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nlexa [21]
3 years ago
8

Calculate the change in time for each quarter of the track. Record the change in time in Table E of your Student Guide. The chan

ge in time for the first quarter is seconds. The change in time for the second quarter is seconds. The change in time for the third quarter is seconds. The change in time for the fourth quarter is seconds.

Physics
2 answers:
kipiarov [429]3 years ago
7 0

Answer:

1.39

0.78

0.64

0.54

Explanation:

The answers are an average of the difference between times. The first one is different because it's the first checkpoint.

For example;

\frac{(2.19 - 1.38) + (2.14 - 1.39) + (2.2 - 1.41)}{3}

\frac{.81 + .75 + .79}{3}

\frac{2.35}{3}

Divide 2.35 by 3, you get the answer, .78 seconds.

RUDIKE [14]3 years ago
3 0

The three pictures you attached represent three different runs of the toy car.

Here's the data for the first run:

The change in time for the first quarter is <em>1.38</em> seconds.

The change in time for the second quarter is (2.19-1.38) = <em>0.81</em> seconds.

The change in time for the third quarter is (2.80 - 2.19) =  <em>0.61</em> seconds.

The change in time for the fourth quarter is (3.31 - 2.80) = <em>0.51</em> seconds.

The quarters are all the same length, but the times are getting shorter.

The car must be getting faster !  

Maybe gravity is pulling it down, do ya reckon ?

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<span>1.) It is 6.00km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10.0km/hr (which uses up energy at the rate of 700W ), or you could walk it leisurely at 3.00km/hr (which uses energy at 290 W). A.)Which choice would burn up more energy? running or walking? b.)How much energy (in joules) would it burn? c.)Why is it that the more intense exercise actually burns up less energy than the less intense one? Follow 2 answers Report Abuse Answers billrussell42 Best Answer: running, at 10 km/hour for 6 km is 6 km / 10 km/hour = 0.6 hour or 36 min energy used is 700 watts or 700 joules/s x 36 min x 60s/min = 1.512e6 joules or 1.5 MJ walking, at 3 km/hour for 6 km 6 km / 3 km/hour = 2 hour or 120 min energy used is 290 watts or 290 joules/s x 120 min x 60s/min = 1.872e6 joules or 1.8 MJ C) should be obvious PS, this has nothing to do with potential energy. billrussell42 · 5 years ago 0 Thumbs up 1 Thumbs down Report Abuse Comment Simon van Dijk I assume the watt consumption is per hour. Then running 6km at 10.0 km/h results in 700*6/10 = 420 w.h and walking in 290*6/3 = 580 w.h So walking would burn up more energy (kwh) b) 1 kilowatt hour = 3 600 000 joules so 420 wh = 0.42 kwh = 1.51.10^6 joule c) when you put more effort in making the distance your energy is used more efficient. Simon van Dijk · 5 years ago 0 Thumbs up 2 Thumbs down Report Abuse Comment</span>
7 0
3 years ago
You are driving at 23 m/s when you see a red light ahead. Your car is capable of decelerating at a rate of 3.41 m/s2 and it take
Pavel [41]

Answer:

39.2 m

Explanation:

i think

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3 years ago
Which of the following is not a component of a normal experiment?
MaRussiya [10]

Answer:

D. Validity group

Explanation:

A validity group is not necessarily a component of a normal experiment.

A normal experiment is made up of the dependent variables, independent variables, and most times the control group.

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3 years ago
What is one possible effect of a hurricane in Florida?
Mice21 [21]

Answer:

The answer is B.

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6 0
3 years ago
Read 2 more answers
A computer hard disk starts from rest, then speeds up with an angular acceleration of 190 rad/s2rad/s2 until it reaches its fina
Otrada [13]

Answer:

962 rpm.

Explanation:

given,

angular acceleration = 190 rad/s²

initial angular speed = 0 rad/s

final angular speed = 7200 rpm

                                 =7200\times\dfrac{2\pi}{60}

                                 =754\ rad/s

we need to calculate the revolution of disk after 10 s.

time taken to reach the final angular velocity

    using equation of angular motion

 \omega_f - \omega_i = \alpha t

 754 - 0 =190\times t

    t = 4 s

rotation of wheel in 4 s

\theta =\omega_i t+  \dfrac{1}{2}\alpha t

\theta = \dfrac{1}{2}\alpha t^2

\theta = \dfrac{1}{2}\times 190 \times 4^2

 θ = 1520 rad

 \theta = \dfrac{1520}{2\pi}

 \theta =242\ rev

now, revolution of the disk in next 6 s

angular velocity is constant

\omega_f = \dfrac{\theta_f-\theta_i}{t_f-t_i}

754 = \dfrac{\theta_f-1520}{10-4}

θ_f = 6044 rad

θ_f = \dfrac{6044}{2\pi}

revolution of the computer hard disk

θ_f =  962 rpm.

total revolution of the computer disk after 10 s is equal to 962 rpm.

3 0
3 years ago
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