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2 years ago

Three 1.5V cells are connected in series in a circuit. What is the total potential difference?

Physics

1 answer:

lorasvet [3.4K]2 years ago

4 0

Answer:

4.5V

Explanation:

1.5x3= 4.5

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How does energy travel in a mechanical wave?

3241004551 [841]

Answer:

In a medium

Explanation:

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3 years ago

Needddd helppppppp!!!

yulyashka [42]

Answer:

2/9 times as strong.

Explanation:

From the question given above, the following assumptions were made:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

Next, we shall obtain an expression to determine the new force. This can be obtained as follow:

F = GMm / r²

Cross multiply

Fr² = GMm

Divide both side by Mn

G = Fr² / Mm

Since G is constant, then we have

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Finally, we shall determine the new force as follow:

Initial mass of 1st planet (M₁ ) = M

Initial mass of 2nd planet (m₁ ) = m

Initial distance apart (r₁) = r

Initial Force of attraction (F₁) = F

Final mass of 1st planet (M₂) = 2M

Final mass of 1st planet (m₂) = constant = m

Final distance apart (r₂) = 3r

Final force of attraction (F₂) =?

F₁r₁² / M₁m₁ = F₂r₂² / M₂m₂

Fr² / Mm = F₂ × (3r)² / 2M × m

Fr² / Mm = F₂ × 9r² / 2Mm

Cross multiply

Fr² × 2Mm = F₂ × 9r² × Mm

Divide both side by 9r² × Mm

F₂ = Fr² × 2Mm / 9r² × Mm

F₂ = F × 2 / 9

F₂ = 2/9 F

Thus, the new force is 2/9 times the original force i.e 2/9 times as strong.

4 0

2 years ago

An initially uncharged air-filled capacitor is connected to a 3.013.01 V charging source. As a result, 9.91×10−59.91×10−5 C of c

Nookie1986 [14]

Answer:

Explanation:

7 0

2 years ago

A real power supply can be modeled as an ideal EMF of 60 Volts in series with an internal resistance. The voltage across the ter

lara31 [8.8K]

Answer:

5 ohms

Explanation:

Given:

EMF of the ideal battery (E) = 60 V

Voltage across the terminals of the battery (V) = 40 V

Current across the terminals (I) = 4 A

Let the internal resistance be 'r'.

Now, we know that, the voltage drop in the battery is given as:

$V_d=Ir$

Therefore, the voltage across the terminals of the battery is given as:

$V= E-V_d\\\\V=E-Ir$

Now, rewriting in terms of 'r', we get:

$Ir=E-V\\\\r=\frac{E-V}{I}$

Plug in the given values and solve for 'r'. This gives,

$r=\frac{60-40}{4}\\\\r=\frac{20}{4}\\\\r=5\ ohms$

Therefore, the internal resistance of the battery is 5 ohms.

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3 years ago

HELP SCIENCE WORK ON MOTION DUE!

Anna71 [15]

Huh? Wheres the question at?

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3 years ago