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3 years ago

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A ring placed along y 2 + z 2 = 4, x = 0 carries a uniform charge of 5 μ C/m. ( a ) Find D at P(3,0,0) . ( b ) If two identical

point charges Q are placed at (0, - 3,0) and (0,3,0) in addition to the ring, find the value of Q such that D = 0 at P.

1 answer:

GuDViN [60]3 years ago

3 0

For right now idk :p

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Left arm has area 10.0 cm2, right has area 5.00 cm2. 100 g of water isadded to right side. a) Determine the length of the water

Masja [62]

Answer:

0.49 cm

Explanation:

We are given that

$A_1=10 cm^2$

$A_2=5 cm^2$

Mass of water,m=100 g

a. $\rho_w=1 g/cm^3$

Volume, $V=\frac{mass}{density}=\frac{100}{1}=100 cm^3$

Length of water column in the right arm= $L=\frac{V}{A_2}=\frac{100}{5}=20 cm$

b. $\rho_m=13.6 g/cm^3$

In equilibrium condition

Pressure at point A=Pressure at point B

$P+\rho_mg(h+h_2)=P+\rho_wgL$

$hA_1=h_2A_2$

$h_2=\frac{hA_1}{A_2}$

$13.6\times 9.8(h+h(\frac{A_1}{A_2}))=1\times 9.8\times 20$

$13.6\times h(\frac{A_2+A_1}{A_2})=20$

$h=\frac{20A_2}{13.6(A_2+A_1)}$

$h=\frac{20\times 5}{13.6\times (10+5)}=0.49 cm$

4 0

4 years ago

Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton

solmaris [256]

Answer:

a

$\lambda = 3.68 *10^{-36} \ m$

b

$\lambda_p = 1.28*10^{-14} \ m$

Explanation:

From the question we are told that

The mass of the person is $m = 180 \ kg$

The speed of the person is $v = 1 \ m/s$

The energy of the proton is $E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J$

Generally the de Broglie wavelength is mathematically represented as

$\lambda = \frac{h}{m * v }$

Here h is the Planck constant with the value

$h = 6.62607015 * 10^{-34} J \cdot s$

So

$\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }$

=> $\lambda = 3.68 *10^{-36} \ m$

Generally the energy of the proton is mathematically represented as

$E_p = \frac{1}{2} * m_p * v^2_p$

Here $m_p$ is the mass of proton with value $m_p = 1.67 *10^{-27} \ kg$

=> $8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2$

=> $v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }$

=> $v = 3.09529 *10^{7} \ m/s$

So

$\lambda_p = \frac{h}{m_p * v_p }$

so $\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }$

=> $\lambda_p = 1.28*10^{-14} \ m$

5 0

3 years ago

If I exert a 203 N force on a 58 kg box, what will the acceleration of the box be?

Anit [1.1K]

Answer:

<h2>3.5 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

m is the mass

f is the force

From the question we have

$a = \frac{203}{58} = \frac{7}{2} \\$

We have the final answer as

<h3>3.5 m/s²</h3>

Hope this helps you

7 0

3 years ago

A 0.20-kg object is attached to the end of an ideal horizontal spring that has a spring constant of 120 N/m. The simple harmonic

miss Akunina [59]

Answer:

0.07756 m

Explanation:

Given mass of object =0.20 kg

spring constant = 120 n/m

maximum speed = 1.9 m/sec

We have to find the amplitude of the motion

We know that maximum speed of the object when it is in harmonic motion is given by $v_{max}=A\omega$ where A is amplitude and $\omega$ is angular velocity

Angular velocity is given by $\omega=\sqrt{\frac{k}{m}}$ where k is spring constant and m is mass

So $v_{max}=A\sqrt{\frac{k}{m}}$

$A=V_{max}\sqrt{\frac{m}{k}}=1.9\times \sqrt{\frac{0.2}{120}}=0.07756 \ m$

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3 years ago

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If gravity did not affect the pain of a horizontally thrown ball the ball would

Drupady [299]

If gravity had no effect on a ball after you threw it ... and there also
were no air to slow it down ... then the ball would continue traveling
in a straight line, in whatever direction you threw it.

That's the heart and soul of Newton's laws of motion ... any object
keeps moving at the same speed, and in a straight line in the same
direction, until a force acts on it to change its speed or direction.\

If you threw the ball horizontally, then it would keep moving in the
same direction you threw it. But don't forget: The Earth is not flat.
The Earth is a sphere. So, as the ball kept going farther and farther
in the same straight line, the Earth would curve away from it, and it
would look like the ball is getting farther and farther from the ground.

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3 years ago

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