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2 years ago

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A uniform charge density of 600 nC/m3 is distributed throughout a spherical volume (radius = 14 cm). Consider a cubical (3.2 cm

along the edge) surface completely inside the sphere. Determine the electric flux through this surface.

1 answer:

jolli1 [7]2 years ago

5 0

Answer: The electric flux through this surface is 2.22 Nm^2/C

Explanation: Please see the attachments below

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Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proport

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Answer:

$[F]=[MLT^{-2}]$

Explanation:

Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :

F = m × a

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m is the mass of the object

a is the acceleration due to gravity

We need to find the dimensions of force. The dimension of force m and a are as follows :

$[m]=[M]$

$[a]=[LT^{-2}]$

So, the dimension of force F is, $[F]=[MLT^{-2}]$ . Hence, this is the required solution.

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If the work done to stretch an ideal spring by 4.0 cm is 6.0 j, what is the spring constant (force constant) of this spring?

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Answer:

The spring constant is 3750 N/m

Explanation:

Use the following two relationships:

(Work) = (Force) x (Displacement)

(Force) = (Spring constant) x (Displacement)

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(Spring constant) = (Force) / (Displacement) = (Work) / (Displacement)^2

(Spring constant) = 6.0 kg.(m^2/s^2) / 0.0016 m^2 = 3750 N/m

The spring constant is 3750 N/m

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3 years ago

The five general principles from the APA are meant to __________. A. be enforceable rules B. be posted in every office C. guide

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2 years ago

An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

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2 years ago

I need help this is extremely hard

telo118 [61]

Answer:

For the first one shown, the answer is Directly Proportional, The second one is Inversly Proportional, and the last is fourtl times the original value

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2 years ago