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3 years ago

Any projectile always has a constant vertical acceleration of _______.

Physics

1 answer:

Mashcka [7]3 years ago

8 0

Answer:

The answer to your question is: C. -9.81 m/s²

Explanation:

A. 9.81 m/s² acceleration is considered positive when it goes to the center of the earth, so this option is incorrect.

B. 0 m/s² This option is incorrect because acceleration is 0 for a linear motion without acceleration.

C. -9.81 m/s² If a projectile goes to the sky, then the acceleration will be negative.

D. It is not constant. Acceleration is constant.

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Which model could represent a neutral atom of boron?

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Answer:

model 3

Explanation:

Boron with atomic number 5 will have 3 valence electrons

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3 years ago

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The speed of a 180-g toy car at the bottom of a vertical circular portion of track is 7.75 m/s. If the radius of curvature of th

Bezzdna [24]

Answer:

11.28 N toward the center of the track

Explanation:

Centripetal force: This is the force that tend to draw a body close to the center of a circle, during circular motion.

The formula for centripetal force is given as,

F = mv²/r................................ Equation 1

Where F = force, m = mass of the toy car, v = velocity, r = radius

Given: m = 108 g = 0.108 kg, v = 7.75 m/s, r = 57.5 cm = 0.575 m

Substitute into equation 1

F = 0.108(7.75²)/0.575

F = 11.28 N

Hence the magnitude and direction of the force = 11.28 N toward the center of the track

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2 years ago

What is the kinetic energy of an object moving 40m/s with a mass of 20kg?

Svet_ta [14]

KE = (1/2) (mass) (speed)²

KE = (1/2) (20 kg) (40 m/s)²

KE = (1/2) (20 kg) (1,600 m²/s²)

KE = (10 kg) (1,600 m²/s²)

KE = 16,000 Joules

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3 years ago

3 light-years =________.

KATRIN_1 [288]

   ( 3 yr) · (186,282.397 mile/s) · (86,400 s/day) · (365 day/yr)

   = (3 · 186,282.397 · 86,400 · 365) mile

   = 1.762380502 x 10¹³ miles

   = 1.8 x 10¹³ miles (rounded to the nearest trillion miles)

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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

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3 years ago

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