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The weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
To find the answer, we have to know about the pressure.
<h3>How to find the weight of a column of air?</h3>- As we know that the expression of pressure as,
where; F is the force, here it is equal to the weight of the air column, and A is the area of cross section.
- It is given that, the air column is extending from earth's surface to the top of the atmosphere, thus, the pressure will be atmospheric pressure,
- From this, the value of weight will be,
Thus, we can conclude that, the weight of a column of air with cross-sectional area 4. 5 m^2 extending from earth's surface to the top of the atmosphere is, 4.56*10^5N.
Learn more about the pressure here:
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Answer:
The acceleration of the refrigerator is
Explanation:
The expression of the equation of the net force acting on the refrigerator is as follows;
F-f= ma
Here, F is the applied force, f is the force of friction, m is the mass and a is the acceleration.
It is given in the problem that you're having a hard time pushing a refrigerator having mass 355 kg across the kitchen floor. The force of your own push is 993 N. The force of friction opposing your own push is 973 N.
Put F= 993, f= 973 N and m = 355 kg in the above expression of the equation to calculate the acceleration of the refrigerator.
993 - 973 = (355)a
20 = 355 a
Therefore, the acceleration of the refrigerator is .
Answer:
5 ft/s²
Explanation:
u = Initial velocity = 100 ft/s
v = Final velocity = 0
s = Displacement = 1000 ft
a = Acceleration
From equation of motion
The airplane's deceleration is 5 ft/s² or 1.524 m/s²