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IRINA_888 [86]
3 years ago
15

What is the equivalent resistance?​

Physics
1 answer:
mote1985 [20]3 years ago
7 0

Answer:

C 3.33 Ω

Explanation:

From the equation given    1 / Rtot = 1/10 + 1/5 =  3/10

 then Rtot = 10/3 = 3.33 Ω

As an aside:

When there are only two resistors in parallel, the equivalent R is

=    R1*R2 / (R1 + R2) =  10*5 / (10+5) = 50 / 15 = 3.33 Ω

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How does climate change affect the ocean??
Debora [2.8K]
It raises the water level because of the melting glaciers. 
6 0
4 years ago
A particle leaves the origin with an initial velocity v⃗ =(2.40 m/s)xˆv→=(2.40 m/s)x^ , and moves with constant acceleration a⃗
shtirl [24]

Answer:

distance stop 1.52m,

velocity  4.0 m/s y^

Explanation:

The movement of the particle is two-dimensional since it has acceleration in the x and y axes, the way to solve it is by working each axis independently.

a) At the point where the particle begins to return its velocity must be zero (Vfx = 0)

     Vfₓ = V₀ₓ + aₓ t  

     t = -  V₀ₓ/aₓ

     t = - 2.4/(-1.9)

     t=  1.26 s

At this time the particle stops, let's find his position

     X1 = V₀ₓ t + ½ aₓ t²

     X1= 2.4 1.26 + ½ (-1.9) 1.26²

     X1= 1.52 m

At this point the particle begins its return

b) The velocity has component x and y

   As a section, the X axis x Vₓ = 0 m/s is stopped, but has a speed on the y axis

    Vfy= Voy + ay t

    Vfy= 0 + 3.2 1.26

    Vfy = 4.0 m/s

the velocity is  

    V = (0 x^ + 4.0 y^) m/s

c) In order to make the graph we create a table of the position x and y for each time, let's start by writing the equations

      X = V₀ₓ t+ ½  aₓ t²

      Y = Voy t + ½  ay t²

      X= 2.4 t + ½ (-1.9) t²

      Y= 0 + ½ 3.2 t²

      X= 2.4 t – 0.95 t²

      Y=   1.6 t²

With these equations we build the table to graph, for clarity we are going to make two distance graph with time, one for the x axis and another for the y axis

                       Chart to graph

              Time (s)     x(m)            y(m)

                 0                0               0

                 0.5             0.960       0.4

          1       1.45          1.6

                 1.50      1.46      3.6

                2.00      1.00      6.4

7 0
3 years ago
Two sound waves, A and B, are traveling at the same speed. Wave A has a wavelength of 50 cm and a frequency of 7000 Hz. Wave B h
vekshin1

Answer:

D. 100 cm

Explanation:

The speed of a wave is the wavelength times the frequency.

v = λf

Wave A and B have the same speed, so:

λf = λf

(50 cm) (7000 Hz) = λ (3500 Hz)

λ = 100 cm

6 0
3 years ago
You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

3 0
4 years ago
KI + Cl2 ---&gt; KCl + I2<br> Balance the single replacement chemical reaction.
snow_lady [41]

Answer:

2KI   +  Cl₂   →   2KCl     +    I₂

Explanation:

The reaction equation is given as:

        KI   +  Cl₂   →    KCl     +    I₂

The problem at hand is to balance this chemical reaction. To solve this problem we use a mathematical approach;

        aKI   +  bCl₂   →    cKCl     +    dI₂

 Conserving K : a = c

                     I :  a  = 2d

                   Cl : 2b = c  

 Now let a = 1, c  = 1 , d  = \frac{1}{2}, b   = \frac{1}{2}, ;

   Multiply through by 2;

             a  = 2, b = 1 , c = 2, d  = 1

 

            2KI   +  Cl₂   →   2KCl     +    I₂

 

5 0
3 years ago
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