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mixer [17]
2 years ago
13

In a series circuit with three resistors connected to a power source, what would happen to the equivalent resistance and current

if you removed any one of the resistors? The equivalent resistance would decrease and the current would increase. The equivalent resistance would increase and the current would increase. The equivalent resistance would decrease and the current would decrease. The equivalent resistance would increase and the current would decrease.
Physics
2 answers:
amm18122 years ago
8 0

Answer:

the answer to this is A

Explanation:

got it right on the Assignment.

igor_vitrenko [27]2 years ago
3 0

<u>Answer:</u>

<h2><em></em></h2>

<em>The equivalent resistance would decrease and the current would increase.</em>

<em></em>

<u><em>Mark me as the brainlyest</em></u>

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A 15.0 kg fish swimming at 1.10 m/s suddenly gobbles up a 4.50 kg fish that is initially stationary. Ignore any drag effects of
Nana76 [90]

Answer

given.

Mass of big fish = 15 Kg

speed of big fish = 1.10 m/s

mass of the small fish = 4.50 Kg

speed of the fish after eating small fish =?

a) using conservation of momentum

m₁v₁ + m₂v₂ = (m₁+m₂) V

15 x 1.10 + 4.50 x 0 = (15 + 4.5)V

16.5 = 19.5 V

V = 0.846 m/s

b) Kinetic energy before collision

KE_1 = \dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2

KE_1 = \dfrac{1}{2}\times 15 \times 1.1^2 + \dfrac{1}{2}m_2\times 0^2

KE₁ = 9.075 J

Kinetic energy after collision

KE_2= \dfrac{1}{2}(15+4.5)\times 0.846^2

KE₂ = 6.98 J

Change in KE = 6.98 - 9.075 = -2.096 J

hence,

mechanical energy was dissipated during this meal = -2.096 J

5 0
3 years ago
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
AURORKA [14]

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

5 0
3 years ago
The op amp in this circuit is ideal. R3 has a maximum value of 100 kΩ and σ is restricted to the range of 0.2 ≤ σ ≤ 1.0. a. Calc
Firlakuza [10]

I have attached the circuit image missing in the question.

Answer:

A) The range of vo is; -6.6V≤ vo ≤-1V

B) σ = 0.1861

Explanation:

A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.

Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0

So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0

Simplifying further; -25 vg - vΔ = 0

From the question, vg = 40mV = 0.04 V

So - 25(0.04) = vΔ

So: vΔ = - 1 V

Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0

So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0

So RΔ = 100kΩ or 100,000Ω from the question.

So, substituting for RΔ, we get,

[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0

Let's put the value of - 1 for vΔ as gotten before.

So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0

Now let's make vo the subject of the equation to get;

-1 - vo = (1 - σ)[2 + (1/σ)]

-1 - vo = 2 - 2σ + (1/σ) - 1

-vo = 1 + 2 - 2σ + (1/σ) - 1

-vo = 2 - 2σ + (1/σ)

vo = - 1 (2 - 2σ + (1/σ))

When σ = 0.2; vo = - 1(2 - 0.4 + 5) =

- 1 x 6.6 = - 6.6V

Also when σ = 1;

vo = - 1(2 - 2 + 1) = - 1V

Therefore, the range of vo is;

- 6.6V ≤ vo ≤ - 1V

B) it will saturate at vo = - 7V

So, from;

vo = - 1 (2 - 2σ + (1/σ))

-7 = - 1 (2 - 2σ + (1/σ))

Divide both sides by (-1)

7 = (2 - 2σ + (1/σ))

Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)

Multiply each term by α to get;

5σ = - 2σ^(2) + 1

So 2σ^(2) + 5σ - 1 = 0

Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861

8 0
4 years ago
4. A disobedient student dropped his Physics textbook (mass 0.1kg) from the window (15m above the ground). How fast was it going
vaieri [72.5K]

Answer:

v= 17.15 m/s

Explanation:

mass of the book=0.1 Kg

height above ground, h= 15 m

Using conservation of energy

Potential energy is converted into kinetic energy

mgh = \frac{1}{2}mv^2

v=\sqrt{2gh}

v=\sqrt{2\times 9.8\times 15}

v=\sqrt{294}

v= 17.15 m/s

Hence, the book will hit the ground at the speed of 17.15 m/s.

8 0
3 years ago
The centers of two 15.0 kg spheres are separated by 3.00 m. The magnitude of the gravitational force between the two spheres is
kompoz [17]
 we have to use newtons law of gravitation which is
F=GMm/r^2 
G=6.67 x 10^<span>-11N kg^2/m^2
</span>M=<span>(15kg)
</span>m=15 kg
r=(3.0m)^2<span> 
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<span>=(6.67 x 10^-11N kg^2/m^2)(15kg)(15kg)/(3.0m)^2 </span>
=1.67 x 10^-9N 
7 0
3 years ago
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