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3 years ago

Which of the following statements is always true?

Physics

1 answer:

Gwar [14]3 years ago

7 0

Answer: B

Explanation:

An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. An unbalanced force is an unopposed force that causes a change in motion.

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Does any body know anything about Criminal justice??

Natasha_Volkova [10]

Criminal justice is the delivery of justice to those who have committed crimes. The criminal justice system is a series of government agencies and institutions whose goal is to identify and catch the law-breakers and to inflict a form of punishment on them.

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2 years ago

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A vehicle moves in a straight line with an

nordsb [41]

The solution to your problem is as follows:

<span>5km/h2 x 1 h/60 min x 1 min/60 sec x 1000m/km = 1.39 meters per second change each second.
</span>
Therefore, the speed change each second is <span>1.39 meters per second change each second.

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3 years ago

Why do we eat today

earnstyle [38]

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3 years ago

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A 6157 N piano is to be pushed up a(n) 2.41 m frictionless plank that makes an angle of 21.9 ◦ with the horizontal. Calculate th

xz_007 [3.2K]

Answer:

Work done, W = 5534.53 J

Explanation:

It is given that,

Force acting on the piano, F = 6157 N

It is pushed up a distance of 2.41 m friction less plank.

Let W is the work done in sliding the piano up the plank at a slow constant rate. It is given by :

$W=Fd\ sin\theta$

Since, $F=F\ sin\theta$ (in vertical direction)

$W=6157\times 2.41\ sin(21.9)$

W = 5534.53 J

So, the work done in sliding the piano up the plank is 5534.53 J. Hence, this is the required solution.

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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

5 0

3 years ago

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