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alisha [4.7K]
2 years ago
10

Q|C (a) You need a 45-ω resistor, but the stockroom has only 20-ω and 50-ω resistors. How can the desired resistance be achieved

under these circumstances?
Physics
1 answer:
NNADVOKAT [17]2 years ago
3 0
  1. In order to achieve the desired resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with the 20 Ohms resistors.
  2. In order to get a 35 Ohms resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with two 20 Ohms resistors that are connected in parallel.

<h3>How to achieve the desired resistance under these circumstances?</h3>

In order to achieve the desired resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with the 20 Ohms resistors.

Mathematically, the total equivalence resistance of two resistors that are connected in parallel is given by:

1/Rt = 1/R₁ + 1/R₂

1/Rt = 1/50 + 1/50

1/Rt = 2/50

1/Rt = 1/25

Rt = 25 Ohms.

Next, we would connect this 25 Ohms resistor in series with the 20 Ohms resistor:

R₃ = 20 + Rt

R₃ = 20 + 25

R₃ = 45 Ohms.

<h3>Part B.</h3>

In order to get a 35 Ohms resistance under the given circumstances, we would connect two 50 Ohms resistors in parallel and then connect it in series with two 20 Ohms resistors that are connected in parallel.

1/Rt = 1/R₁ + 1/R₂

1/Rt = 1/50 + 1/50

1/Rt = 2/50

1/Rt = 1/25

Rt = 25 Ohms.

1/R't = 1/R₁ + 1/R₂

1/R't = 1/20 + 1/20

1/R't = 2/20

1/R't = 1/10

R't = 10 Ohms.

Next, we would connect the 25 Ohms resistor in series with the 10 Ohms resistor:

R₃ = 10 + Rt

R₃ = 10 + 25

R₃ = 35 Ohms.

Read more on resistors in parallel here: brainly.com/question/15121871

#SPJ4

Complete Question:

You need a 45-ω resistor, but the stockroom has only 20-ω and 50-ω resistors.

(a) How can the desired resistance be achieved under these circumstances?

(b) What can you do if you need a 35-ω resistor?

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Delicious77 [7]

(a) The required magnitude of the electric field when the point charge is an electron is 5.57 x 10⁻¹¹ N/C.

(b) The required magnitude of the electric field when the point charge is an proton is 1.02 x 10⁻⁷ N/C.

<h3>Magnitude of electric field </h3>

The magnitude of electric field is given by the following equation.

F = qE

But F = mg

mg = qE

E = mg/q

where;

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  • q is charge of the particle
<h3>For an electron</h3>

E = (9.11 x 10⁻³¹ x 9.8)/(1.602 x 10⁻¹⁹)

E = 5.57 x 10⁻¹¹ N/C

<h3>For proton</h3>

E = (1.67 x 10⁻²⁷ x 9.8)/(1.602 x 10⁻¹⁹)

E = 1.02 x 10⁻⁷ N/C

Thus, the required vertical electric field is greater when the charge is proton.

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4 0
2 years ago
The scale is 1:30 and their are 10 boulders. How many boulders are their in the real pond
Alenkasestr [34]

Answer:

the correct answer is C

Explanation:

When we express that the scale is 1:30 we mean that the objects of the realization are reduced by a factor of 30 in the graph, for example a distance of 30 cm in the graph is represented by a distance of 1 cm.

Therefore something that in the graph has n value to bring it to real size must be multiplied by the scale.

Applying this to our case if there is

               10 boulder on the chart

in reality there are #_boulder = 10 30

                                #_boulder = 300 boulder

so the correct answer is C

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4 years ago
An airplane flying at an altitude of 6 miles passes directly over a radar antenna. When the airplane is 10 miles away (s = 10),
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Answer:

Explanation:

Given

altitude of the Plane h=6\ miles

When Airplane is s=10\ miles away

Distance is changing at the rate of \frac{\mathrm{d} s}{\mathrm{d} t}=290\ mph

From diagram we can write as

h^2+x^2=s^2

differentiate above equation w.r.t time

2h\frac{\mathrm{d} h}{\mathrm{d} t}+2x\frac{\mathrm{d} x}{\mathrm{d} t}=2s\frac{\mathrm{d} s}{\mathrm{d} t}

as altitude is not changing therefore \frac{\mathrm{d} h}{\mathrm{d} t}=0

0+x\frac{\mathrm{d} x}{\mathrm{d} t}=s\frac{\mathrm{d} s}{\mathrm{d} t}

at s=10\ miles\ and\ h=6\ miles

substitute the value we get x=\sqrt{10^2-6^2}=8\ miles

8\times \frac{\mathrm{d} x}{\mathrm{d} t}=10\times 290

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7 0
3 years ago
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