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Ierofanga [76]
3 years ago
7

Describe how does the temperature relate to density and the movement of atmospheric gases? A) As the air warms expanding its vol

ume, decreasing its density, and the air molecules rise. B) UV light causes air to warm its volume declines, density increases, and the air molecules sink. C) The volume of air distends as it cools, while its density increases and the air molecules descend. D) Radiation cools the air and its volume extends, density diminishes, and the air molecules accelerate.
Physics
1 answer:
cluponka [151]3 years ago
8 0
The correct answer is A) As the air warms expanding its volume, decreasing its density, and the air molecules rise

The air molecules speed up and spread. If the container is closed then it may pop because the air expands. Hot air goes up and cold air goes down.

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You do 45.0 joules of work and 3.00 seconds how much power do you use​
Arada [10]

Answer:

15 watt

Explanation:

Power is the rate at which work is done.

This means you divide the work done with the amount of time used to perform the work.

The formula for Power is : P = W/t  where;

W= work done in J = 45

t= time in seconds = 3 sec

P= 45/ 3 = 15 watt

8 0
3 years ago
A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi
vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

V=\frac {C_2V_2-C_1V_1}{C_1+C_2}

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}

Therefore

V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437

Charge, Q is given by CV hence for the first capacitor charge will be Q_1=C_1V

Here, Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C

8 0
3 years ago
How quickly do muscles become fatigued?​
agasfer [191]

Answer:

Fatigue is usually defined as the reversible decline of performance during activity, and most recovery occurs within the first hour. However, there is also a slowly reversible component that can take several days to reverse (155). Muscle injury also causes a decline in performance that reverses only very slowly.

5 0
3 years ago
In a double slit experiment, 450 nm light passes through two slits producing an interference pattern where the first bright frin
musickatia [10]

Answer:

1.3 x 10⁻⁴ m

Explanation:

\lambda = wavelength of the light = 450 nm = 450 x 10⁻⁹ m

n = order of the bright fringe = 1

θ = angle = 0.2°

d = separation between the slits

For bright fringe, Using the equation

d Sinθ = n \lambda

Inserting the values

d Sin0.2° = (1) (450 x 10⁻⁹)

d (0.003491) = (450 x 10⁻⁹)

d = 1.3 x 10⁻⁴ m

6 0
3 years ago
A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box  = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0
3 years ago
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