To solve the problem it is necessary to use Newton's second law and statistical equilibrium equations.
According to Newton's second law we have to
where,
m= mass
g = gravitational acceleration
For the balance to break, there must be a mass M located at the right end.
We will define the mass m as the mass of the body, located in an equidistant center of the corners equal to 4m.
In this way, applying the static equilibrium equations, we have to sum up torques at point B,
Regarding the forces we have,
Re-arrange to find M,
Therefore the maximum additional mass you could place on the right hand end of the plank and have the plank still be at rest is 16.67Kg
Answer: The area of the parking lot is 14,400 meters squared.
Explanation:
We have the dimensions of the parking lot.
60m by 240m
The units used here are meters.
Now, if we want to know the area of the parking lot is equal to the product between the length and the width:
A = 60m*240m = 14,400 m^2
The area of the parking lot is 14,400 meters squared.
The mass of the aeroplane is 300,000 kg.
<h3>What is Newton's second law of motion?</h3>
It states that the force F is directly proportional to the acceleration a of the body and its mass.
The law is represented as
F =ma
where acceleration a = velocity change v / time interval t
Given is the aeroplane lands at a speed of 80 m/s. After landing, the aeroplane takes 28 s to decelerate to a speed of 10 m/s. The mean resultant force on the aeroplane as it decelerates is 750 000 N.
The force expression will be
F = mv/t
Substitute the values and we have
750000 = m x (80 -10)/ 28
750,000 = m x 2.5
m = 300,000 kg
Thus, the mass of the aeroplane is 300,000 kg.
Learn more about Newton's second law of motion.
brainly.com/question/13447525
#SPJ1
Answer:
The speed velocity and acceleration escape room is an immersive experience for your students. It allows them to demonstrate their knowledge of these principles in a fun and engaging way. Speed Velocity Acceleration Escape Room - Single Student DIGITAL Version
Explanation:
Answer:
Explanation:
Force between two charges of q₁ and q₂ at distance d is given by the expression
F = k q₁ q₂ / d₂
Here force between charge q₁ = - 15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = (1.66 - 1.24 ) = .42 mm
k = 1/ 4π x 8.85 x 10⁻¹²
putting the values in the expression
F = 1/ 4π x 8.85 x 10⁻¹² x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 9 x 10⁹ x - 15 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 35969.4 x 10⁻³ N .
force between charge q₂ = 34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C when distance between them d = ( 1.24 - 0 ) = 1.24 mm .
putting the values in the expression
F = 1/ 4π x 8.85 x 10⁻¹² x 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 9 x 10⁹ x - 34.5 x 10⁻⁹ x 47 x 10⁻⁹ /( .42 x 10⁻³)²
= 82729.6 x 10⁻³ N
Both these forces will act in the same direction towards the left (away from the origin towards - ve x axis)
Total force = 118699 x 10⁻³
= 118.7 N.
