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2 years ago

Higher frequency electromagnetic waves have ?

Physics

1 answer:

Sergeeva-Olga [200]2 years ago

3 0

High frequency for the range of radio frequency electronmagnetic waves between 3 and 30 megahertz (MHz)

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A small ball of charge Q and mass m has a velocity v at infinity. It collides head-on with a ball of the same charge and mass wh

Hatshy [7]

Answer:

Explanation:

Kinetic energy of ball in motion = 1/2 m v² . Potential energy = 0

Let the minimum distance between the balls be d on collision.

Electric potential energy at that time= k Q²/d , Here kinetic energy is converted into potential energy . So

1/2 m v² = kQ²/d

d =2 k Q² / mv²,= 18 x 10⁹ x Q²/ m v².

8 0

3 years ago

Which formula can be used to solve problems related to the first law of thermodynamics?

docker41 [41]

Answer: The formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

Explanation:

First law of thermodynamics states that the total energy of the system remains conserved. Energy can neither be destroyed, nor be created but it can only be transformed into one form to another.

Its implication is any change in the internal energy will be either due to heat energy or work energy.

Mathematically,

$\Delta U=Q+W$

where, Q = heat energy

W = work energy

$\Delta U$ = Change in internal energy

Sign convention for these energies:

For Q: Heat absorbed will be positive and heat released will be negative.

For W: Work done by the system is negative and work done on the system is positive.

For $\Delta U$ : When negative, internal energy is decreasing and when positive, internal energy is increasing.

Hence, the formula used to solve the problems related to first law of thermodynamics is $\Delta U=Q+W$

8 0

3 years ago

Read 2 more answers

Which of the following has to happen for Earth's shadow to fall on the moon?

Ronch [10]

The correct answer is a, lunar eclipse

6 0

2 years ago

What<br> must always be<br> included on the<br> graph

Nookie1986 [14]

Clearly visible data points and appropriate labels on each access that include units

4 0

2 years ago

The Earth’s radius is 6378.1 kilometers. If you were standing at the equator, you are essentially undergoing uniform circular mo

emmainna [20.7K]

Answer:

b. 84 minutes

Explanation:

$a_c=g$ = Centripetal acceleration = 9.81 m/s²

r = Radius of Earth = 6378.1 km

v = Velocity

Centripetal acceleration is given by

$a_c=\dfrac{v^2}{r}\\\Rightarrow v=\sqrt{a_cr}\\\Rightarrow v=\sqrt{9.81\times 6378100}\\\Rightarrow v=7910.06706\ m/s$

Time period is given by

$T=\dfrac{2\pi r}{v60}\\\Rightarrow T=\dfrac{2\pi 6378.1\times 10^3}{7910.06706\times 60}\\\Rightarrow T=84.43835\ minutes$

The time period of Earth’s rotation would be 84.43835 minutes

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2 years ago