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3 years ago

11

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.032.03 times a second. A tack is stuck in the ti

re at a distance of 0.357 m0.357 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed.

1 answer:

Luda [366]3 years ago

6 0

Answer:

4.55 m/s

Explanation:

The frequency is defined as the number of rotations in one second

So, f = 2.03 Hz

r = 0.357 m

The tangential speed is

v = r ω = r x 2 x 3.14 x f = 0.357 x 2 x 3.14 x 2.03 = 4.55 m/s

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Explanation:

Work = force × distance

W = (900 N) (100 m)

W = 90,000 J

W = 90 kJ

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The climate of an area is based on A) Rivers and lakes B) natural resources. C) mountain and valleys. D) Precipitation and tempe

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It is D as it is the only answer referring to weather and climate.

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13. A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V a

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Answer:

The output power is 2 kW

Explanation:

It is given that,

Number of turns in primary coil, $N_p=250$

Number of turns in secondary coil, $N_s=500$

Voltage of primary coil, $V_p=200\ V$

Current drawn from secondary coil, $I_s=5\ A$

We need to find the power output. It is equal to the product of voltage and current. Firstly, we will find the voltage of secondary coil as :

$\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}$

$\dfrac{250}{500}=\dfrac{200}{V_s}$

$V_s=400\ V$

So, the power output is :

$P_s=V_s\times I_s$

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3 years ago

Assume that the operating cost of a certain truck (excluding driver's wages) is 12+x/6 cents per mile when the truck travels at

podryga [215]

Answer:

$x = 60 mph$

Explanation:

Given that the operating cost is

$c = 12 + \frac{x}{6}$ cents per mile

total miles covered is given as

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so total cost of drive is given as

$C = (12 + \frac{x}{6})(4)$ $

time taken by the truck to move the distance is given as

$t = \frac{400}{x}$

So total earnings of the driver is given as

$E = \frac{400}{x} \times 6$ $

now total profit of the driver is given as

$P = \frac{2400}{x} - (48 + \frac{2x}{3})$ $

to maximize the profit we have

$\frac{dP}{dx} = 0$

$-\frac{2400}{x^2} + \frac{2}{3} = 0$

so we have

$x = 60 mph$

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3 years ago

A basketball weighing 0.63 kg is dropped from a height of 6.0 meters onto a court. Use the conservation of energy equation to de

frutty [35]

Answer:

The velocity of the ball at a height of 2.0 meters above the court is approximately 8.85 m/s

Explanation:

The given parameters of the ball are;

The mass of the ball, m = 0.63 kg

The height from which the ball is dropped, h₁ = 6.0 meters

The height at which the velocity of the ball is sought, h₂ = 2 meters

The initial potential energy of the ball, P.E. = m·g·h₁ = 0.63 × 9.8 × 6.0 = 37.044

The initial potential energy of the ball, P.E.₁ = 37.044 J

The potential energy of the ball, when the ball is at 2 meters above the court, P.E.₂ = m·g·h₂ = 0.63 × 9.8 × 2.0 = 12.348

The potential energy of the ball, when the ball is at 2 meters above the court, P.E.₂ = 12.348 J

From M.E> = P.E. + K.E.

Where;

M.E = The total mechanical energy of the ball = Constant

P.E. = The potential energy of the ball

K.E. = The kinetic energy of the ball

By the conservation of energy principle, we have;

The potential energy lost by the ball = The kinetic energy gained by the ball

The potential energy lost by the ball = P.E.₁ - P.E.₂ = 37.044 - 12.348 = 24.696

The potential energy lost by the ball = 24.696 J

The kinetic energy gained by the ball = 1/2·m·v² = 1/2×0.63×v²

Where;

v = The velocity of the ball

∴ The potential energy lost by the ball at 2.0 meters above the court = 24.696 J = The kinetic energy gained by the ball at 2.0 meters above the court = 1/2×0.63×v²

24.696 J = 1/2×0.63 kg ×v²

v² = 24.696 J / (1/2×0.63 kg) = 78.4 m²/s²

∴ v = √(78.4 m²/s²) = 8.85437744847 m/s

The velocity of the ball at a height of 2.0 meters above the court, v ≈ 8.85 m/s.

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3 years ago