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Arturiano [62]
3 years ago
7

Make a graph of the data. You may use a graphing program. Think about what data should be on the y-axis and the x-axis. Be sure

to label each axis and note the units used in the measurements. Be sure to draw a smooth curve through the points. Do not just connect the dots. Upload your data and graph in the essay box below and answer the following questions. Did the car travel at a constant speed? What was the average speed of the car? What are some practical applications for determining the motion of an object?
Physics
1 answer:
Lynna [10]3 years ago
8 0

Answer:

yes it was a constant speed and the car traveled 10 meters in 20 seconds.

Explanation:

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Ya because they’re are both 50 liters
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The forklift exerts a 1,500.0 N force on the box and moves it 3.00 m forward to the stack. How much work does the forklift do ag
allochka39001 [22]
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 <span>work = force times distance
</span>replacing data you will get:
W = (1.500) (3)
W =  4.500 NM
The answer should be in NM. So it will be 4500 NM againts the force of gravity
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Clouds form from _____ temperature change, which occurs when an expanding gas cools.
IrinaVladis [17]
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5.Acarexhibitsaconstantaccelerationof0.300m/s2paralleltotheroadway.Thecar
lana [24]

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6 0
2 years ago
Two small, identical metal balls with charges 7.0 µC and 14.0 µC are held in place 1.9 m apart. In an experiment, they are conne
Nostrana [21]

The new charge on 7.0 µC ball is 10.5 µC and the new charge on 14.0 µC ball is 10.5 µC.

The change in the electrostatic force after the experiment is 0.031 N.

The given parameters:

  • Charge on first metal ball, q1 = 7.0 µC
  • Charge on second metal ball, q2 = 14 µC
  • Distance between the charges, r = 1.9 m

<h3>Charge on each ball after experiment</h3>

After the experiment the charges will be at equilibrium, and the charge on each metal ball will be equal.

Q_t = q_1 + q_2\\\\Q_t = 7 \mu C + 14 \mu C\\\\Q_t = 21 \ \mu C

Q_1 = Q_2 = \frac{Q_t}{2} = \frac{21 \mu C}{2} = 10.5 \ \mu C

The new charge on 7.0 µC ball = 10.5 µC

The new charge on 14.0 µC ball = 10.5 µC

<h3>Change in electrostatic force</h3>

F_1 =  \frac{kQ_1Q_2}{r^2} \\\\F_1 = \frac{9\times 10^9}{1.9^2} (7 \times 10^{-6} \times 14 \times 10^{-6}) \ \\\\ F_1 =  0.244 \ N\\\\for \ new \ charges;\\\\F_2 =  \frac{kQ_1Q_2}{r^2} \\\\F_2 = \frac{9\times 10^9}{1.9^2} (10.5 \times 10^{-6} \times 10.5 \times 10^{-6}) \ \\\\ F_2 =  0.275 \ N\\\\\Delta F = F_2 - F_2\\\\\Delta F = 0.275 \ N \ - \ 0.244 \ N\\\\\Delta F = 0.031 \ N

Lear more about electrostatic force here: brainly.com/question/17692887

8 0
2 years ago
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