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mart [117]
3 years ago
8

Air pressure is gravity pulling down on air molecules. a. True b. False

Physics
1 answer:
s344n2d4d5 [400]3 years ago
8 0
<span>Air pressure is gravity pulling down on air molecules. Like us, air has also mass and the pressure is brought down by the earth’s gravity causing an increase in weight exerted on you as you descend lower into the atmosphere. So, as you enter into the other layers of atmosphere above the troposphere, the air pressure starts to decrease. <span>Below the atmosphere is the hydrosphere. This is where all liquid forms are located. And since the seawater has a greater mass than air, it has the greatest pressure. The answer is true.</span></span>
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7. If the magnitude of the gravitational force of
JulijaS [17]

Newton's Third Law states that for every action there is an opposite and equal reaction:

If the gravitational force of the Earth on the Moon is F then the gravitational force of the Moon on the Earth is also F

3 0
3 years ago
A 6000 kg roller coaster goes around a loop of radius 30 m at 6 m/s. What is the centripetal acceleration?
Margarita [4]

Answer:

The answer to the question is 7200

7 0
3 years ago
When two oceanic plates converge, the plate with a greater density will subduct. This plate will partially melt, causing the for
N76 [4]

Answer:

Island arc

Explanation:

When two oceanic plates share a convergent type of plate boundary, the denser oceanic plate will subduct below the less dense oceanic plate. This will result in the formation of the subduction zone, where the rocks are being pulled down to the mantle. This subduction zone is typically marked by the presence of a narrow depression commonly known as an oceanic trench, that lies just above the zone.

The rocks of the subducting plate undergo partial melting and mix up with the magma that rises upwards towards the surface due to the force exerted by the convection currents. This later gives rise to the formation of volcanoes or a chain of volcanoes which are commonly known as an island arc.

3 0
3 years ago
A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),
labwork [276]

Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

<u>FOR FORCE (P)</u>:

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

W = Pd

<u>Therefore, work done by force (P) is Positive.</u>

<u></u>

<u>FOR NORMAL FORCE (Fn) AND WEIGHT (W)</u>:

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

W = Fn d(0) = mg d (0)

W = 0

<u>Therefore, work done by Normal Force (Fn) and Weight (mg) is Zero.</u>

<u></u>

<u>FOR KINETIC FRICTIONAL FORCE (fk)</u>:

Since, kinetic frictional force acts in the opposite direction of motion of the box. Therefore, θ = 180°

Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

<u>Therefore, work done by Kinetic Frictional Force (fk) is Negative.</u>

<u></u>

8 0
4 years ago
A mass moves back and forth in simple harmonic motion with amplitude A and period T.
Sever21 [200]

a. 0.5 T

- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position

- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

1 T : 4 A = t : 2 A

and solving for t we find

t=\frac{(1T)(2 A)}{4A}=0.5 T

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

- the mass takes a time of 1 T to cover a distance of 4A

we can set the following proportion:

1 T : 4 A = t : 5 A

And by solving for t, we find

t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T

6 0
3 years ago
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