Ask question

Ask question

All categories

3 years ago

5

A sphere with radius 10 cm is filled with a uniform charge distribution. The magnitude of the electric field at a point 5 cm fro

m the center of the sphere is 3014 N/coul. Use this fact to calculate the charge density rho inside the sphere.

1 answer:

Rom4ik [11]3 years ago

4 0

Answer:15 cm

Explanation: u had 10cm. them u had another 5cm

You might be interested in

What is the displacement of the runner, whose velocity versus time graph is shown in the Figure, in the first 15.5 s?

muminat

Answer:

10 displacement of the runner

6 0

3 years ago

Aliun [14]

Answer:

the correct answer is c, they will accelerate away from each other at different speeds. the 80kg will go faster due to less mass

6 0

3 years ago

Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),

salantis [7]

Answer:

The drag coefficient is $D_z = 1.30512$

Explanation:

From the question we are told that

The density of air is $\rho_a = 1.21 \ kg/m^3$

The diameter of bottom part is $d = 0.15 \ m$

The power trend-line equation is mathematically represented as

$F_{\alpha } = 0.9226 * v^{0.5737}$

let assume that the velocity is 20 m/s

Then

$F_{\alpha } = 0.9226 * 20^{0.5737}$

$F_{\alpha } = 5.1453 \ N$

The drag coefficient is mathematically represented as

$D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

Where

$F_{\alpha }$ is the drag force

$\rho$ is the density of the fluid

$v$ is the flow velocity

A is the area which mathematically evaluated as

$A = \pi r^2 = \pi \frac{d^2}{4}$

substituting values

$A = 3.142 * \frac{(0.15)^2}{4}$

$A = 0.0176 \ m^2$

Then

$D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

$D_z = 1.30512$

3 0

3 years ago

A mass of 100 g stretches a spring 5 cm. If the mass is set in motion from its equilibrium position with a downward velocity of

Rudiy27

Answer:

Explanation:

Given

mass of spring $m=100\ gm$

extension in spring $x=5\ cm$

downward velocity $v=70\ cm/s$

Position in undamped free vibration is given by

$u(t)=A\cos \omega _0t+B\sin \omega _0t$

where $\omega _0^2=\frac{k}{m}$

also $\frac{k}{m}=\frac{g}{L}$

$\omega _0^2=\frac{k}{m}=\frac{9.8}{0.05}$

$\omega _0=14$

$u(t)=A\cos(14t)+B\sin(14t)$

it is given

$u(0)=0$

$u'(0)=70\ cm/s$

substituting values we get

$A=0$

$u(t)=B\sin (14t)$

$u'(t)=14B\cos (14t)$

$70=14B$

$B=\frac{10}{2}$

$B=5$

$u(t)=5\sin (14t)$

3 0

4 years ago

If an 800 kg roller coaster is at the top of its 50 m high track, it will have a potential energy pf 392,000 J and a kinetic ene

Viefleur [7K]

I just woke him crying crying laughing crying and laughing I’m so mad he got me blocked him lol I got a hold on her phone the answer is 0 m

5 0

3 years ago