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Artemon [7]
2 years ago
15

At room temperature, an oxygen molecule, with mass of 5.31x10-26kg , typically has a kinetic energy of about 6.21x10-21J.How fas

t is it moving?
Physics
1 answer:
alex41 [277]2 years ago
3 0
<h2>Oxygen is moving at 483.63 m/s</h2>

Explanation:

Kinetic energy = 0.5 x Mass x Velocity²

Kinetic energy = 6.21 x 10⁻²¹ J

Mass = 5.31 x 10⁻²⁶ kg

Substituting

        Kinetic energy = 0.5 x Mass x Velocity²

        6.21 x 10⁻²¹ = 0.5 x 5.31 x 10⁻²⁶ x Velocity²

        Velocity² = 2.34 x 10⁵

        Velocity = 483.63 m/s

Oxygen is moving at 483.63 m/s  

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An uncharged 30.0-µF capacitor is connected in series with a 25.0-Ω resistor, a DC battery, and an open switch. The battery has
kolezko [41]

Answer:

(a) Maximum current through resistor is 1.43 A

(b) Maximum charge capacitor receives is 1.50\times 10^{-3}\text{ C}.

Explanation:

(a)

In an RC (resistor-capacitor) DC circuit, when charging, the current at any time, <em>t</em>, is given by

I(t) = I_0e^{-t/\tau}

Here, I_0 is the maximum current and <em>τ</em> represents time constant which is given by RC (the product of the resistance and capacitance).

The maximum current is given by

I = \dfrac{V}{R_\text{eff}}

<em>V</em> is the emf of the battery and R_\text{eff} is the effective resistance.

In this question, R_\text{eff} = 10.0 Ω + 25.0 Ω = 35.0 Ω

I = \dfrac{50.0\text{ V}}{35.0\ \Omega} = 1.43\text{ A}

(b) The maximum charge is given

<em>Q</em> = <em>CV</em>

where <em>C</em> is the capacitance of the capacitor

Q = (30.0\times10^{-6}\text{ F})(50.0\text{ V}) = 1.50\times 10^{-3}\text{ C}

3 0
2 years ago
Read 2 more answers
Three balls with the same radius 21 cm are in water. Ball 1 floats, with
Masteriza [31]

Answer:

Explanation:

A )

The ball floats with half of it exposed above the water level . So it must have density half that of water . In other words its density must have been 500 kg / m³

B )

Tension in the ball will be equal to net force acting on the ball

Net force on the ball = buoyant force - weight .

4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1000 - 893 )

= 40.65 x 10⁻⁶ N .

C )Tension in the 3 rd  ball will be equal to net force acting on the ball

Net force on the ball =  weight  - buoyant force

= 4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1320 - 1000  )

=  121.6  x 10⁻⁶ N .

7 0
2 years ago
Ronnie kicks a playground ball with an initial velocity of 16 m/s at an angle of 40° relative to the ground. What is the approxi
AleksandrR [38]
The horizontal component is calculated as:
Vhorizontal = V · cos(angle)

In your case Vhoriontal = 16 · cos(40) = 12.3 m/s

Answer: 12.3 m/s
7 0
2 years ago
Read 2 more answers
Global Precipitation Measurement (GPM) is a tool scientists use to forecast weather. Which statements describe GPM? Select three
lyudmila [28]

Answer:

B.It is a satellite that collects data about rain and snow

C.Its orbit covers 90 percent of Earth’s surface

F.The sensors measure microwaves

5 0
2 years ago
The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.
Reika [66]

Answer:

tan \theta = \mu_s

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

mg sin \theta - \mu_s R =0 (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, \theta the angle of the slope

\mu_s R is the frictional force, with \mu_s being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

R-mg cos \theta = 0

where

R is the normal reaction

mg cos \theta is the component of the weight perpendicular to the slope

Solving for R,

R=mg cos \theta

And substituting into (1)

mg sin \theta - \mu_s mg cos \theta = 0

Re-arranging the equation,

sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of \mu_s, the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0
2 years ago
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