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bagirrra123 [75]
3 years ago
13

The independent variable is

Physics
1 answer:
Katarina [22]3 years ago
7 0

Answer:

An independent variable is exactly what it sounds like. It is a variable that stands alone and isn't changed by the other variables you are trying to measure.

Explanation:

someone's age might be an independent variable.

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I have an algerbra 2 question.
Andreyy89

In order to find out if the exponential function represents a growth or a decay, let's look at the number that is base to the exponent x.

If the number is greater than 1, so the function represents a growth, and if the number is less than 1, the function represents a decay.

Since the number is 1.075, the function represents a growth.

To find the % increase, first let's convert the number to percentage, and then subtract 100%:

\begin{gathered} 1.075=1.075\cdot100\text{\%}=107.5\text{\%} \\ 107.5\text{\%}-100\text{\%}=7.5\text{\%} \end{gathered}

So the percent increase is 7.5%.

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1 year ago
Please answer!!! I’m desperate
rodikova [14]

a is the answer hope this helps

8 0
4 years ago
In a traffic Jam when drivers can't get where they are going on time or at the expected speed of travel this is referred to as w
zvonat [6]
A.) restriction because your can’t get to your place!
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3 years ago
Read 2 more answers
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, v_{d} , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

v_{d} = \frac{I}{qnA}             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

v_{d} = \frac{E}{qnp}      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

4 0
3 years ago
Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at se
adoni [48]

Answer:

(a)\ F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

(b)\ F_{No} = 771.125N

Explanation:

Given

d_D = 6000ft ---- Altitude of container in Denver

A = 0.0155m^2 -- Surface Area of the container lid

P_D = 79000Pa --- Air pressure in Denver

P_{No} = 100250Pa --- Air pressure in New Orleans

<em>See comment for complete question</em>

Solving (a): The expression for F_{No

Force is calculated as:

F = \triangle P * A

The force in New Orleans is:

F_{No} = \triangle P * A

Since the inside pressure is half the pressure at sea level, then:

\triangle P = P_{No} - \frac{P_{area}}{2}

Where

P_{area} = 101000Pa --- Standard Pressure

Recall that:

F_{No} = \triangle P * A

This gives:

F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

Solving (b): The value of F_{No

In (a), we have:

F_{No} = [P_{No} - \frac{P_{area}}{2}]* A

Where

A = 0.0155m^2

P_{No} = 100250Pa

P_{area} = 101000Pa

So, we have:

F_{No} = [100250 - \frac{101000}{2}] * 0.0155

F_{No} = [100250 - 50500] * 0.0155

F_{No} = 49750* 0.0155

F_{No} = 771.125N

4 0
3 years ago
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