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3 years ago

Which type of electromagnetic radiation cannot be focused?

Physics

2 answers:

Irina18 [472]3 years ago

7 0

The answer is A. Hope this helps. :)

dalvyx [7]3 years ago

5 0

(A) Gamma rays

Focussing is a process in which a beam of light is passed and concentrated on the particular point.

Gamma rays are the type of elecromagnetic radiation that cannot of focused. Gamma rays has high frequency and also are quite energetic due to which when the beam of light is passed through it, it becomes too difficult to focus on a particular point as they interacts strongly with the matter and destroys itself. Hence, Gamma rays are not easily focused.

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On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed

JulsSmile [24]

Answer: 1.25 m

Explanation:

Given

initial velocity $(v_i )=3 m/s$

Final velocity $(v_f)=0.55\times 3=1.65 m/s$

Change in kinetic Energy =work done by Friction

change in Kinetic Energy $=\frac{m}{2}\left ( v_i^2-v_f^2\right )$

work done by friction $=\mu mgL$

$\frac{m}{2}\left ( 3^2-1.65^2\right )=0.25\cdot mg\times L$

$3.135=0.25\times 9.8\times L$

L=1.25

7 0

3 years ago

Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite

-Dominant- [34]

Answer:

The required work done is $6.5\times10^{9}\ J$

Explanation:

Given that,

Mass of each satellites = 940 kg

Altitude of A = 4500 km

Altitude of B = 11100 km

We need to calculate the potential energy

Using formula of potential

$U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}$

Put the value into the formula

$U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}$

$U_{A}=-3.44\times10^{10}\ J$

We need to calculate the potential energy

Using formula of potential

$U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}$

Put the value into the formula

$U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}$

$U_{B}=-2.14\times10^{10}\ J$

We need to calculate the value of $k_{A}$

Using formula of $k_{A}$

$k_{A}=-\dfrac{1}{2}U_{A}$

Put the value into the formula

$k_{A}=\dfrac{1}{2}\times3.44\times10^{10}$

$k_{A}=1.72\times10^{10}\ J$

We need to calculate the value of $k_{B}$

Using formula of $k_{B}$

$k_{B}=-\dfrac{1}{2}U_{B}$

Put the value into the formula

$k_{B}=\dfrac{1}{2}\times2.14\times10^{10}$

$k_{B}=1.07\times10^{10}\ J$

We need to calculate the work done

Using formula of work done

$W=\Delta K+\Delta U$

$W=(k_{B}-k_{A})+(U_{B}-U_{A})$

$W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})$

$W=\dfrac{1}{2}(U_{B}-U_{A})$

Put the value into the formula

$W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})$

$W=6.5\times10^{9}\ J$

Hence, The required work done is $6.5\times10^{9}\ J$

7 0

3 years ago

A motorcycle of 170 kg is moving with a velocity 25m/ . Calculate the kinetic energy

topjm [15]

Answer: 53125joules

Explanation: The formula for kinetic energy is 1/2mv^2.

So we would have

1/2[(170)(25x25)]=53125

3 0

3 years ago

Why does Mila’s phone start playing music again when she plugged her phone into the battery pack ?

denis-greek [22]

Answer:

cause it got charged

Explanation:

3 0

3 years ago

a resistor, of resistance 47 , is damaged when it dissipates a power greater than 0.5W. determine the maximum voltage that can b

harkovskaia [24]

Answer:

<em>The maximum voltage that can be applied without damaging the resistor is 4.85 V</em>

Explanation:

<u>Electric Power in a Resistor</u>

Given a resistor or resistance R connected to a circuit of voltage V carrying a current I. The relation between these three magnitudes is given by Ohm's Law:

V = R.I

The dissipated power P of a resistor can be calculated by the following equation, known as Joule's first law:

$P = I^2.R$