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Tasya [4]
3 years ago
15

3.Three resistors of 25.0Ω, 30.0Ω, and 40.0Ω are in a series circuit with a 6.0-volt battery. What is the current in the circuit

?
please answer by steps
Physics
1 answer:
AURORKA [14]3 years ago
5 0

Answer:

Current = 0.063 Amperes

Explanation:

Let the three resistors be R1, R2, and R3 respectively.

Given the following data;

R1 = 25.0Ω,

R2 = 30.0Ω

R3 = 40.0Ω

Voltage = 6 Volts

First of all, we would determine the equivalent or total resistance;

Total resistance (in series) = R1 + R2 + R3

Total resistance = 25.0Ω + 30.0Ω + 40.0Ω

Total resistance = 95 Ω

Next, we find the current flowing through the circuit;

Voltage = current * resistance

Substituting into the formula, we have;

6 = current * 95

Current = 6/95

Current = 0.063 Amperes

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February. we will be in the future types, I

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A parachutist of mass 100 kg falls from a height of 500 m. Under realistic conditions, she experiences air resistance. Based on
antoniya [11.8K]
Given:
mass: 100 kg
height: 500 m
1 kJ = 1000 J
gravity = 9.8 m/s²

velocity before impact: v = √2gh ; v = √2 * 9.8 m/s² * 500 m ; v = 98.99494 m/s

KE = 1/2 m v²
KE = 1/2 * 100 kg * (98.99494 m/s)²
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5 0
3 years ago
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Chemical weathering is greatest under conditions of
Pavel [41]
<h2>Answer: higher mean annual rainfall and temperatures. </h2>

Explanation:

Chemical weathering is the set of destructive processes through which rocky materials go trhough. These processes cause changes in the color, texture, composition, firmness and shape of the material.

It should be noted that this happens when the rocks come into contact with atmospheric agents such as oxygen and carbon dioxide.

Another important aspect is that rocks are able to break up more easily thanks to this type of weathering, since <u>the mineral grains within the rock lose adherence and dissolve better under the action of some physical agents</u>, such as <u>humidity (rainfall included) and temperature</u>.

Therefore:

Chemical weathering is greatest under conditions of <u>higher mean annual rainfall and temperatures. </u>

5 0
3 years ago
The angular momentum of a flywheel having a rotational inertia of 0.140 kg ·m2 about its central axis decreases from 3.00 to 0.8
Rasek [7]

Answer

given,

I = 0.140 kg ·m²

decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.

a) \tau = \dfrac{\Delta L}{\Delta t}

   \tau = \dfrac{0.8-3}{1.5}

        τ = -1.467 N m

b) angle at which fly wheel will turn

   \theta= \omega t +\dfrac{1}{2}\alpha t^2

   \theta= \dfrac{L}{I} t +\dfrac{1}{2}\dfrac{\tau}{I}t^2

   \theta= \dfrac{3}{0.14}\times 1.5+\dfrac{1}{2}\dfrac{-1.467}{0.14}\times 1.5^2

        θ = 20.35 rad

c) work done on the wheel

     W = τ x θ

     W = -1.467 x 20.35 rad

    W = -29.86 J

d) average power of wheel

    P_{av} =-\dfrac{W}{t}

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7 0
3 years ago
Read 2 more answers
A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos
lukranit [14]

Answer:

The inductor contains N = 523962.32 loops  

Explanation:

From the question we are told that

     The capacitance of the capacitor is  C =  286nF = 286 * 10^{-9} \  F

      The resonance frequency is  f = 18.0 kHz =  18*10^{3} Hz

       The diameter is  d =  1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m

       The  of the air-core inductor is l = 12 \ m

        The permeability of free space is  \mu_o = 4 \pi *10^{-7} \ T \cdot m/A

 

Generally the inductance of this air-core inductor is mathematically represented as

              L =  \frac{\mu_o * N^2 \pi d^2}{4 l}

This inductance can also be mathematically represented as

               L = \frac{1}{w^2}

Where w is the angular speed mathematically given as

             w = 2 \pi f

So

            L =  \frac{1}{4 \pi ^2 f^2}

Now equating the both formulas for inductance

         \frac{\mu_o * N^2 \pi d^2}{4 l}  =  \frac{1}{4 \pi ^2 f^2}

making N the subject of  the formula

              N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C}  }

              N =  \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }

             

 Substituting value

            N =  \frac{1}{ 3.142  * 18*10^{3} * 0.00011 }  \sqrt{\frac{12}{ 3.142  * 4 \pi *10^{-7}* 286 *10^{-9}} }

              N = 523962.32 loops  

4 0
3 years ago
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