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finlep [7]
3 years ago
15

Which changes would most likely limit the growth of a plant-eating animal population

Physics
1 answer:
BabaBlast [244]3 years ago
8 0
Probably the changes in environment....
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static electricity and friction

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In the circuit diagram, what does the symbol made of two long lines and two short lines with a positive and a negative sign at e
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For those seeking for the answer, its a source of electrical energy.

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Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential
mestny [16]

Answer:

Resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

Explanation:

When the two resistors are in series, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B.

Given that V₁ + V₂ = 6.0 V and V₁ = 4.0 V,

V₂ = 6.0 V - V₁ = 6.0 V - 4.0 V = 2.0 V

Also, let the current in series be I.

So, V₁ = IR₁ and V₂ = IR₂

I = V₁/R₁ and I = V₂/R₂

equating both expressions, we have

V₁/R₁ = V₂/R₂

4.0 V/R₁ = 2.0 V/R₂

dividing through by 2.0 V, we have

2/R₁ = 1/R₂

taking the reciprocal, we have

R₂ = R₁/2

R₁ = 2R₂

From the parallel connection, let V₁ = voltage in resistor A and R₁ = resistance of resistor A and V₂ = voltage in resistor B and R₂ = resistance of resistor B. Since it is parallel, V₁ = V₂ = V = 6.0 V

Also, V₂ = I₂R₂ where I₂ = current in resistor B = 2.0 A and R₂ = resistance of resistor B

So, R₂ = V₂/I₂

= 6.0 V/2.0 A

= 3.0 Ω

R₁ = 2R₂

= 2(3.0 Ω)

= 6.0 Ω

So, resistance of resistor A = 6.0 Ω and resistance of resistor B = 3.0 Ω

6 0
3 years ago
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Fittoniya [83]

Answer:

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Explanation:

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2 years ago
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The magnetic field at the center of a 1.0-cm-diameter loop is 2.5 mT.
olga nikolaevna [1]

Answer:

(a) The current in the wire is 19.89 A

(b) The distance from the wire is 0.159 cm

Explanation:

Given;

magnetic field, B = 2.5 mT

diameter of the wire, d = 1 cm

radius of the wire, r = 0.5 cm = 0.005 m

(a) The current in the wire is calculated as;

I = \frac{2Br}{\mu_0} \\\\I = \frac{2\times 2.5 \times 10^{-3} \times 0.005 }{4\pi \times 10^{-7} } \\\\I = 19.89 \ A

(b) The distance from the wire where the magnetic field is 2.5 mT is calculated as;

B = \frac{\mu_0 I}{2\pi d} \\\\where;\\\\d \ is \ the \ distance \ from \ the \  wire\\\\d = \frac{\mu_0 I}{2\pi B} = \frac{4 \pi \times 10^{-7} \times 19.89}{2\pi \times 2.5 \times 10^{-3}}  = 0.00159 \ m = 0.159 \ cm

6 0
3 years ago
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