Answer:
Francium has fewer valence electrons, but they are in a higher energy level
Answer:
Explanation:
Magnets are of two major forms namely the permanent magnet and the temporary magnets. Temporary magnets magnetizes and demagnetize easily while permanent magnets does not magnetizes and demagnetize easily.
This permanents magnets are applicable in loudspeakers, generators, induction motor etc.
To increase the
The following will tend to increase the magnetic force acting on the rotor in an induction motor.
1. Increasing the strength of the bar magnet. Increase in strength of the magnet will lead to increase in the magnetic force acting on the rotor.
2. Increase in the magnetic line of force also known as the magnetic flux around the magnet will also increase the magnetic force acting on the rotor.
Answer:
Friction 100 j
Explanation:
Friction causes the inequality between PE and KE
100 J
Answer:
Q = 7272 Kilojoules.
Explanation:
<u>Given the following data;</u>
Mass = 2.0*101kg = 202kg
Initial temperature, T1 = 10°C
Final temperature, T2 = 90°C
We know that the specific heat capacity of iron = 450J/kg°C
*To find the quantity of heat*
Heat capacity is given by the formula;
Where;
- Q represents the heat capacity or quantity of heat.
- m represents the mass of an object.
- c represents the specific heat capacity of water.
- dt represents the change in temperature.
dt = T2 - T1
dt = 90 - 10
dt = 80°C
Substituting the values into the equation, we have;
Q = 7272KJ or 7272000 Joules.
Answer:
Minimum work = 5060 J
Explanation:
Given:
Mass of the bucket (m) = 20.0 kg
Initial speed of the bucket (u) = 0 m/s
Final speed of the bucket (v) = 4.0 m/s
Displacement of the bucket (h) = 25.0 m
Let 'W' be the work done by the worker in lifting the bucket.
So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.
Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

Therefore, the work done by the worker in lifting the bucket is given as:

Now, plug in the values given and solve for 'W'. This gives,

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.