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3 years ago

For a growing quantity to reach a value 32 times its initial value, how many doubling times are required? A) 4

Physics

1 answer:

Arisa [49]3 years ago

4 0

Answer:

B) n=5

Explanation:

We call the initial value Xo. We start to double this initial value

X1=2*Xo n=1

We double again:

X2=2*X1=2*(2Xo) n=2

X2=2*X1=2^2*Xo n=2

In general:

Xn=(2^n)*Xo

If we want to reach a value 32 times its initial value:

2^n=32

then: n=5

2^5=2*2*2*2*2=32

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Kipish [7]

Answer:

Velocity

Explanation:

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^^This explanation is from physicsclassroom.com

3 0

3 years ago

You will now examine the relationship between the number of field lines through a surface and the tangle betwcen A and E) angle

nikitadnepr [17]

Answer:

a. cosθ b. E.A

Explanation:

a.The electric flux, Φ passing through a given area is directly proportional to the number of electric field , E, the area it passes through A and the cosine of the angle between E and A. So, if we have a surface, S of surface area A and an area vector dA normal to the surface S and electric field lines of field strength E passing through it, the component of the electric field in the direction of the area vector produces the electric flux through the area. If θ the angle between the electric field E and the area vector dA is zero ,that is θ = 0, the flux through the area is maximum. If θ = 90 (perpendicular) the flux is zero. If θ = 180 the flux is negative. Also, as A or E increase or decrease, the electric flux increases or decreases respectively. From our trigonometric functions, we know that 0 ≤ cos θ ≤ 1 for 90 ≤ θ ≤ 0 and -1 ≤ cos θ ≤ 0 for 180 ≤ θ ≤ 90. Since these satisfy the limiting conditions for the values of our electric flux, then cos θ is the required trigonometric function. In the attachment, there is a graph which shows the relationship between electric flux and the angle between the electric field lines and the area. It is a cosine function

b. From above, we have established that our electric flux, Ф = EAcosθ. Since this is the expression for the dot product of two vectors E and A where E is the number of electric field lines passing through the surface and A is the area of the surface and θ the angle between them, we write the electric flux as Ф = E.A

4 0

4 years ago

A 14.0 gauge copper wire of diameter 1.628 mm carries a current of 12.0 mA . A) What is the potential difference across a 1.80 m

Serga [27]

Answer: a) 139.4 μV; b) 129.6 μV

Explanation: In order to solve this problem we have to use the Ohm law given by:

V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.

Then we have:

for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω

and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω

Finalle we calculate the potential difference (V) for both wires:

Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V

V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V

8 0

4 years ago

Read 2 more answers

When does a rubber band, which has been shot at a wall, have the most potential energy?

Yuri [45]

D When it is stretched ready to shoot at the wall

4 0

3 years ago

Read 2 more answers

In a two-source circuit, one source acting alone produces 10 ma through a given branch. the other source acting alone produces 8

pashok25 [27]

Refer to the figure below.
R = resistance.

Case 1:
The voltage source is V₁ and the current is 10 mA. Therefore
V₁ = (10 mA)R

Case 2:
The voltage source is V₂ and the current is 8 mA. Therefore
V₂ = (8 mA)R

Case 3:
The voltage across the resistance is V₁ - V₂. Therefore the current I is given by
V₁ - V₂ = IR
10R - 8R = (I mA)R
2 = I
The current is 2 mA.

Answer: 2 mA

6 0

3 years ago