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2 years ago

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Figure I shows the position of the moon at local time 10:00 P. M with respect to point A on earth’s surface. Figure II shows the

position of the moon 24 hours later with respect to the same place. Which phenomenon will a person at point A observe?

1 answer:

PtichkaEL [24]2 years ago

7 0

Answer:

A

Explanation:

i just did this on usatestprep the answer A

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How to solve part (b)?

maria [59]

Answer:

<em>1.11m</em>

Explanation:

From the diagram we are given the following forces;

F1 = 24.3N

F3 = 30N

Since the sum of upward forces is equal to that of downward force, then;

F2 = F1 + F3

F2 = 24.3N + 30N

F2 = 54.3N

Required

Distance between B and C

First we need to get Length of AC

Take moment about A

Anticlockwise moment = F3 cos20 * AC

Anticlockwise moment = 30ACcos 20

Clockwise moment = 1.2 * F2

Clockwise moment = 1.2(54.3) = 65.16Nm

Applying the principle of moment;

Sum of ACW moment = Sum of CW moments

30ACcos 20 = 65.16

AC = 65.16/30cos20

AC = 65.16/28.19

AC = 2.31m

Get the distance BC

AC = AB + BC

BC = AC-AB

BC = 2.31 - 1.2

BC = 1.11m

Hence the separation between B and C is 1.11m

<em>Note that the force F1 got in (a) was the value used in the calculation.</em>

<em></em>

8 0

3 years ago

Brent made the table shown to discribe two different relationships between animals.

grigory [225]

i think it is the second one due to the preditory descritpion of relationship A

5 0

4 years ago

A pipe open only at one end has a fundamental frequency of 266 Hz. A second pipe, initially identical to the first pipe, is shor

Alika [10]

Answer:

1.16cm were cut off the end of the second pipe

Explanation:

The fundamental frequency in the first pipe is,

<em><u>Since the speed of sound is not given in the question, we would assume it to be 340m/s</u></em>

f1 = v/4L, where v is the speed of sound and L is the length of the pipe

266 = 340/4L

L = 0.31954 m = 0.32 m

It is given that the second pipe is identical to the first pipe by cutting off a portion of the open end. So, consider L’ be the length that was cut from the first pipe.

<u>So, the length of the second pipe is L – L’</u>

Then, the fundamental frequency in the second pipe is

f2 = v/4(L - L’)

<u>The beat frequency due to the fundamental frequencies of the first and second pipe is</u>

f2 – f1 = 10hz

[v/4(L - L’)] – 266 = 10

[v/4(L – L’)] = 10 + 266

[v/4(L – L’)] = 276

(L - L’) = v/(4 x 276)

(L – L’) = 340/(4 x 276)

(L – L’) = 0.30797

L’ = 0.31954 – 0.30797

L’ = 0.01157 m = 1.157 cm ≅ 1.16cm

Hence, 1.16 cm were cut from the end of the second pipe

6 0

3 years ago

A person travels by car from one city to another with different constant speeds between pairs of cities. she drives for 50.0 min

liraira [26]

I think it's 45 miles. Don't know for sure though

3 0

3 years ago

One point of the circuit is grounded (V = 0). What are the (a) size and (b) direction (up or down) of the current through resist

Svetach [21]

<h3>Answer:</h3>

(a) <u>i₁ = 0.03818 A = 38.18 mA</u>

(b) downward

(c) <u>i₂ = 0.01091 A = 10.91 mA</u>

(d) rightward

(e) <u>i₃ = 0.02727 A = 27.27 mA</u>

(f) leftward

(g) <u>Eₐ = 3.818 Volts</u>

<h3>Question:</h3>

The complete question is stated below and the figure is provided in the attachment:

In Fig. 27-47, E 1 = 6.00 V, E 2 = 12.0 V, R1 = 100 Ω,

R2 = 200 Ω, and R3 = 300 Ω. One point of the circuit is grounded

(V = 0). What are the (a) size and (b) direction (up or down) of the

current through resistance 1, the (c) size and (d) direction

(left or right) of the current through resistance 2, and the

(e) size and (f) direction of the current through resistance 3?

(g) What is the electric potential at point A?

<h3></h3><h3>Explanation:</h3>

Applying Kirchoff's voltage law in the loops of botg E₁ and E₂, in the clockwise and anti clockwise direction:

E₁ - i₂R₂ - i₁R₁ = 0

E₂ - i₃R₃ - i₁R₁ = 0

If, we apply Kirchhoff's current law at junction A, we get:

i₁ = i₂ + i₃

Using these relations in loop equations, and re-arranging:

E₁ - i₂R₂ - (i₂ + i₃) R₁ = 0 ___________ eqn (1)

E₂ - i₃R₃ - (i₂ + i₃) R₁ = 0 ___________ eqn (2)

Eqn (1) implies:

6 - 200 i₂ - 100 i₂ - 100 i₃ = 0

i₂ = (6 - 100i₃)/300

Eqn (2) implies:

12 - 300 i₃ - 100 i₂ - 100 i₃ = 0

12 - 400 i₃ = 100 i₂

using value of i₃ from eqn (1)

12 - 400 i₃ = (1/3)(6 - 100 i₃)

36 - 1200 i₃ = 6 - 100 i₃

1100 i₃ = 30

<u>i₃ = 0.02727 A</u>

using this value in eqn of i₂:

i₂ = [6 - 100(0.02727)]/300

i₂ = (6 - 2.727)/300

<u>i₂ = 0.01091 A</u>

Since:

i₁ = i₂ + i₃

i₁ = 0.01091 A + 0.02727 A

<u>i₁ = 0.03818 A</u>

<u></u>

(a)

<u>i₁ = 0.03818 A = 38.18 mA</u>

(b)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>downward</u>

(c)

<u>i₂ = 0.01091 A = 10.91 mA</u>

(d)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>rightward</u>

(e)

<u>i₃ = 0.02727 A = 27.27 mA</u>

(f)

Since, the value of current is positive, thus it will have the direction that was assumed.

Therefore, its direction will <u>leftward</u>

<u>(g)</u>

With respect to the grounded portion, the potential drop at the resistance 1 will be equal to the potential at A Eₐ.

Therefore,

Eₐ = i₁R₁

Eₐ = (0.03818 A)(100 Ω)

<u>Eₐ = 3.818 Volts</u>

3 0

4 years ago