Answer:
The lose of thermal energy is, Q = 22500 J
Explanation:
Given data,
The mass of aluminium block, m = 1.0 kg
The initial temperature of block, T = 50° C
The final temperature of the block, T' = 25° C
The change in temperature, ΔT = 50° C - 25° C
= 25° C
The specific heat capacity of aluminium, c = 900 J/kg°C
The formula for thermal energy,
<em>Q = mcΔT</em>
= 1.0 x 900 x 25
= 22500 J
Hence, the lose of thermal energy is, Q = 22500 J
I could use math to help with baking and measuring how much I should use for a cake or stuff like that, they also show us how to add our money together to help us with stuff.
Math, is helpful with many jobs, stuff learned in school can help you strategize and school also makes you think outside of the box sometimes, which can help with jobs. Also if you want to do a career with fashion designing or engineering, you can use math for that too. School also gives you the ability to read.
Answer:
0.2448 point²
Explanation:
1 gry = 1/10 line
1 line = 1/12 inch
=> 1 gry in inches = 1/10 * 1/12 = 1/120 inch
=> 1 inch = 120 gry
1 point = 1/72 inch
=> 1 inch = 72 points
Therefore,
120 gry = 72 points
=> 1 gry = 3/5 point
Therefore,
1 gry² = (3/5)² point²
1 gry² = 9/25 point²
This means that 0.68 gry² will be:
0.68 gry² = 0.68 * 9/25 point²
=> 0.68 gry² = 0.2448 point²
The second object, the one that had twice the force applied to it, would move twice as far, I believe.
Answer:
Explanation:
Sam mass=75kg
Height is 50m
20° frictionless slope
Horizontal force on Sam is 200N
According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.
Therefore
Wg - Ww =∆K.E
Note initial the body was at rest at top of the slope.
Then, ∆K.E is K.E(final) - K.E(initial)
K.E Is given as ½mv²
Since initial velocity is zero then, K.E(initial ) is zero
Therefore, ∆K.E=½mVf²
Wg is work done by gravity and it is given by using P.E formulas
Wg=mgh
Wg=75×9.8×50
Wg=36750J
Ww is work done by wind and it's is given by using formulae for work
Work=force × distance
Ww=horizontal force × horizontal distance
Using Trig.
TanX=opposite/adjacent
Tan20=h/x
x=h/tan20
x=50/tan20
x=137.37m
Then,
Ww=F×x
Ww=200×137.37
We=27474J
Now applying the formula
Wg - Ww =∆K.E
36750 - 27474 =½×75×Vf²
9276=37.5Vf²
Vf²=9275/37.5
Vf²= 247.36
Vf=√247.36
Vf=15.73m/s
