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Setler [38]
3 years ago
8

How can the motion of an object that is NOT moving change?​

Physics
1 answer:
Semenov [28]3 years ago
6 0
Only if a force acts upon it, it can move.
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When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelera
adell [148]

Answer:

4.9 eV .

Explanation:

It is the case of discharge through mercury tube light . In it , mercury  atoms are exited due to which electrons are sent to higher energy  level . Here current drops to zero because electrons are excited to higher level . Energy are absorbed in quantised manner . Energy absorbed by electrons will be 4.9 eV . That means , difference in energy  between two energy level is  4.9 eV .

3 0
3 years ago
How does work affect energy between objects so it can cause a change in the form of energy?
mihalych1998 [28]

Answer:

Doing work' is a way of transferring energy from one object to another, energy is transferred when a force moves through a distance.

Explanation:  So more energy, more work done bc u transferred more energy to move the object and doing the work. and if you only use a little of energy, the work done also only a little.

5 0
3 years ago
Read 2 more answers
What is the total momentum of a 30 kg object traveling left at 3 m/s and a 50 kg object traveling at 2 m/s to the right?
madam [21]

Answer:

there are 25 kg objective travelling at 2m/s to the right.

4 0
2 years ago
What is the lowest possible temperature
HACTEHA [7]
The lowest possible temperature is absolute zero. However scientists have not reached this temperature, rather they have come very close to absolute zero.

6 0
3 years ago
Read 2 more answers
A rectangular copper strip 1.5cm wide and 0.10cn thick carries a current of 5.0A. Find the Hall voltage for a 1.2T magnetic fiel
Scrat [10]

Answer:

4.4345× 10^-7V

Explanation:

The computation of the half voltage for a 1.2T magnetic field applied is shown below

The volume of one mole of copper is

v = m ÷p

= 63.5 ÷ 8.92

= 7.12cm

Now the density of free electrons in copper is

n = Na ÷ V

= 6.02 × 10^23 ÷ 7.12

= 8.456× 10^28/m^3

Now the half voltage is

= IB ÷ nqt

= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)

= 4.4345× 10^-7V

7 0
3 years ago
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