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3 years ago

Which bond(s) shown are double bonds?

Physics

1 answer:

Nikitich [7]3 years ago

3 0

Answer:

I'm pretty sure its the 2nd one.

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Give you point but please i need help in physcis

Anastasy [175]

Explanation:

BRAINILIEST PLEASE

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3 years ago

Do you divide mass by volume to get density

bogdanovich [222]

Density is the mass of an object divided by its volume. So the answer would be Yes. Hope it helps! (:

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3 years ago

An inductor has an inductance of 0.12 mH. Calculate the number of turns per unit length of this inductor if it has a total of 50

SSSSS [86.1K]

Answer:

Explanation:

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3 years ago

Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

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4 years ago

An electric field from a charge has a magnitude of 1.5 × 104 N/C at a certain location that points inward. If another charge wit

zysi [14]

Answer:

-0.045 N, they will attract each other

Explanation:

The strength of the electrostatic force exerted on a charge is given by

$F=qE$

where

q is the magnitude of the charge

E is the electric field magnitude

In this problem,

$q=3.0\cdot 10^{-6}C$

$E=-1.5\cdot 10^4 N/C$ (negative because inward)

So the strength of the electrostatic force is

$F=(-3.0\cdot 10^{-6}C)(1.5\cdot 10^4 N/C)=-0.045 N$

Moreover, the charge will be attracted towards the source of the electric field. In fact, the text says that the electric field points inward: this means that the source charge is negative, so the other charge (which is positive) is attracted towards it.

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4 years ago

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