Answer:
Explanation:
Given
Ball of mass m
maximum Bearable Tension in string is F
Let length of the cord be L m and moving at a speed of v m/s
Here Tension will Provide Centripetal Force
T=Centripetal Force


Answer:
Newton's third law is: For every action, there is an equal and opposite reaction.
Explanation:
The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.
Answer:
1000N
Explanation:
Based on force=mass*acceleration, if the acceleration is constant at 2 metres per second squared, 1,000kg*2m/s^2=2,000N of force.
If the acceleration steadily increases to 2m/s^2 in 20 seconds, take the average which is 1m/s^2 therefore force=1,000N
Answer: 2561.7 pounds
Explanation:
If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:
(1)
Where:
is the slope of the line
is the airplane weight with 51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)
is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)
This means we already have one point of the graph, which coordinate is:

Rewritting (1):
(2)
As Y is a function of X:
(3)
Substituting the known values:
(4)
(5)
(6)
Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):
(7)
(8) This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m