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Afina-wow [57]
3 years ago
10

Where do subduction zones often occur

Physics
1 answer:
vodka [1.7K]3 years ago
4 0

Subductions occur at convergent boundaries. Its where one tectonic plate moves under another.

-Steel jelly

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What are the largest and smallest resistances you can obtain by connecting a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together?
myrzilka [38]
<h2>Answer:</h2>

786Ω and 20.32Ω respectively.

<h2>Explanation:</h2>

(a) Given a number of resistors each with its own resistance, the largest resistance can be obtained when these resistors are connected in series.

From the question, the resistors have the following resistances;

36.0-Ω, 50.0-Ω , and 700-Ω

Now, when they are connected in series, the total resistance (R) obtainable is given by the sum of these individual resistances as follows;

R = 36.0-Ω + 50.0-Ω + 700-Ω

R = 786Ω

Therefore, the largest resistance that can be obtained by connecting  a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together is 786Ω

(b) Similarly, given a number of resistors each with its own resistance, the smallest resistance can be obtained when these resistors are connected in parallel.

From the question, the resistors have the following resistances;

36.0-Ω, 50.0-Ω , and 700-Ω

Now, when they are connected in parallel, the total resistance (R) obtainable is given by using the relation as follows;

\frac{1}{R} = \frac{1}{36.0} + \frac{1}{50.0} + \frac{1}{700.0}

\frac{1}{R} = \frac{35000+25200+1800}{1260000}

\frac{1}{R} = \frac{62000}{1260000}

\frac{1}{R} = \frac{62}{1260}

R = \frac{1260}{62}

R = 20.32Ω

Therefore, the smallest resistance that can be obtained by connecting  a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together is 20.32Ω

7 0
3 years ago
Why are images reflected from a rough surface not as clear as those reflected from a smooth
ioda

Answer:

Just as images are reflected from the surface of a mirror, light reflected from a smooth water surface also produced a clear image. ... Consequently, the outgoing rays are reflected at many different angles and the image is disrupted. Reflection from such a rough surface is called diffuse reflection and appears matte.

Explanation:

hi po I hope it's help you

7 0
3 years ago
Which planet has the most gravity and why
djyliett [7]

heya mate

the planet Jupiter has most gravity due to its size

7 0
3 years ago
Read 2 more answers
Which of the following describes resistance force?
Verdich [7]
Force applied by the machine to over come resistance
5 0
3 years ago
Read 2 more answers
Two charges, each of 2.9 microC are placed at two corners of a square 50cm on a side, If the charges are on one side of the squa
anyanavicka [17]

Answer:

The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

Explanation:

Given that,

First charge q_{1}= 2.9\mu C

Second chargeq_{2}= 2.9\mu C

Distance between two corners r= 50 cm

We need to calculate the electric field due to other charges at one corner

For E₁

Using formula of electric field

E_{1}=\dfrac{kq}{r'^2}

Put the value into the formula

E_{1}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\sqrt{2}\times10^{-2})^2}

E_{1}=52200=52.2\times10^{3}\ N/C

For E₂,

Using formula of electric field

E_{1}=\dfrac{kq}{r^2}

Put the value into the formula

E_{2}=\dfrac{9\times10^{9}\times2.9\times10^{-6}}{(50\times10^{-2})^2}

E_{2}=104400=104.4\times10^{3}\ N/C

We need to calculate the horizontal electric field

E_{x}=E_{1}\cos\theta

E_{x}=52.2\times10^{3}\times\cos45

E_{x}=36910.97=36.9\times10^{3}\ N/C

We need to calculate the vertical electric field

E_{y}=E_{2}+E_{1}\sin\theta

E_{y}=104.4\times10^{3}+52.2\times10^{3}\sin45

E_{y}=141310.97=141.3\times10^{3}\ N/C

We need to calculate the net electric field

E_{net}=\sqrt{E_{x}^2+E_{y}^2}

Put the value into the formula

E_{net}=\sqrt{(36.9\times10^{3})^2+(141.3\times10^{3})^2}

E_{net}=146038.69\ N/C

E_{net}=146.03\times10^{3}\ N/C

We need to calculate the direction of electric field

Using formula of direction

\tan\theta=\dfrac{141.3\times10^{3}}{36.9\times10^{3}}

\theta=\tan^{-1}(\dfrac{141.3\times10^{3}}{36.9\times10^{3}})

\theta=75.36^{\circ}

Hence, The magnitude of the electric field and direction of electric field are 146.03\times10^{3}\ N/C and 75.36°.

4 0
3 years ago
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