The three longest wavelengths for the standing waves on a 264-cm long string that is fixed at both ends are:
- 5.2 meters.
- 2.6 meters.
- 1.7meters.
Given data:
Length of the fixed string = 264cms = 2.64 meters
The wavelength for standing waves is given by:
λ = 2L/n
where,
- λ is the wavelength
- L is the length of the string
For n = 1,
= 5.2 meters
For n = 2,
= 2.6 meters
For n = 3,
= 1.7 meters
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Answer:
The energy absorbed by the atomic electrons in the mercury atom is
J
Explanation:
Given:
Potential
V
According to the conservation law,
Loss in kinetic energy = Gain in potential energy
Here, energy absorbed by the atomic electrons is given by,

Where
( charge of electron )

J
Therefore, the energy absorbed by the atomic electrons in the mercury atom is
J
The centripetal force (of gravity) on a satellite in orbit is an
unbalanced force (because there's no equal force pulling
the satellite away from Earth), changes the direction of the
satellite (into a closed orbit instead of a straight line), and
always acts toward the center of whatever curve the satellite
happens to be on at the moment.
According to x-ray observations, the space between galaxies in a galaxy cluster is very hot. It is because the matter between galaxies (often called the intergalactic medium) is mostly hot, ionized hydrogen with bits of heavier elements such as carbon, oxygen and silicon thrown in.
Massive structures are collapsing than at earlier times. Large collapsing structures lead to higher velocity intergalactic shocks and, as a result, significant intergalactic shock heating, with some gas heated well above the
K temperatures.
Heating also occurs as galaxies expel out most of the gas that fell into them. The final product is a warm/hot phase, with temperatures of >
K.
Now, Let's know how do you use X-rays to make space observations?
X-radiation is absorbed by the Earth's atmosphere, so instruments to detect X-rays must be taken to high altitude by balloons, sounding rockets, and satellites.
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Total distance = 36500 m
The average velocity = 19.73 m/s
<h3>Further explanation</h3>
Given
vo=initial velocity=0(from rest)
a=acceleration= 1 m/s²
t₁ = 20 s
t₂ = 0.5 hr = 1800 s
t₃= 30 s
Required
Total distance
Solution
State 1 : acceleration


State 2 : constant speed

State 3 : deceleration


Total distance : state 1+ state 2+state 3

the average velocity = total distance : total time
