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Suppose you are in total equilibrium in water (in other words you are floating and not moving in any direction)What could you do

Eva8 [605]

Explanation:

According to newtons first law of motion:

'' a body will continue in its state of rest or uniform motion along a path unless it is acted upon by an external force".

A body in equilibrium that is floating will be stable and not move in any direction. Even if it moves, the motion will be constant wouldn't change.

To move the body in any direction, one has to swim.
Swimming is the application of an external force to counter the balanced forces at equilibrium on a body.
This works when the net external force is greater than that the balanced forces.

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Newton brainly.com/question/11411375

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5 0

3 years ago

PHYSICS PLEASE HELP

WINSTONCH [101]

Explanation:

Newton’s First Law of Motion - if an object is at rest, it takes un-

balanced forces to make it move. Conversely, if an object is moving

it takes an unbalanced force to make it change it’s direction or speed.

Newton was the first to see that such apparently diverse phenomena as a satellite moving near the Earth's surface and the planets orbiting the Sun operate by the same principle: Force equals mass multiplied by acceleration, or F=ma.

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7 0

3 years ago

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What causes a compass needle to point to geographic north?

morpeh [17]

I believe it’s “C”
Hope that’s help you(:

8 0

3 years ago

A nucleus with a mass number of 64 has a mean radius of about: _________.

dangina [55]

Answer:

The correct option is A

Explanation:

From the question we are told that

The mass number is $A= 64$

Generally the mean radius is mathematically evaluated as

$R = R_o A^{\frac{1}{3} }$

Here $R_o$ is a constant with a value $R_o =1.2*10^{-15}$

So

$R = 1.2*10^{-15} * 64^{\frac{1}{3} }$

$R = 4.8 *10^{-15}$

$R = 4.8\ fm$

5 0

3 years ago

A battery with an emf of 12.0 V shows a terminal voltage of 11.7 V when operating in a circuit with two lightbulbs, each rated a

wariber [46]

<h2>Answer:</h2>

0.46Ω

<h2>Explanation:</h2>

The electromotive force (E) in the circuit is related to the terminal voltage(V), of the circuit and the internal resistance (r) of the battery as follows;

E = V + Ir --------------------(a)

Where;

I = current flowing through the circuit

But;

V = I x Rₓ ---------------------(b)

Where;

Rₓ = effective or total resistance in the circuit.

<em>First, let's calculate the effective resistance in the circuit:</em>

The effective resistance (Rₓ) in the circuit is the one due to the resistances in the two lightbulbs.

Let;

R₁ = resistance in the first bulb

R₂ = resistance in the second bulb

Since the two bulbs are both rated at 4.0W ( at 12.0V), their resistance values (R₁ and R₂) are the same and will be given by the power formula;

P = $\frac{V^{2} }{R}$

=> R = $\frac{V^{2} }{P}$ -------------------(ii)

Where;

P = Power of the bulb

V = voltage across the bulb

R = resistance of the bulb

To get R₁, equation (ii) can be written as;

R₁ = $\frac{V^{2} }{P}$ --------------------------------(iii)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iii) as follows;

R₁ = $\frac{12.0^{2} }{4}$

R₁ = $\frac{144}{4}$

R₁ = 36Ω

Following the same approach, to get R₂, equation (ii) can be written as;

R₂ = $\frac{V^{2} }{P}$ --------------------------------(iv)

Where;

V = 12.0V

P = 4.0W

Substitute these values into equation (iv) as follows;

R₂ = $\frac{12.0^{2} }{4}$

R₂ = $\frac{144}{4}$

R₂ = 36Ω

Now, since the bulbs are connected in parallel, the effective resistance (Rₓ) is given by;

$\frac{1}{R_{X} }$ = $\frac{1}{R_1}$ + $\frac{1}{R_2}$ -----------------(v)

Substitute the values of R₁ and R₂ into equation (v) as follows;

$\frac{1}{R_X}$ = $\frac{1}{36}$ + $\frac{1}{36}$

$\frac{1}{R_X}$ = $\frac{2}{36}$

Rₓ = $\frac{36}{2}$

Rₓ = 18Ω

The effective resistance (Rₓ) is therefore, 18Ω

<em>Now calculate the current I, flowing in the circuit:</em>

Substitute the values of V = 11.7V and Rₓ = 18Ω into equation (b) as follows;

11.7 = I x 18

I = $\frac{11.7}{18}$

I = 0.65A

<em>Now calculate the battery's internal resistance:</em>

Substitute the values of E = 12.0, V = 11.7V and I = 0.65A into equation (a) as follows;

12.0 = 11.7 + 0.65r

0.65r = 12.0 - 11.7

0.65r = 0.3

r = $\frac{0.3}{0.65}$

r = 0.46Ω

Therefore, the internal resistance of the battery is 0.46Ω

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4 years ago

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