Answer:
The objects morion will remain the same
Explanation:
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 strength of the magnetic field is
.
Explanation:
Given that,
Length of the rod, L = 1.01 m
Speed with which the rod is moving, v = 3.47 m/s
We need to find the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.265 mV across the rod. When the rod is moving with some speed, an emf gets induced and it is given by :

B is magnetic field

So, the strength of the magnetic field is
.
Answer:
3.75s
Explanation:
a = 8.0 m/s V = 30 m/s U = 0 m/s t = ?
t = V - U/a
t = 30 - 0/8
t = 30/8 = 15/4
t = 3.75s