Answer:
F = 2.6692 x 10⁻⁹ N
Explanation:
Given,
The mass of the rock, m = 10 kg
The mass of the boulder, M = 100 kg
The distance between them, d = 5 m
The gravitational force between the two bodies is proportional to the product of their masses and inversely proportional to the square of the distance between them. It is given by the formula
<em> F = GMm/d² newton</em>
Where,
G - Universal gravitational constant
Substituting the given values,
F = 6.673 x 10⁻¹¹ x 100 x 10 / 5²
F = 2.6692 X 10⁻⁹ N
Hence, the force between the two bodies is, F = 2.6692 X 10⁻⁹ N
You would know a decomposition reaction occurred if the reactants separated. For example from AB → A+B.
Now if you look at your options only 1 works out for that equation. Letter A.
From the compound K2CO3 it split up to K2O +CO2
It cannot be letter B because synthesis/combination occurred. The same goes for letter C. Letter D, single displacement occurred.
Again, the answer is A.
<h3>
Answer:</h3>
49 N
<h3>
Explanation:</h3>
<u>We are given;</u>
- Mass of the brick as 3 kg
- The coefficient of friction as 0.6
We are required to determine the force that must be applied by the woman so the brick does not fall.
- We need to importantly note that;
- For the brick not to fall the, the force due to gravity is equal to the friction force acting on the brick.
- That is; Friction force = Mg
But; Friction force = μ F
Therefore;
μ F = mg
0.6 F = 3 × 9.8
0.6 F = 29.4
F = 49 N
Therefore, she must use a force of 49 N
media.discordapp.net/attachments/782414373888458783/826224189828366377/video0.mp4
Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
k = 1.4
Work done is given as;
inlet velocity is negligible;
Therefore, the exit velocity is 629.41 m/s
