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Lady_Fox [76]
2 years ago
10

For tax and accounting purposes, corporations depreciate the value of equipment each year. One method used is called "linear dep

reciation," where the value decreases over time in a linear manner. Suppose that two years after purchase, an industrial milling machine is worth $820,000, and five years after purchase, the machine is worth $430,000. Find a formula for the machine value V.
Physics
1 answer:
Sladkaya [172]2 years ago
7 0

Answer:

V=-130000x+1080000

Explanation:

<u>Linear Dependence </u>

Some variables are known or assumed to have linear dependence which means the graph of the ordered pairs (x,V) is a straight line.

If we know two points of the line, we can come up with the exact equation and therefore make predictions for other values of x

The linear depreciation gives us these points (2,820000) and (5,430000)

The general equation of the line is

V=mx+b

Where V is the machine value and x is the  number of years after purchase. We need to find the values of m and b.

Replacing the first point

820000=m(2)+b

2m+b=820000

Replacing the second point

5m+b=430000

Subtracting them  

-3m=390000

m=-130000

Replacing in any of the equations, say, the first one

2(-130000)+b=820000

Solving for b

b=820000+260000

b=1080000

The formula for the machine value V is  

\boxed{V=-130000x+1080000}

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2 years ago
1. A2 .7-kg copper block is given an initial speed of 4.0m/s on a rough horizontal surface. Because of friction, the block final
tatyana61 [14]

Answer:

A. Increase in temperature is 0.0176 degree Celsius. b. the remaining energy will be lost.

Explanation:

The mass of copper block = 7kg

Initial speed = 4.0 m/s

Specific heat of copper = 0.385 j/g degree Celcius.

a. The increase in temperature is calculated below:

\text{Change in kinetic energy} = \frac{1}{2}mv^{2} \\=  \frac{1}{2} \times 2.7 \times 4^{2} = 21.6 J \\

85% of energy is converted into internal energy.

mc\DeltaT = 21.6 \times 0.85 \\2.7 \times 385 \times \Delta T = 21.6 \times 0.85 \\\Delta T = 0.0176 degree \ celsius

b. The remaining  15 per cent of kinetic energy will be lost and it will be changed into other forms.

7 0
2 years ago
A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer
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The initial volume of the gas is
V_i = 600 cm^3
while its final volume is
V_f = 200 cm^3
so its variation of volume is
\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3

The pressure is constant, and it is
p=400 kPa = 400 \cdot 10^3 Pa

Therefore the work done by the gas is
W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J
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Q=+100 J

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\Delta U = Q-W = 100 J - (-160 J)=+260 J
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The maximum efficiency of a heat engine operating between 2 degrees c and 200 degrees c will be about:
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2 years ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

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<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
2 years ago
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