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2 years ago

What is the equivalent resistance of a circuit that contains three 10.0 Ω

Physics

2 answers:

yanalaym [24]2 years ago

8 0

Answer:

O D.30.0 Ω

Explanation:

this is the correct answer!

Cerrena [4.2K]2 years ago

8 0

Question :

What is the equivalent resistance of a circuit that contains three 10.0 Ω resistors connected in series to a 6.0 V battery?

Options given :

A. 0.600 Ω
Β. 60.0 Ω
C. 16.0 Ω
D. 30.0 Ω

Solution :

As we know that,

When a circuit is connected in series resistance is calculated by :

Rs = R1 + R2 + R3 .... Rn

Here,

R is resistance

As a circuit contains three resistors. So R1 , R2 and R3 would be :

R1 = 10 Ω
R2 = 10 Ω
R3 = 10 Ω

Now,

>> Rs = 10 + 10 + 10

>> Rs = 30 Ω

Therefore,

Option D is correct.

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3 years ago

Read 2 more answers

A horizontal pipe of inner diameter 2.2 cm carries water with a density of 1000.0 kg/m3 flowing at a rate of 1.5 kg/s. If the pi

EleoNora [17]

The speed of the water in the wider part will be 1.194 m/sec. Speed is a time-based quantity. Its SI unit is m/sec.

<h3> What is speed?</h3>

Speed is defined as the rate of change of the distance or the height attained.

The given data in the problem is;

The initial diameter is, $\rm d_1 = 2.2 \ cm$

initial radius,

$r_1 = \frac{d_1}{2} \\\\ r_1 = \frac{2.2}{2} \\\\ r_1 = 1.1\ cm$

The initial crossection area;

$\rm A_1 = \pi r_1^2 \\\\ \rm A_1 = 3.14 \times (1.1\times 10^{-2})^2 \\\\ \rm A_1 =3.8 \times 10^{-4} \ m^2$

The final crossection area;

$\rm A_2 = \pi r_2^2 \\\\ \rm A_2 = 3.14 \times ( 2 \times 10^{-2})^2 \\\\ \rm A_2 = 12.56 \ m^2$

The initial flow rate is;

R = density ×velocity ×area

$\rm R = \rho A V \\\\ 1.5 = 1000 \times V_1 \times 3.8 \times 10^{-4} \\\\ V_1 = 3.947 \ m/sec$

The speed of the water in the wider part will be;

From the continuity equation;

$\rm A_1 V_1 = A_2V_2 \\\\\ 3.8 \times 10^{-4} \times 3.947 = 12.56 \times 10^{-4} \times V_2 \\\\ V_2= 1.194 \ m/sec$

Hence, the speed of the water in the wider part will be 1.194 m/sec.

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2 years ago

A mug rests on an inclined surface, as shown in (Figure 1) , θ=17∘.

Anna [14]

Refer to the figure shown below.

g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.

Let μ = the coefficient of static friction.

The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N

For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058

The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306

Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306