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4 years ago

A clock on a moving train runs __________ an identical clock at rest.

Physics

1 answer:

katovenus [111]4 years ago

4 0

It depends who is watching it.

-- A passenger on the train says that his clock is normal, but the clock at the station he just passed is running slow.

-- A ticket agent at the station says that his clock is normal, but the clock on the train that just rolled through is running slow.

-- They're both correct, but ...

-- The difference is too small for either of them to notice, unless the train is going faster than about half the speed of light.

-- It isn't.

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Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7

sleet_krkn [62]

Answer : $\underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Explanation :

Given that,

Charge of particle 1 = $-1.50\times10^{-7} c$

Distance x = 6 cm

Charge of particle 2 = $1.50\times10^{-7} c$

Distance x = 27 cm

Total distance = $\dfrac{6+27}{2}$

$r = 16.5\ cm$

Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, $r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm$

Formula of electric field

$E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}$

Now, the the electric field due to particle 1

$\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Similarly, the electric field due to particle 2

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Resultant Electric field

$\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}$

$\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Hence, this is the required answer.

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3 years ago

How long would it take light from the sun to reach mercury?

BARSIC [14]

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4 years ago

What is Albert Einstein’s contribution to the understanding of nuclear energy?

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3 years ago

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A small rubber wheel is used to drive a large pottery wheel. The two wheels are mounted so that their circular edges touch. The

Goshia [24]

Answer:

$0.629\ \text{rad/s}^2$ counterclockwise

$9.98\ \text{s}$

Explanation:

$r_1$ = Small drive wheel radius = 2.2 cm

$\alpha_1$ = Angular acceleration of the small drive wheel = $8\ \text{rad/s}^2$

$r_2$ = Radius of pottery wheel = 28 cm

$\alpha_2$ = Angular acceleration of pottery wheel

As the linear acceleration of the system is conserved we have

$r_1\alpha_1=r_2\alpha_2\\\Rightarrow \alpha_2=\dfrac{r_1\alpha_1}{r_2}\\\Rightarrow \alpha_2=\dfrac{2.2\times 8}{28}\\\Rightarrow \alpha_2=0.629\ \text{rad/s}^2$

The angular acceleration of the pottery wheel is $0.629\ \text{rad/s}^2$ .

The rubber drive wheel is rotating in clockwise direction so the pottery wheel will rotate counterclockwise.

$\omega_i$ = Initial angular velocity = 0

$\omega_f$ = Final angular velocity = $60\ \text{rpm}\times \dfrac{2\pi}{60}=6.28\ \text{rad/s}$

t = Time taken

From the kinematic equations of linear motion we have

$\omega_f=\omega_i+\alpha_2t\\\Rightarrow t=\dfrac{\omega_f-\omega_i}{\alpha_2}\\\Rightarrow t=\dfrac{6.28-0}{0.629}\\\Rightarrow t=9.98\ \text{s}$

The time it takes the pottery wheel to reach the required speed is $9.98\ \text{s}$

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3 years ago

An adult with a BMI of 25 to 30 is considered

soldier1979 [14.2K]

Option A overweight

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3 years ago

Read 2 more answers