

Angle between beam and the horizontal component = 31°

As the system of the beam, hi_nge and cable are in equilibrium.

The magnitude of the force that the beam exerts on the hi_nge can be calculated by -

F =The horizontal component of force + the vertical component of force

F = 86.62 N + 40 × 9.8 × sin 31°

F =86.62 N + 201.89 N

F = 288.51 N

Hence, the magnitude of the force that the beam exerts on the hi_nge is 288.51 N.

Learn more about components of forces here brainly.com/question/26446720

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2 years ago

Jane, looking for tarzan, is running at top speed (6.8 m/s) and grabs a vine hanging vertically from a tall tree in the jungle.

erica [24]

Jane's mechanical energy at any time is
$E=U+K$
where $U=mgh$ is the potential energy, while $K= \frac{1}{2} mv^2$ is the kinetic energy.

Initially, Jane is on the ground, so the altitude is h=0 and the potential energy is zero: U=0. She's running with speed v, so she has kinetic energy only:
$E=K= \frac{1}{2} mv^2$
Then she grabs the vine, and when she reaches the maximum height h, her speed is zero: v=0, and so the kinetic energy becomes zero: K=0. So now her mechanical energy is just potential energy:
$E=U=mgh$

But E must be conserved, so the initial kinetic energy must be equal to the final potential energy:
$\frac{1}{2}mv^2=mgh$
from which we can find h, the maximum height Jane can reach:
$h= \frac{v^2}{2g}= \frac{(6.8 m/s)^2}{2\cdot 9.81 m/s^2}=2.36 m$

6 0

3 years ago