Answer:
Assuming air resistance is negligible, all of the potential energy that the object has at the top of the ramp is converted into kinetic energy by the time it gets to the bottom of the ramp. This is because no matter what path the object takes to move the 5m vertically (ie. falling straight down v. sliding on the ramp), gravity does the same amount of work on it.
Thus, calculate the total amount of potential energy at the top of the ramp:
Ep=mgh
Ep=4(9.81)5
Ep=196.2 Joules
Because all of this potential energy is converted into kinetic energy in the object by the bottom of the ramp, the object hits the spring with 196.2J of energy.
By using the formula for elastic potential energy, you can calculate exactly how far the spring compresses.
196.2=(1/2)k(x^2)
392.4=(350)(x^2)
1.1211=x^2
sqrt(1.1211)=x
x=1.059m
As for the last part of the question, after the object compresses the spring fully and stops momentarily, the spring converts it's elastic potential energy back into kinetic energy in the object and pushes it away again.
Explanation:
Water goes through evaporation.
Answer:
u = 104.68 m/s
Explanation:
given,
horizontal distance = 150 m
elevation of 12.4 m
angle = 8.6°
horizontal motion = x = u cos θ. t .............(1)
vertical motion =
................(2)
from equation(1) and (2)
..........{3}
u = 104.68 m/s
The initial speed of the ball is u = 104.68 m/s
Answer:
D
Explanation:
6CO² + 6H²O > sunlight, chlorophyll, enzymes > C⁶H¹²O⁶ + 6O²
Incomplete Question.The Complete question is
The Earth spins on its axis and also orbits around the Sun. For this problem use the following constants. Mass of the Earth: 5.97 × 10^24 kg (assume a uniform mass distribution) Radius of the Earth: 6371 km Distance of Earth from Sun: 149,600,000 km
(i)Calculate the rotational kinetic energy of the Earth due to rotation about its axis, in joules.
(ii)What is the rotational kinetic energy of the Earth due to its orbit around the Sun, in joules?
Answer:
(i) KE= 2.56e29 J
(ii) KE= 2.65e33 J
Explanation:
i) Treating the Earth as a solid sphere, its moment of inertia about its axis is
I = (2/5)mr² = (2/5) * 5.97e24kg * (6.371e6m)²
I = 9.69e37 kg·m²
About its axis,
ω = 2π rads/day * 1day/24h * 1h/3600s
ω= 7.27e-5 rad/s,
so its rotational kinetic energy
KE = ½Iω² = ½ * 9.69e37kg·m² * (7.27e-5rad/s)²
KE= 2.56e29 J
(ii) About the sun,
I = mR²
I= 5.97e24kg * (1.496e11m)²
I= 1.336e47 kg·m²
and the angular velocity
ω = 2π rad/yr * 1yr/365.25day * 1day/24h * 1h/3600s
ω= 1.99e-7 rad/s
so
KE = ½ * 1.336e47kg·m² * (1.99e-7rad/s)²
KE= 2.65e33 J
