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3 years ago

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Can someone please describe surface tension to me. Please don't give me the dictionary definition. Thank you

1 answer:

11111nata11111 [884]3 years ago

6 0

Surface tension - My definition -

It's exactly what it says - The tension of a surface with a liquid (such as water), caused by the attraction of the surfaces layer ---- I hope this helps ---- I actually did research it and got some of this from a dictionary, but I changed some of it, too.... Sorry if this doesn't help :)

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An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the

Wewaii [24]

Answer:

$\frac{dB}{dt} = 3.03 \times 10^6 T/s$

Explanation:

As we know that the power emitted by the source is given as

$P = 170 W$

now we know that

$P = \frac{N}{t} (\frac{hc}{\lambda})$

now we know that energy density is given as

$u = \frac{B^2}{2\mu_0} + \frac{\epsilon_0 E^2}{2}$

now we have

$E = B c$

$u = \frac{B^2}{2\mu_0}$

intensity is defined as

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

now we have

$\frac{I}{c} = u = \frac{B^2}{2\mu_0}$ [/tex]

now we have

$\frac{dB}{dt} = \omega B$

$\frac{dB}{dt} = \frac{2\pi c B}{\lambda}$

$\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}$

here we have

$I = \frac{P}{4\pi r^2}$

$I = \frac{170}{4\pi (410)^2}$

$I = 8.05 \times 10^{-5}$

now we have

$\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}$

$\frac{dB}{dt} = 3.03 \times 10^6 T/s$

4 0

3 years ago

An ant crawls along a sidewalk with a velocity of 0.1 m/s in a direction that is 45° relative to edge of the sidewalk. If it has

Marat540 [252]

Answer:14 s

Explanation:

Given

Velocity of ant is 0.1 m/s in a direction $45^{\circ}$

if it has traveled 1 m perpendicular to the edge of the sidewalk

i.e. from diagram

$\sin 45=\frac{1}{L}$

$L=\sqrt{2}$

$time=\frac{Distance}{speed}$

$t=\frac{\sqrt{2}}{0.1}$

$t=14.14 s$

3 0

3 years ago

¿Por qué crees que se debe estudiar la física?

Akimi4 [234]

No se porque?

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6 0

3 years ago

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My Notes A container is divided into two equal compartments by a partition. One compartment is initially filled with helium at a

Ostrovityanka [42]

Answer:

$final-temperature = T_{f} = 252.51K$

Explanation:

we can solve this problem by using the first law of thermodynamics.

$\Delta U= Q-W$

Q= heat added

U= internal energy

W= work done by system

$E_{final}= E_{initial}$

<u> $C_{v} (N_{2}) C_{v}(He) T_{f} =C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}$ (1)</u>

$C_{v}(N_{2})=1.04\frac{KJ}{Kg K}$

$C_{v}(He)=5.193\frac{KJ}{Kg K}$

now

From equation 1

$T_{f}=\frac{C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}}{C_{v} (N_{2}) C_{v}(He)}$

$T_{f} = \frac{315\times1.04+5.193\times240}{1.04+5.193}$

$T_{f} = 252.51K$

4 0

4 years ago

The position x of a particle moving along x axis varies with time t as x = Asin(wt) , where A and w are constants . The accelera

Elodia [21]

Let's see

$\\ \rm\Rrightarrow x=Asin(\omega t)\dots(1)$

Now we know the formula of acceleration

$\\ \rm\Rrightarrow \alpha=-A\omega^2sin(\omega t)$

$\\ \rm\Rrightarrow \alpha=-Asin(\omega t)\times \omega^2$

From eq(1)

$\\ \rm\Rrightarrow \alpha=-x\omega^2$

Or

$\\ \rm\Rrightarrow \alpha=-\omega^2x$

6 0

2 years ago

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