The energy that transforms into kinetic energy is the Potential Energy. It happens that objects can store energy as a result of its position. Image for example a slingshot. When you stretch the slingshot, it stores energy, this energy would be the energy you used to stretch the slingshot, the material aborbs it and then release to throw the projectile.
Now, on earth and everywhere in the universe where you are close to an object with mass, it exists a force called gravity that attracts you towards that object. Every object that has mass exercises gravitational attration towards the other objects. It just happens that Earth is has so much mass that its gravitational pull is way stronger that the gravitational pull of another object on its surface. This means things will tend to be as close as earth as possible, and in order to move something away from earth, you will have to perform a force in the opposite direction to Earth and, therefore, consume energy. This energy will be store as potential energy, and when you drop the object, the potential energy will be the energy that will transform to kinetic energy.
The greater the mass the greater is inertia.
Answer:
4500.5 nutritional calories per gram
Explanation:
Heat lost by the new candy = heat gained by the bomb calorimeter.
Heat gained by the bomb calorimeter = c×ΔT
where c = heat capacity of the calorimeter = 32.20 KJ/K = 32200 J/K
ΔT = change in temperature = 2.69°C = 2.69 K.
Heat gained by the bomb calorimeter = 32200 × 2.69 = 86618 J
Heat lost by the new candy = heat gained by the bomb calorimeter = 86618 J = 20702.2 calories
4.60 g of the new candy lost this amount of calories by undergoing combustion,
The amount of calories per g = 20702.2 calories/4.6 g = 4500.5 calories per gram
Answer:
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write
where
is the distance of the new object from the sun (orbital radius)
is the orbital period of the object
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for , we find
![r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m](https://tex.z-dn.net/?f=r_o%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7DT_o%5E2%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.50%5Ccdot%2010%5E%7B11%7Dm%29%5E3%7D%7B%28365%20d%29%5E2%7D%28180%20d%29%5E2%7D%3D9.4%5Ccdot%2010%5E%7B10%7D%20m)
