Answer:
0.686 g of ice melts each second.
Solution:
As per the question:
Cross-sectional Area of the Copper Rod, A = 
Length of the rod, L = 19.6 cm = 0.196 m
Thermal conductivity of Copper, K = 
Conduction of heat from the rod per second is given by:

where
= temperature difference between the two ends of the rod.
Thus

Now,
To calculate the mass, M of the ice melted per sec:

where
= Latent heat of fusion of water = 333 kJ/kg

Acceleration is a change in *speed* over time. In this case, the speed of the car increased by 90 km/hr in 6 s, giving it a rate of 90 km/hr/6s, or 15 km/hr/s. We’re asked for the acceleration in m/s^2, though, so we’ll need to do a few conversions to get our units straight.
There are 1000 m in 1 km, 60 min, or 60 * 60 = 3600 s in 1 hr, so we can change our rate to:
(15 x 1000)m/3600s/s, or (15 x 1000)m/3600 s^2
We can reduce this to:
(15 x 10)m/36 s^2 = 150 m/36 s^2
Which, dividing numerator and denominator by 36, gets us a final answer of roughly 4.17 m/s^2
Answer:
The work is -67.76 J
Explanation:
The law of conservation of energy is considered one of one of the fundamental laws of physics and states that the total energy of an isolated system remains constant. except when it is transformed into other types of energy.
This is summed up in the principle that energy can neither be created nor destroyed in the universe, only transformed into other forms of energy.
In this case you must calculate the loss of kinetic energy. This loss is actually the work done against the resistive force in the air. Friction is the only force other than gravity that acts on the ball.
So, the loss of kinetic energy is 
You know:
- mass=m=0.22 kg
- Initial velocity of the ball:

Final velocity of the ball: 
Replacing:
= -67.76 J
Friction work is always negative because friction is always against displacement.
<u><em>The work is -67.76 J</em></u>
speed = 86km/hr
= 86000/3600
= 23.889m/s
time = 4.6s
acceleration = 23.889/4.6
a = 5.19m/s^2
The outside magnetic field of a solenoid is similar to that of a bar magnet even though the inside field is strong and uniform.