Answer:
Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.
Explanation:
Energy conservation law: In isolated system the amount of total energy remains constant.
The types of energy are
- Kinetic energy.
- Potential energy.
Kinetic energy 
Potential energy =
Here, q₁= +5.00×10⁻⁴C
q₂=-3.00×10⁻⁴C
d= distance = 4.00 m
V = velocity = 800 m/s
Total energy(E) =Kinetic energy+Potential energy
+
=(1280-337.5)J
=942.5 J
Total energy of a system remains constant.
Therefore,
E +
m/s
Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.
Answer:
181.54 K
Explanation:
From gas laws, we know that v1/t1= v2/t2 where v and t represent volume and temperatures, 1 and 2 for the first and second container. Making t2 the subject of the formula then
T2=v2t1/ v1
Given information
V1 435 ml
V2 265 ml
T1 298K
Substituting the given values then
T2=265*298/435=181.54 K
Answer:
2. You must be able to precisely measure variations in the star's brightness with time.
5. As seen from Earth, the planet's orbit must be seen nearly edge–on (in the plane of our line-of-sight).
6. You must repeatedly obtain spectra of the star that the planet orbits.
Explanation:
The transit method is a very important and effective tool for discovering new exoplanets (the planets orbiting other stars out of the solar system). In this method the stars are observed for a long duration. When the exoplanet will cross in front of theses stars as seen from Earth, the brightness of the star will dip. To observe this dip following conditions must be met:
1. The orbit of the planet should be co-planar with the plane of our line of sight. Then only its transition can be observed.
2. The brightness of the star must be observed precisely as the period of transit can be less than a second as seen from Earth. Also the dip in brightness depends on the size of the planet. If the planet is not that big the intensity dip will be very less.
3. The spectrum of the star needs to be studied and observe during the transit and normally to find out the details about the planets.
4. Also, the orbital period should be less than the period of observation for the transit to occur at least once.
What that means is the atom is so radioactive that the nucleus is unstable.
Answer:
–735.17 N
The negative sign indicate that the force is acting in opposition direction to the car.
Explanation:
The following data were obtained from the question:
Mass (m) of car = 782.10 kg
Initial velocity (u) = 7.60 m/s
Final velocity (v) = 3.61 m/s
Time (t) = 4.23 s
Force (F) =?
Next, we shall determine the acceleration of the car. This can be obtained as follow:
Initial velocity (u) = 7.60 m/s
Final velocity (v) = 3.61 m/s
Time (t) = 4.23 s
Acceleration (a) =?
a = (v – u) / t
a = (3.61 – 7.60) / 4.23
a = –3.99 / 4.23
a = –0.94 m/s²
Finally, we shall determine the force experienced by the car as shown below:
Mass (m) of car = 782.10 kg
Acceleration (a) = –0.94 m/s²
Force (F) =?
F = ma
F = 782.10 × –0.94
F = –735.17 N
The negative sign indicate that the force is acting in opposition direction to the car.
