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nata0808 [166]
4 years ago
15

A used car already has 40,000 miles on its odometer. The average owner will drive 12,000 miles per year. Which function best mod

els the linear relationship? y = 12,000x + 40,000 y = 40,000x + 12,000 y = 52,000x + 12,000 y = 12,000x - 40,000
Physics
1 answer:
bekas [8.4K]4 years ago
5 0

Answer:

y = 12,000x + 40,000

Explanation:

A linear relationship that would model the mileage of the car in this example is:

y=mx+q

where

y is the number of miles

x is the number of years

m is the number of miles driven per year

q is the number of miles already in the car at x=0

In this problem, we have

m = 12,000 (number of miles driven per year)

q = 40,000 (number of miles already in the car at x=0)

So substituting into the equation we get

y=12,000 x + 40,000

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A hockey puck is traveling to the left with a velocity of v=10 when it is struck by the hockey stick
Lera25 [3.4K]
We have to calculate the impulse of a hockey puck.
Imp = m * ( v 1 - v 2 ) = m * Δ v
v 1 = - 10 i m/s,
v 2 = ( 20 * cos 40° ) i + ( 20 * sin 40° ) j =
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Δ v = ( 15.32 i + 12.855 j ) - ( - 10 i ) =
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| Δv | = √ ( 25.32² + 12.855²) = √806.35 = 28.4 m/s
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Answer: D ) 5.68 N-s. 
 
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3 years ago
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3 0
3 years ago
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A student measures the speed of yellow light in water to be 2.00x10^8
max2010maxim [7]

NOTE: The given question is incomplete.

<u>The complete question is given below.</u>

A student measures the speed of yellow light in water to be 2.00 x 10⁸ m/s. Calculate the speed of light in air.

Solution:

Speed of yellow light in water (v) = 2.00 x 10⁸ m/s

Refractive Index of water with respect to air (μ) = 4/3

Refractive Index = Speed of yellow light in air / Speed of yellow light in water

Or,  The speed of yellow light in air = Refractive Index × Speed of yellow light in water

or,                                           = (4/3) × 2.00 x 10⁸ m/s

or,                                           = 2.67 × 10⁸ m/s ≈ 3.0 × 10⁸ m/s

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7 0
3 years ago
3.25 kcal is the same amount of energy as
PIT_PIT [208]
<span>1 cal = 4,185 J
1 kcal = 1*10^3 cal
or
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8 0
3 years ago
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What is the magnitude of the electric field at a distance of 89 cm from a 27 μC charge, in units of N/C?
GuDViN [60]

Answer:

306500 N/C

Explanation:

The magnitude of an electric field around a single charge is calculated with this equation:

E(r) = \frac{1}{4 \pi *\epsilon 0} \frac{q}{r^2}

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