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Elena-2011 [213]
3 years ago
6

To verify that this expression for (vf)α has the correct units of velocity, you need to perform some unit analysis. begin by fin

ding the equivalent of a volt in terms of basic si units. what is a volt in terms of meters (m), seconds (s), kilograms (kg), and coulombs (c)?
Physics
1 answer:
Ket [755]3 years ago
7 0
<h2>Answer</h2>

1 V = 1 kg m2 s−² C

1 V = 1 kg m2 s−³ A−1

(one kilogram meter squared per second cubed per ampere)

<h2>Explanation</h2>

<em>The work required to move an electric charge of one coulomb through an electrical potential difference of one volt,</em>

<h2> Volts = Joules / Coulombs</h2><h2 />

    Joules = Newton x metre

     Volts = Nm / coulombs

    Newton = kg x metre / second ²

    Volts = kg x metre / second ² coulombs

Volt in terms of meters (m), seconds (s), kilograms (kg), and coulombs (c)

<h3>For further:</h3>

Coulomb = Amp * second

Volts = kg x metre / second ³ Amp

<h2 />

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You have a rod with a length of 146.4 cm. You prop up one end on a brick which is 3.8 cm thick. Your uncertainty in measuring th
AfilCa [17]

Answer:

\partial \theta = 0.003

Explanation:

we know that

sin\theta = \frac{3.8}{146.4}

\theta = sin^{-1} \frac{3.8}{146.4}

\theta = 1.484°

\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians

as we see that sin\theta = \theta

relative error\frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}

Where X_1 IS HEIGHT OF ROCK

X_2 IS THE HEIGHT OF ROAD

\partial X = uncertainity in measuring  distance

\partial X = 0.05

Putting all value to get uncertainity in angle

\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}

solving for \partial \theta we get

\partial \theta = 0.003

3 0
3 years ago
Tammy is a forensic investigator examining a body at a crime scene. She notes that the body is stiff but still flexible. What do
wlad13 [49]
I may be wrong but does it mean it was revent? because i know shortly after someone dies your body becomes fully stiff so maybe it was recent and it's in the process off stiffening up
8 0
2 years ago
An x-ray photon is scattered by an originally stationary electron. how does the frequency of the scattered photon compare relati
Viefleur [7K]

The frequency of the scattered photon decreases or it will be lower compare to the frequency of incident photon. An x-ray photon scatters in one direction after a collision and some energy is transferred to the electron as it recoils in another direction resulting to have less energy in the scattered photon. In addition, the frequencies will also depend on the differences of the angle at which the scattered photon leaves the collision and this incident is called Compton Effect.

8 0
3 years ago
Your companion on a train ride through Illinois notices that telephone poles near the tracks appear to be passing by very quickl
Nikitich [7]

Answer: Relative motion

Explanation: If two objects are moving either towards or away from each other with both having their velocities in a reference frame and someone is outside this reference frame seeing the motion of the two objects.

The observer ( in his own frame of reference) will measure a different velocity as opposed to the velocities of the two object in their own reference frame. p

Both the velocity measured by the observer in his own reference frame and the velocity of both object in their reference is correct.

Velocities of this nature that have varying values based on motion referenced to another body is known as relative velocity.

Motion of this nature is known as relative motion.

<em>Note that the word reference frame is simply any where the motion is occurring and the specified laws of motion is valid</em>

<em />

For this example of ours, the reference frame of the companion is the train and the telephone poles has their reference frame as the earth.

The companion will measure the velocity of the telephone poles relative to him and the velocity of the telephone pole relative to an observer outside the train will be of a different value.

6 0
2 years ago
How much heat (in calories) are needed to raise the temperature of 60 g of water
laila [671]

Answer:

Well, each ml of water requires one calorie to go up 1 degree Celsius, so this liter of water takes 1000 calories to go up 1 degree Celsius.

Explanation:

3 0
2 years ago
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