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Zina [86]
3 years ago
14

Help wit these questions someone.

Physics
1 answer:
romanna [79]3 years ago
7 0

In series circuit, Req = R₁ + R₂ + R₃ + ···

In parallel circuit, \frac{1}{Req}  = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} +...

<h3>Q7.</h3>

total resistance in the upper branch = R₂ + R₃ = R₂ + 2

\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}

R₂ + 2 = 12

R₂ = 10Ω

<h3>Q8.</h3>

\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

\frac{1}{Req} = \frac{1}{2+1} + \frac{1}{4}

Req = 1.7Ω

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The whole body metabolic rate of a bottlenose dolphin is 8000kcalsday. The dolphin weighs 190kg. What is the mass specific metab
Komok [63]

Answer: 42.1

Explanation:

Mass specific metabolic rate of a dolphin can be defined as the rate at which the dolphin consume energy per unit mass of body weight.

R = E/M

Where R = mass specific metabolic rate

E = Energy consumption = 8000kcalsday

M = mass = 190kg

R = 8000kcalsday/190kg

R = 42.1

4 0
3 years ago
The human nerve cells have a net negative charge and the material in the interior of the cell is a good conductor. if the cell h
mash [69]

The magnitude and direction (inward or outward) of the net flux through the cell boundary is - 0.887 wb.m².

<h3>What is flux?</h3>

Flux describes any effect that appears to pass or travel through a surface or substance.

The magnitude and direction (inward or outward) of the net flux through the cell boundary is calculated as follows;

Ф = Q/ε

where;

  • Q is net charge
  • ε is permittivity of free space

Φ = (-7.85 x 10⁻¹²)/(8.85 x 10⁻¹²)

Φ = - 0.887 wb.m²

Learn more about flux here: brainly.com/question/10736183

#SPJ1

6 0
2 years ago
If it takes 726 watts of power to move a mass 36 meters in 14 seconds, then what is the magnitude of the object’s mass?
Colt1911 [192]

Answer:

20.2

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4 0
2 years ago
A 3.6-volt battery is used to operate a cell phone
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Explanation :

It is given that,

Potential energy, V=3.6\ V

Power dissipated,  P=0.064\ Watt

We know that the power dissipated is given by :

P=VI

I is the current passing through the phone.

I=\dfrac{P}{V}

I=\dfrac{0.064\ W}{3.6\ V}

I=0.017\ A

or

I = 0.018 A

Hence, the current that passes through the phone is (1) 0.018 A.

3 0
3 years ago
Read 2 more answers
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