Explanation:
a. A grocery bag as you lift it up
b. A crane moving dirt......
d. A crate as you push it along the floor
Answer:
2274 J/kg ∙ K
Explanation:
The complete statement of the question is :
A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.
= mass of metal = 400 g
= specific heat of metal = ?
= initial temperature of metal = 100 °C
= mass of aluminum cup = 100 g
= specific heat of aluminum cup = 900.0 J/kg ∙ K
= initial temperature of aluminum cup = 15 °C
= mass of water = 500 g
= specific heat of water = 4186 J/kg ∙ K
= initial temperature of water = 15 °C
= Final equilibrium temperature = 40 °C
Using conservation of energy
heat lost by metal = heat gained by aluminum cup + heat gained by water
Answer:
a = 9.8 m/s²
Explanation:
Acceleration due to gravity on Earth is constant, which is 9.8 m/s²
Answer:
52 rad
Explanation:
Using
Ф = ω't +1/2αt²................... Equation 1
Where Ф = angular displacement of the object, t = time, ω' = initial angular velocity, α = angular acceleration.
Since the object states from rest, ω' = 0 rad/s.
Therefore,
Ф = 1/2αt²................ Equation 2
make α the subject of the equation
α = 2Ф/t².................. Equation 3
Given: Ф = 13 rad, t = 2.5 s
Substitute into equation 3
α = 2(13)/2.5²
α = 26/2.5
α = 4.16 rad/s².
using equation 2,
Ф = 1/2αt²
Given: t = 5 s, α = 4.16 rad/s²
Substitute into equation 2
Ф = 1/2(4.16)(5²)
Ф = 52 rad.
Answer:
Zero
Explanation:
As force acting on the body is equal to the product of mass and acceleration.
Acceleration is equal to rate of change in velocity.
Here velocity is constant so acceleration is zero.
It means the net force acting on the vehicle is zero.
