Answer:
* Experiment with a higher range of materials
* Use a galvanometer.
* Vary in number of coils of the electromagnet
Explanation:
This is an experiment of electricity and magnetism, in general the best way to improve the results are:
* Experiment with a higher range of materials
allowing to know the scope of the initial assumptions
* Use a galvanometer.
The more accurate the readings the error of the derived quantities is the less which will improve the precision of the experiment.
* Vary in number of coils of the electromagnet
Since it allows to have greater magnetic fields and therefore expand the range of measurements
Answer:
elastic partial width is 2.49 eV
Explanation:
given data
ER E = 250 eV
spin J = 0
cross-section magnitude σ = 1300 barns
peak P = 20ev
to find out
elastic partial width W
solution
we know here that
σ = λ²× W / ( E × π × P ) ...................1
put here all value
σ = (0.286)² × W × 
/ ( 250 × π × 20 )
1300 × 
= (0.286)² × W × / ( 250 × π × 20 )
solve it and we get W
W = 249.56 × 
so elastic partial width is 2.49 eV
Ice melts due to the chemical properties of water. There are more hydrogen bonds between the molecules of ice than in water.
Water boils when the thermal energy in the water, which is a type of kinetic energy which causes the water molecules to move around, exceeds the strength of the hydrogen bonds between the other molecules.
Answer:
The least number of forces required to stretch a spring is one.
Explanation:
Let suppose that spring is ideal, that is, that effects from its mass can be neglected since it is insignificant in comparison with external forces. In addition, let the spring have a linear behavior, meaning that net external longitudinal force exerted on spring is directly proportional to defomation. (Hooke's Law) That is:
(1)
(2)
Where:
- Net external force, measured in newtons.
- Spring constant, measured in newtons per meter.
- Deformation of spring, measured in meters.
Hence, the least number of forces required to stretch a spring is one.
Answer:
a) 37.70 m/s
b)710.6 m/s²
Explanation:
Given that ;
Mass of object = 2 kg
Radius of the motion = 2m
Frequency of motion = 3 rev/s
The formula to apply is;
v= 2πrf where v is linear speed
v = 2×π×2×3 =12π = 37.70 m/s
Centripetal acceleration is given as;
a= 4×π²×r×f²
a= 4×π²×2×3²
a=710.6 m/s²
