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Otrada [13]
3 years ago
9

What has highest frequency in electromagnetic spectrum color?

Physics
1 answer:
natulia [17]3 years ago
6 0
In Electromagnetic spectrum, X-rays has very high frequency but gamma-rays has more than that. If you consider cosmic rays as an electromagnetic wave, then it would be highest energetic and has highest frequency.

In short, Your Answer is "Cosmic rays"

Hope this helps!
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Is the refraction different entering medium that has a higher index of refraction compared to entering amedium that has a lower
oee [108]

Answer:

Yes

Explanation:

The speed of light when it travels through glass, diamond, etc, the light travels at different speed from the speed of light. Speed of the light in material is related to the index of refraction.

The change in speed which occurs when the light passes from one medium to the another is responsible for bending of the light which is called as refraction.

<u>When the light goes into a medium with the higher index of the refraction, light bends towards normal. Conversely, if the light traveling goes from higher refractive index to lower refractive index, it will bend away from the normal. </u>

<u>Hence, the refraction is different in both the scenario.</u>

6 0
3 years ago
What is your opinion on Moonman? ( Do not delete this saying it is "racist". it is simply a character from a commercial called M
Leni [432]

Answer:

I think hes cool

Explanation:

8 0
3 years ago
What is the kinetic energy of a 1130 kg truck that is moving with a velocity of 40 m/s?
Makovka662 [10]

Answer:

The answer is 904,000.

Kinetic energy=1/2mv^2.

1/2×1130×40^2.

1/2×1808000=904,000Joules.

8 0
3 years ago
Two light bulbs have resistances of 400 Ω and 800 ΩThe two light bulbs are connected in series across a 120- V line. Find the cu
Natasha2012 [34]

1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega

And so, the current through the circuit is (using Ohm's law):

I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A

And since the two bulbs are connected in series, the current through each bulb is the same.

2) 4 W and 8 W

The power dissipated by each bulb is given by the formula:

P=I^2 R

where I is the current and R is the resistance.

For the first bulb:

P_1 = (0.1 A)^2 (400 \Omega)=4 W

For the second bulb:

P_1 = (0.1 A)^2 (800 \Omega)=8 W

3) 12 W

The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:

P_{tot} = P_1 + P_2 = 4 W + 8 W=12 W

3 0
3 years ago
The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho= rho0[1+α(T−T0)] , where
Stolb23 [73]

Answer:

At 81. 52 Deg C its resistance will be 0.31 Ω.

Explanation:

The resistance of wire =R_T =\frac{\rho_T \ l}{A}

Where R_T =Resistance of wire at Temperature T

\rho_T = Resistivity at temperature T =\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]

Where T_0 =20\ Deg\ C , \  \rho_0 = Constant,  \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)

l=Length of the wire

& A = Area of cross section of wire

For long and thin wire the resistance & resistivity relation will be as follows

\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}

\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}

1.24=1+\alpha (T-20)

0.24=\alpha(\ T -20 )

Putting\ the\ value\ of \alpha = 3.9 \times 10^-^3 DegC^-1

T = 81.52 Deg C

4 0
3 years ago
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