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4 years ago

5

Assuming that voltage remains constant, what happens to the current in a wire as its diameter decreases? A. The current is not a

ffected by a change in wire diameter. B. The current alternates between high and low values. C. The current increases. D. The current decreases.

1 answer:

kondaur [170]4 years ago

5 0

Answer:

<em> The current decreases.</em>

Explanation:

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The mechanical advantage of an inclined plane can be increased by increasing the ramp's _____. slope length friction height

artcher [175]

Answer:

height is the answer i'm pretty sure.

Explanation:

8 0

4 years ago

A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 285-mL mark with 40.5°C glycerin. After the flask

Charra [1.4K]

Answer:

V_f = 287.04 mL

Explanation:

We are given the initial/original volume of the glycerine as 285 mL.

Now, after it is finally cooled back to 20.0 °C , its volume is given by the formula;

V_f = V_i (1 + βΔT)

Where;

V_f is the final volume

V_i is the original volume = 285 mL

β is the coefficient of expansion of glycerine and from online tables, it has a value of 5.97 × 10^(-4) °C^(−1)

Δt is change in temperature = final temperature - initial temperature = 32 - 20 = 12 °C

Thus, plugging in relevant values;

V_f = 285(1 + (5.97 × 10^(-4) × 12))

V_f = 287.04 mL

7 0

3 years ago

5.1C A fluid flows steadily through a pipe with a uniform crosssectional area. The density of the fluid decreases to half its in

prisoha [69]

The options are;

a. V2 equals 2V1.

b. V2 equals (V1)/2.

c. V2 equals V1.

d. V2 equals (V1)/4.

e. V2 equals 4V1.

Answer:

Option A: V2 equals 2V1

Explanation:

Since the flow is steady, then we can say;

mass flow rate at input = mass flow rate at output.

Formula for mass flow rate is;

m' = ρVA

Thus;

At input;

m'1 = ρ1•V1•A1

At output;

m'2 = ρ2•V2•A2

So, m'1 = m'2

Now, we are told that the density of the fluid decreases to half its initial value.

Thus; ρ2 = (ρ1)/2

Since m'1 = m'2, then;

ρ1•V1•A1 = (ρ1)/2•V2•A2

Now, the pipe is uniform and thus the cross section doesn't change. Thus;

A1 = A2

We now have;

ρ1•V1•A1 = (ρ1)/2•V2•A1

A1 and ρ1 will cancel out to give;

V1 = (V2)/2

Thus, V2 = 2V1

5 0

3 years ago

Three charges, q1 = +2.06 x 10-9 C, q2 = -3.27 x 10-9 C, and q3 = +1.05 x 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0

lisov135 [29]

Answer:

The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)

Explanation:

Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.

Step 2: I must calculate the magnitude of the forces acting on the third charge.

F13: Force exerted by charge 1 on charge 3.

F23: Force exerted by charge 2 on charge 3.

K: Constant of Coulomb's law.

d13: distance from charge 1 to charge 3.

d23: distance from charge 2 to charge 3

Fr: Resulting force.

q1=+2.06 x 10-9 C

q2= -3.27 x 10-9 C

q3= +1.05 x 10-9 C

K=9-10^9 N-m^2/C^2

d13= 0,20 m

d23= 0,10 m

F13= K * (q1 * q3)/(d13)^2

F13=9,7335*10^(-8) N

F23=K * (q2 * q3)/(d23)^2

F23= -3,09 * 10^(-7)

Step 3: We calculate the resultant force on charge 3.

Fr=F13+F23= -2,11665 * 10^(-7)

3 0

3 years ago

According to the inverse square law of light, how will the apparent brightness of an object change if its distance to us triples

aleksley [76]

According to the inverse square law of light, <span>apparent brightness will decrease by a factor of 9. Use the formula </span> $B=L/(4*pD^2)$ , to check it.

4 0

3 years ago