Your potential energy at the top of the hill was (mass) x (gravity) x (height) .
Your kinetic energy at the bottom of the hill is (1/2) x (mass) x (speed)² .
If there was no loss of energy on the way down, then your kinetic energy
at the bottom will be equal to your potential energy at the top.
(1/2) x (mass) x (speed)² = (mass) x (gravity) x (height)
Divide each side by 'mass' :
(1/2) x (speed)² = (gravity) x (height) . . . The answer we get
will be the same for every skater, fat or skinny, heavy or light.
The skater's mass doesn't appear in the equation any more.
Multiply each side by 2 :
(speed)² = 2 x (gravity) x (height)
Take the square root of each side:
<u>Speed at the bottom = square root of(2 x gravity x height of the hill)</u>
We could go one step further, since we know the acceleration of gravity on Earth:
Speed at the bottom = 4.43 x square root of (height of the hill)
This is interesting, because it says that a hill twice as high won't give you
twice the speed at the bottom. The final speed is only proportional to the
<em>square root </em>of the height, so in order to double your speed, you need to
find a hill that's <em>4 times</em> as high.
Answer:
92704.5 J
596.44737 N
Explanation:
m = Mass of person + bicycle = 75 kg
g = Acceleration due to gravity = 9.81 m/s²
h = Vertical height = 126 m
= Angle = 7.7°
d = Diameter = 0.388 m
Work done against gravity is given by

Work done is 92704.5 J
Force required is given by

The force is 596.44737 N
Answer: 1.95s
Explanation:
Given
ma = 290 cos 34.9 - fk
fk = 290 cos 34.9 - ma
fn = mg + 400 sin Φ
fn = 290 + 400 sin 34.9
fn = 290 + 228.9
fn = 518.9
fk = fn * uk
uk = 0.57
290 cos 34.9 - ma = 518.9 * 0.57
290 cos 34.9 - ma = 295.8
290 cos 34.9 - 295.8 = ma
ma = -58
m = 290/10 = 29
a = 58/29
a = 2
Using equation of motion
S = ut + .5at²
3.8 = 0 + .5*2*t²
3.8 = t²
t = 1.95s
Answer:
C₂ = 2.22 KJ/kg °C
Explanation:
Since, both objects are in thermal contact. Therefore, the law of conservation of energy tells us that:

where,
m₁ = mass of ice = 1 kg
C₁ = specific heat of ice = 2.04 KJ/kg.°C
ΔT₁ = Change in Temperature of Ice = -8.88°C - (-24°C) = 15.12°C
m₂ = mass of metal block = 1 kg
C₂ = specific heat of metal = ?
ΔT₂ = Change in Temperature of Metal Block = 5°C - (-8.88°C) = 13.88°C
Therefore, using these values in the equation, we get:

<u>C₂ = 2.22 KJ/kg °C</u>