The simple machine in a catapult is a lever, although there can alos be a pulley included. Because t<span>he load and effort are on opposite sides of the fulcrum, they move in opposite directions. This is why catapults can launch objects.</span>
Answer: 4.41 joules
Explanation:
Given that:
work is done by gravity = ?
Mass of pine cone (m) = 50g
Convert mass in grams to kilograms
Since 1000g = 1kg
50g = (50/1000) = 0.05kg
Height of tree (h) = 9 m
Apply the formula for work,
Work = mass x acceleration due to gravity x height
Acceleration due to gravity has a value of 9.8m/s^2
Work = mgh
Work = 0.05kg x 9.8ms^2 x 9m
Work = 4.41J
Thus, 4.41 joules of work is done by gravity.
Answer:potential energy=810000j
Kinetic energy=810000j
Mechanical energy=810000j
Explanation:
Mass(m)=5000kg
Velocity(v)=18m/s
Kinetic energy=(mxv^2) ➗ 2
Kinetic energy =(5000x18^2) ➗ 2
Kinetic energy =(5000x18x18) ➗ 2
Kinetic energy =1620000 ➗ 2
Kinetic energy =810000j
Kinetic energy = potential energy
Since energy can neither be created nor destroyed
Answer:
Velocity of skater after throwing the snowball is 2.57 m/s
Explanation:
Given :
Mass of skater, M = 62.2 kg
Mass of snowball, m = 0.145 kg
Velocity of snowball relative to ground, v = 39.3 m/s
Consider v₁ be the velocity of skater after throwing the snowball.
According to the problem, initially the velocity of skater and snowball is same. So,
Velocity of skater before throwing snowball, u = 2.66 m/s
Applying conservation of momentum,
Momentum before throwing snowball = Momentum after throwing snowball
(M + m) u = Mv₁ + mv
Substitute the suitable values in the above equation.
v₁ = 2.57 m/s
Answer:
d = 136.7 ft
Explanation:
Because the truck move with uniformly accelerated movement we apply the following formula:
vf²=v₀²+2*a*d Formula (1)
Where:
d:displacement in meters (ft)
v₀: initial speed in ft/s
vf: final speed in ft/s
a: acceleration in ft/s²
Data
v₀ = 44.0 mi/h
1milla = 5280 ft
1h = 3600 s
v₀ = 44*(5280 ft) / (3600 s) = 64.5 ft/s
vf = 0
d = 47.0 ft
Calculation of the acceleration of the truck
We replace data in the formula (1) :
vf²=v₀²+2*a*d
0 = (64.5)²+2*a*(47)
-(64.5)² = (94)*a
a = -(64.5)² / 94
a = - 44.26 ft/s²
The acceleration (a) it's negative (-) because the truck is braking
Calculation of the minimum stopping distance of the truck to v₀ = 75.0 mi/h
v₀ = 75 mi/h = 75* (5280 ft) / (3600s) = 110 ft/s
We replace v₀ = 110 ft/s and a = - 44.26 ft/s² in the formula (1):
vf²=v₀²+2*a*d
0 = (110)²+2*(-44.26)*d
88.52*d = (110)²
d = (110)² / (88.52)
d = 136.7 ft