Kinetic energy = (1/2) (mass) x (speed)²
At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules
At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules
The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.
That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
Work = force x distance
F= 2.5
D= 3
Work = 2.5 x 3 =7.5
Work = 7.5 J
J=Jules (Jules is the unit uses to calculate work)
A 15.75-g<span> piece of iron absorbs 1086.75 </span>joules<span> of </span>heat<span> energy, and its ... </span>How many joules<span> of </span>heat<span> are </span>needed<span> to raise the temperature of 10.0 </span>g<span> of </span>aluminum<span> from 22°C to 55°C, if the specific </span>heat<span> of </span>aluminum<span> is o.90 J/</span>g<span>”C2 .</span>
Answer:
Explanation:
Even though the minimum temperature is more than the freezing point, Frost was observed on the ground because the ground will cool rapidly as cool air tends to move towards the ground, the temperature of the ground is lower than the atmosphere a few feet above it.
As the thermometer is kept some feet above the ground so ground temperature may be lower than the minimum recorded temperature.
