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GaryK [48]
3 years ago
14

How and why dose air pressure change with altitude in the atmosphere

Physics
1 answer:
arlik [135]3 years ago
8 0
<span>Air pressure changes with altitude because of issues related to gravity molecules have more weight the closer they are to the Earth and more of them move to lower elevations as a result this causes increased pressure because there are more molecules in number and proximity</span>
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Which is an example of a trace fossil?
hodyreva [135]

Answer: The correct answer is A). Animal burrow because burrow fossils represent the preserved byproducts of behavior rather than physical remains, they are considered a kind of trace fossil. One common kind of burrow fossil is known as Skolithos, and the similar Trypanites, Ophiomorpha and Diplocraterion.

4 0
3 years ago
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A crate is pushed up each of the two ramps shown in the diagram below. Based on the concept that simple machines make work easie
Naddik [55]

Answer:

ramp b requires less force than ramp a

Explanation:

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3 years ago
What are the answers to the question
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7 0
3 years ago
A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c
ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

\int _{0}^{i}di =9\int _{0}^{t}dt

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

B = \frac{\mu_{0}i}{2\pi r}

B = \frac{\mu_{0}\times 9t}{2\pi r}

(b)

Magnetic flux,

\phi=\int B\times a dr

\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr

\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )

\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)

\phi = 1.89 \times 10^{-7}t

(c)

R = 3 ohm

e = -\frac{d\phi}{dt}

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0
3 years ago
(15 Points)
oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

Read more about vertical weight here:

brainly.com/question/15244771

#SPJ1

5 0
2 years ago
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