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3241004551 [841]
3 years ago
11

The distance vs. time graph of a car moving at constant speed should be a straight line. Why do the data points in the graph plo

tted from observing the motion not fall right on the line? A graph with 7 points and a line segment representing a straight line. Horizontal axis is labeled Time, in seconds, and extends from 0 to 7. Vertical axis is labeled Distance, in meters, and extends from 0 to 50. Approximate coordinates of plotted points are begin ordered pair 1 comma 8 end ordered pair comma begin ordered pair 2 comma 13 end ordered pair comma begin ordered pair 3 comma 21 end ordered pair comma begin ordered pair 4 comma 28 end ordered pair comma begin ordered pair 5 comma 35 end ordered pair comma begin ordered pair 6 comma 39 end ordered pair comma begin ordered pair 7 comma 49 end ordered pair. Line segment has positive slope, and it begins at the point begin ordered pair 0 comma 0 end ordered pair, and it ends at the point begin ordered pair 7 comma 49 end ordered pair. The measurements are inexact. Data points can never look like they fall right on a straight line. The distance is not exactly proportional to time even if the speed is exactly constant. The distance and time are proportional only if the correct units of meters and seconds are used.
Physics
1 answer:
iragen [17]3 years ago
6 0

Answer:

The measurements are inexact

Explanation:

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How would you find the average speed of a cyclist throughout an entire race
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Speed = distance / time
If you input your numbers into this equation you will be able to find the cyclists average speed
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3 years ago
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
bekas [8.4K]

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

5 0
3 years ago
How you do this resistor in circuit question?
lawyer [7]
If Resistors are in series= The equivalent is the sum.

E.g R1 and R2 in series,  R = R1 + R2.

If in Parallel, equivalent is Product/sum.

E.g If R1 and R2 in parallel,  R = (R1*R2)/(R1+R2)

1.)    60 is parallel with 40 and both are then in series with 20.

60//40   = (60*40)/(60+40) = 2400/100 = 24 

Now the 24 is in series with the 20 

R = 24 + 20 = 44 ohms.

2.)    80 is in series with 40 and both are then in parallel with 40.

Solving the series,  R = 80 + 40 =120.

Parallel:   120//40  =     (120*40)/(120+40) =  4800/160 = 30 
Equivalent Resistance = 30 ohms.

3.) 100 is in parallel with 100 and both are then in series with the parallel of 50 and 50.

The 1st parallel  =  (100*100)/(100+100) = 10000/200 = 50

The 2nd parallel  =  (50*50)/(50+50)  = 2500/100   = 25.

Solving the series  = 50 + 25 =75 ohms.

Cheers.
3 0
3 years ago
Estimate the kinetic energy (in GJ) of a 98,000 metric ton aircraft carrier moving at a speed of at 32 knots. You will need to l
kiruha [24]

Answer:

K = 13.276 GJ

Explanation:

Given:

Mass of the aircraft (m) = 98,000 metric ton

Speed of the aircraft (v) = 32 knots

Kinetic energy of the aircraft (K) = ?

We know that,

1 metric ton = 1000 kg

∴ 98,000 metric ton = 98,000 × 1000 = 9.8 × 10⁷ kg

Also, 1 knot = 1 nautical mile per hour

So, 32 knots = 32 nautical miles per hour

Converting nautical miles per hour to meter per second using the following conversions, we get:

1 nautical mile = 1852 m

1 hour = 3600 s

Therefore, 1 nautical mile per hour = \frac{1852}{3600}\ m/s

So, 32 nautical miles per hour = 32\times \frac{1852}{3600}=16.46\ m/s

Therefore, the mass and speed of the aircraft are:

m=9.8\times 10^{7}\ kg\\v=16.46\ m/s

Now, kinetic energy is given as:

K=\frac{1}{2}mv^2

Plug in the above values and solve for 'K'. This gives,

K=\frac{1}{2}\times 9.8\times 10^7\times (16.46)^2\\\\K=4.9\times 270.9316\times 10^7\ J\\\\K=13.276\times 10^2\times 10^7\ J\\\\K=13.276\times 10^9\ J

Now, we know that, 1 GJ = 10⁹ J

Therefore, the value of 'K' in terms of GJ is given as:

K = 13.276 GJ

8 0
3 years ago
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