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mestny [16]
3 years ago
11

The three notches in the graph occur where the driver changed gears. Describe the changes in velocity and acceleration of the ca

r while in first gear. Is the acceleration just before a gear change larger or smaller than the acceleration just after the change?
Explain your answer
Physics
1 answer:
kaheart [24]3 years ago
6 0
<h3>Answer : </h3><h3 /><h3>A ) The larger gear can be moved by applying a relatively small force on the smaller gear.</h3>

<h3>B ) The force applied on the smaller gear is transmitted without any loss to the larger gear .</h3><h3 /><h3>C ) the direction of motion can be changed without changing the direction of the applied force .</h3>

D ) the system would continue to move without any further, after and initial force has set in motion.

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A mercury thermometer is constructed as
Tamiku [17]

Answer:

The change in height of the mercury is approximately  2.981 cm

Explanation:

Recall that the formula for thermal expansion in volume is:

\frac{\Delta V}{V_0} =\alpha_V\, \Delta\, T\\\Delta V = V_0\,\, \alpha_V\,\,\Delta C

from which we solved for the change in volume \Delta V due to a given change in temperature \Delta T

We can estimate the initial volume of the mercury in the spherical bulb of diameter 0.24 cm ( radius R = 0.12 cm) using the formula for the volume of a sphere:

V_0=\frac{4}{3} \pi \, R^3\\V_0=\frac{4}{3} \pi \, (0.12\,cm)^3\\V_0=0.007238\,cm^3

Therefore, the change in volume with a change in temperature of 36°C becomes:

\Delta V = V_0\,\, \alpha_V\,\,\Delta C\\\Delta V = 0.007238229\, cm^3\,(0.000182)\,(36)\\\Delta V=0.0000474248\, cm^3

Now, we can use this difference in volume, to estimate the height of the cylinder of mercury with diameter 0.0045 cm (radius r= 0.00225 cm):

V_{cyl}=\pi r^2\,h\\h =\frac{V_{cyl}}{\pi r^2} \\h=\frac{0.0000474248\, cm^3}{\pi \, (0.00225\,cm)^2} \\h=2.98188 \,cm

8 0
3 years ago
Ummm Help? What best describes the expectancy theory of motivation?
iogann1982 [59]

Answer:

I strongly feel its 1.

Explanation:

the expectancy theory of motivation is basically getting an reward for a great accomplishment so 1 would make the most sense in this question, lmk if its wrong

4 0
3 years ago
An example of a stable system
yulyashka [42]
Here's a perfect example of a stable and unstable system.

3 0
3 years ago
If the acceleration of a moving object is zero, which of the following best describes its motion
podryga [215]
If the acceleration of a moving object is zero, the object remains at a constant speed (without friction)
7 0
3 years ago
Read 2 more answers
At a particular instant, a proton at the origin has velocity &lt; 5e4, -2e4, 0&gt; m/s. You need to calculate the magnetic field
vesna_86 [32]

Answer:

9.7\times 10^{-5} T

Explanation:

Velocity =5\times 10^4i-2\times 10^4j

r=0.03i+0.05j

r=\mid r\mid=\sqrt{(0.03)^2+(0.05)^2}=0.058

v=\mid V\mid=\sqrt{(5\times 10^4)^2+(-2\times 10^{4})^2}=5.39\times 10^{2}

We know that

B=\frac{mv}{qr}

Where q=1.6\times 10^{-19} C

Mass of proton=1.67\times 10^{-27} kg

Using the formula

B=\frac{1.67\times 10^{-27}\times 5.39\times 10^2}{1.6\times 10^{-19}\times 0.058}

B=9.7\times 10^{-5} T

3 0
3 years ago
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