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Rudik [331]
2 years ago
10

Groups of students are making measurements of small cars rolling across a table. the class has measuring tapes with markings in

inches and centimeters. the teacher asks that students make all the measurements in centimeters. why would the teacher make a decision like this?
Physics
1 answer:
TiliK225 [7]2 years ago
4 0

To find the accurate measurement of small cars, the teacher asks students to make all the measurements in centimeters.

Centimeters Measurements:

  • A centimeter is a metric unit of measurement used for measuring the length of an object, It is written as cm
  • Centimeter is one hundredth of a meter, 1 meter is 0.01 cm.

Inches measurements:

  • An inch can be defined as a unit of length in the customary system of measurement. Length in inches is either represented by in or ''.
  • 1 meter is equal to 39.37 inches

here, the cars are small objects.

The number of centimeters is always bigger,

because a centimeter unit is smaller than an inch unit, and it takes more of them when we are measuring.

Hence,

To find the accurate measurement of small cars, the teacher asks students to make all the measurements in centimeters.

Learn more about accurate measurement here:

<u>brainly.com/question/4119127</u>

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#SPJ4

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Answer:

a) 2nd case rate of rotation gives the greater speed for the ball

b) 1534.98 m/s^2

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Explanation:

(a) v = ωR

when R = 0.60, ω = 8.05×2π

v = 0.60×8.05×2π = 30.34 m/s

Now in 2nd case

when R = 0.90, ω = 6.53×2π

v = 0.90×6.53×2π = 36.92 m/s

6.35 rev/s gives greater speed for the ball.

(b) a = ω^2 R = (8.05×2π)^2 )(0.60) = 1534.98 m/s^2

(c) a = ω^2 R = (6.53×2π)^2 )(0.90) = 1515.05 m/s^2

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3 years ago
A 20 cm-radius ball is uniformly charged to 71 nC.
artcher [175]

Answer:

Part a)

\rho = 2.12\mu C/m^3

Part b)

q_1 = 1.11 nC

q_2 = 8.88 nC

q_3 = 71 nC

Part c)

E_1 = 3996 N/C

E_2 = 7992 N/C

E_3 = 15975 N/C

Explanation:

Part a)

As we know that charge density is the ratio of total charge and total volume

So here the volume of the charge ball is given as

V = \frac{4}{3}\pi R^3

V = \frac{4}{3}\pi(0.20)^3

V = 0.0335 m^3

now the charge density of the ball is given as

\rho = \frac{71 nC}{0.0335} = 2.12\mu C/m^3

Part b)

Now the charge enclosed by the surface is given as

q = \rho V

at radius of 5 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.05)^3

q = 1.11 nC

at radius of 10 cm

q = (2.12 \mu C/m^3)(\frac{4}{3}\pi(0.10)^3

q = 8.88 nC

at radius of 20 cm

q = 71 nC

Part c)

As we know that electric field is given as

E = \frac{kq}{r^2}

so we have electric field at r = 5 cm

E_1 = \frac{(9\times 10^9)(1.11 nC)}{0.05^2}

E_1 = 3996 N/C

electric field at r = 10 cm

E_2 = \frac{(9\times 10^9)(8.88 nC)}{0.10^2}

E_2 = 7992 N/C

electric field at r = 20 cm

E_3 = \frac{(9\times 10^9)(71 nC)}{0.20^2}

E_3 = 15975 N/C

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Answer:

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The answer would be:

A space station orbiting Earth.

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3 years ago
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