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3 years ago

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A thin ring of charge of radius R lies in the x-y plane and is centered on the z-axis. The linear charge density, l, of the loop

depends on the in-plane angle with respect to the x-axis, f. It is given by ______________.

1 answer:

Kruka [31]3 years ago

7 0

Answer:

Linear charge density (I) = Q/2πR

Explanation:

Linear charge density (I) = charge (Q) per unit length(L)

I = Q/L

For a thin ring of charge with radius R, the length will be equal to the circumference of a circle.

Circumference of a circle = 2πR

Then, the length of the thin ring of charge is 2πR

Linear charge density (I) = Q/2πR

Therefore, for a thin ring of charge of radius R, which lies in the x-y plane and is centered on the z-axis. The linear charge density, l, of the loop is given by Q/2πR

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Find the average force exerted by the bat on the ball if the two are in contact for 0.00129 s. Answer in units of N.

Semmy [17]

Answer:

Explanation:

Given

time of contact between bat and ball is $t=0.00129 s$

suppose u is the incoming velocity and v is the final velocity after collision

$m=mass\ of\ ball$

Impulse exerted is given by change in momentum of the particle.

Initial momentum $P_i=m\times u$

Final momentum $P_f=m\times v$

Change in momentum $\Delta P=P_i-P_f$

Impulse $J=F_{avg}\cdot t=\Delta P$

$J=F_{avg}\cdot t=m(u-v)$

$F_{avg}=\frac{m(v-u)}{t}$

$F_{avg}=775.2\times m(v-u)\ N$

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4 years ago

A spaceship whose rest length is 350m has a speed of .82c

igomit [66]

Answer:

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

$u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}$

where:

v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

$u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }$

$u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }$

u'=0.9806c

Time for micrometeorite to pass spaceship can be calculated as:

$t'=\frac{length}{Relatie seed (u')}$

$t'=\frac{350}{0.9806c}$ (c = 3*10^8 m/s)

$t'=\frac{350}{0.9806* 3.0*10^{8} }$

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

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3 years ago

when an object falls to the ground, only the object moves down but the earth's motion is not noticeable.why?

vlada-n [284]

Both the object and earth pulls each other towards itself but since the mass and pulling force of objects are very small the pulling force of objects are negligible.

7 0

3 years ago

A 2.00 kg ball is thrown upward at Some unknown angle from the top of a 20.0 M high building If the initial magnitude of the vel

mixer [17]

Answer:

792 J

Explanation:

The total energy of the ball is E = U + K where U = potential energy = mgh and K = kinetic energy = 1/2mv²

E = mgh + 1/2mv² where m = mass of ball = 2.0 kg, g = acceleration due to gravity = 9.8 m/s², h = height of building = 20.0 m, v = initial velocity of ball = 20.0 m/s.

So, substituting the values of the variables into E, we have

E = mgh + 1/2mv²

= 2.00 kg × 9.8 m/s² × 20.0 m + 1/2 × 2.00 kg × (20.0 m/s)²

= 392 J + 400 J

= 792 J

6 0

3 years ago

You are trying to overhear a juicy conversation, but from your distance of 24.0m , it sounds like only an average whisper of 40.

Neporo4naja [7]

Answer:

The distance is $r_2 = 0.24 \ m$

Explanation:

From the question we are told that

The distance from the conversation is $r_1 = 24.0 \ m$

The intensity of the sound at your position is $\beta _1 = 40 dB$

The intensity at the sound at the new position is $\beta_2 = 80.0dB$

Generally the intensity in decibel is is mathematically represented as

$\beta = 10dB log_{10}[\frac{d}{d_o} ]$

The intensity is also mathematically represented as

$d = \frac{P}{A}$

So

$\beta = 10dB * log_{10}[\frac{P}{A* d_o} ]$

=> $\frac{\beta}{10} = log_{10} [\frac{P}{A (l_o)} ]$

From the logarithm definition

=> $\frac{P}{A * d_o} = 10^{\frac{\beta}{10} }$

=> $P = A (d_o ) [10^{\frac{\beta }{ 10} } ]$

Here P is the power of the sound wave

and A is the cross-sectional area of the sound wave which is generally in spherical form

Now the power of the sound wave at the first position is mathematically represented as

$P_1 = A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ]$

Now the power of the sound wave at the second position is mathematically represented as

$P_2 = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

Generally power of the wave is constant at both positions so

$A_1 (d_o ) [10^{\frac{\beta_1 }{ 10} } ] = A_2 (d_o ) [10^{\frac{\beta_2 }{ 10} } ]$

$4 \pi r_1 ^2 [10^{\frac{\beta_1 }{ 10} } ] = 4 \pi r_2 ^2 [10^{\frac{\beta_2 }{ 10} } ]$

$r_2 = \sqrt{r_1 ^2 [\frac{10^{\frac{\beta_1}{10} }}{ 10^{\frac{\beta_2}{10} }} ]}$

substituting value

$r_2 = \sqrt{ 24^2 [\frac{10^{\frac{ 40}{10} }}{10^{\frac{80}{10} }} ]}$

$r_2 = 0.24 \ m$

7 0

3 years ago