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insens350 [35]
2 years ago
6

Your best friend has very low energy levels and complains of not being able to sleep at night. Determine the BMI score that woul

d most likely contribute to this problem.
14
20
22
23
Physics
1 answer:
morpeh [17]2 years ago
4 0

Answer: 14

Explanation: 20-23 is normal.

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3 years ago
a race car is traveling at a speed of 80.0m/s on a circular race track of radius 450m what is the centripetal acceleration
slega [8]

Answer:

The answer of this question is =1.258*10-4

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2 years ago
Which of these is an experiment generally regarded as being first carried out by James Joule?
kipiarov [429]
The correct answer for this question is this one: "measuring the temperature increase of water from doing work stirring it." This experiment is generally regarded as being first carried out by James Joule is this one, <span>measuring the temperature increase of water from doing work stirring it.</span>
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Even though the Sun has a greater mass than Earth, the Moon orbits Earth because it's ___________________ to Earth than to the S
Pani-rosa [81]
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3 0
3 years ago
Read 2 more answers
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

3 0
3 years ago
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