1 hour = 3,600 seconds
1 km = 1,000 meters
75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s
The mean speed during the deceleration is
(1/2)(20-5/6 + 0) = 10-5/12 m/s.
Traveling at this average speed for 21 seconds,
the bus covers
(10-5/12) × (21) = 218.75 meters.
The formula for range is:
Given values are:
where θ equals 14.1 degrees
Using the equation above,
The calculated range is 66.7 meters.
Therefore, the range is approximately 66.1 meters.
Answer:
F=126339.5N
Explanation:
To compute the force required to escape, a free-body diagram for the hatch must be drawn. We will equate the downward and upward forces, thus applying the following equation:
Fw=W+Fi+F
where
Fw= force or weight exerted by the water column above the submarine.
To calculate Fw, we can use:
Fw=h. γ. A
h=height
γ=
specific weight of seawater = 10074N / m ^ 3
A=Area
Fw=28x10074x0.7=197467N
w represents the hatch weight = 200N
Fi denotes the internal pressure force in the submarine, which is 1 atm = 101325Pa. We can calculate this force using:
Fi=PA=101325x0.7=70927.5N
Finally, the force needed to open the hatch is determined by the original equation:
Fw=W+Fi+F
F=Fw-W+Fi
F=197467N-200N-70927.5N
F=126339.5N
The string does not experience any force of tension, as it balances two forces acting in the same direction. Hence, the tension is zero.
Explanation:
If tension existed in the string, it would mean that two equal but opposite forces are exerting pull in contrary directions.
When a force of f newtons is applied from the right and another force of f newtons from the left, the resulting action occurs through one force. Because there is action on the same string in opposing directions, the tension in the string can only be equal to the magnitude of the string itself.
Therefore, the string indeed has no tension since it is dealing with two forces acting in the same direction. Thus, the tension is zero.
I will assume the girl is on the right while the boy is on the left.
The net force represents the total of all forces acting on an object, factoring in negatives.
Let the force from the boy be denoted as b. We’ll apply the formula F = ma.
b + 3.5 = 0.2(2.5)
This reduces to a straightforward algebraic problem. By solving, we find that the boy is applying a force of -3N to the left.
