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3 years ago

I need answers for Part 2 of 2 please Help its UT homework On ramps & My HS teacher doesnt go over it

Physics

1 answer:

lora16 [44]3 years ago

3 0

OK. "Linear model" just means a straight-line graph. That's no problem here, because we know two points on the line, and when we know two points, we know (??) how to find the equation of the line.

The equation is going to be

Y = m t + b .

Y is the chicken consumption in pounds

m is the slope of the line

t is the number of years

b is the y-intercept of the line

One point: 1983. t=3, y=30.9 pounds

Other point, 1989: t=9, y=47 pounds

Slope = (47-30.9) / (9-3)

Slope = 16.1/6 = 2.683 pounds per year

Plug that number into the equation for either of the points that we know, and you find that the y-intercept b is 22.85 pounds. So ...

Y = 2.683 t + 22.85 .

(and t is the number of years after 1980)

In 1995, t=15 . Stick that into the equation, and you'll get the estimate for chicken consumption in 1995, IF your model ... based on changes from 1983 to 1989 ... is still accurate for 6 more years after 1989.

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A radar used to detect the presence of aircraft receives a pulse that has reflected off an object 5 ✕ 10−5 s after it was transm

Sunny_sXe [5.5K]

Answer:

7500 m

Explanation:

The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.

We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so

d = 2L

In a time of

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$d=vt$ (1)

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Re-arranging eq(1) and substituting, we find L:

$L=\frac{vt}{2}=\frac{(3.0\cdot 10^8)(5\cdot 10^{-5})}{2}=7500 m$

7 0

4 years ago

A block of mass

Gennadij [26K]

(a) The work done by the applied force is 26.65 J.

(b) The work done by the normal force exerted by the table is 0.

(d) The work done by the net force on the block is 26.65 J.

<h3>Work done by the applied force</h3>

W = Fdcosθ

W = 14 x 2.1 x cos25

W = 26.65 J

<h3>Work done by the normal force</h3>

W = Fₙd

W = mg cosθ x d

W = (2.5 x 9.8) x cos(90) x 2.1

W = 0 J

<h3>Work done force of gravity</h3>

The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.

<h3> Work done by the net force on the block</h3>

∑W = 0 + 26.65 J = 26.65 J

Thus, the work done by the applied force is 26.65 J.

The work done by the normal force exerted by the table is 0.

The work done by the force of gravity is 0.

The work done by the net force on the block is 26.65 J.

Learn more about work done here: brainly.com/question/8119756

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2 years ago

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nevsk [136]

Answer:

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