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Helga [31]
2 years ago
13

Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit

Physics
2 answers:
stepan [7]2 years ago
8 0

Answer:

0.03 A

Explanation:

Explain, step by step, how to calculate the amount of current (I) that will go through the resistor in this circuit

<u>0.03 A</u>

anygoal [31]2 years ago
3 0

Answer:

0.03 A

Explanation:

From the question given above, the following data were obtained:

Voltage (V) = 12 V

Resistor (R) = 470 Ω

Current (I) =?

From ohm's law, the voltage, current and resistor are related by the following formula:

Voltage = current × resistor

V = IR

With the above formula, we can obtain the current in the circuit as follow:

Voltage (V) = 12 V

Resistor (R) = 470 Ω

Current (I) =?

V = IR

12 = I × 470

Divide both side by 470

I = 12 / 470

I = 0.03 A

Thus, the current in the circuit is 0.03 A

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Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

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3 years ago
PLEASE HELP ME IM TIMED
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Yeah it’s the mantle
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3 years ago
How much work is done when a porcelain vase with mass of 2.5 kg is lifted up 3 m to a shelf? (HINT- Weight can be used as Force
alexdok [17]
Work = force x distance
F= 2.5
D= 3
Work = 2.5 x 3 =7.5
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8 0
3 years ago
A hiker walks 11 km due north from camp and then turns and walks 11 km due east. What is the magnitude of the displacement (on a
sattari [20]

Answer:

16 km

Explanation:

Drawing a right triangle to model the problem helps. I started by drawing the lines of the triangle to model the hiker's journey- a vertical straight line for 11 km north and then a horizontal line connected to the top of it for 11 km east; I then drew the hypothenuse to connect the two lines.

The hypothenuse is what we have to solve for, so we will use the Pythagorean Theorem, a^2 + b^2 = c^2. Since both distances are 11 km both a and b in the equation are 11.

11^2 + 11^2 = c^2

121 + 121 = c^2

242 = c^2

c = 15.56

Rounding the answer makes it 16 km for the hiker's magnitude of displacement.

5 0
3 years ago
A meter stick moves parallel to its axis with speed of 0.96 c relative to you. What would you measure for the length of the stic
Fed [463]

Answer:

The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

Explanation:

Given that,

Relative speed of stick v= 0.96 c

Speed of light c= 2.99793\times10^{8}\ m/s

Proper length of stick = 1 m

We need to calculate the length of the stick

Using formula of length

\Delta l=\Delta l_{0}\sqrt{(1-\dfrac{v^2}{c^2})}

Put the value into the formula

\Delta l=1\sqrt{1-\dfrac{(0.96)^2c^2}{c^2}}

\Delta l=1\sqrt{1-(0.96)^2}

\Delta l=0.28\ m

We need to calculate the time the stick take to move

Using formula of time

t=\dfrac{\Delta l}{v}

Put the value into the formula

t=\dfrac{0.28}{0.96\times(2.99793\times10^{8})}

t=9.72\times10^{-10}\ sec

t=0.97\ ns

Hence, The length of the stick is 0.28 m.

The time the stick take to move is 0.97 ns.

7 0
3 years ago
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