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3 years ago

A skateboard has a mass of 2kg and accelerates a rate of 6 m /s 2 what is the amount of UN balanced force

Physics

2 answers:

mr Goodwill [35]3 years ago

8 0

F = m • a

The NET force on the skateboard is 12 N.

8090 [49]3 years ago

7 0

Answer:

12n

Explanation:

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The speed of light in vacuum is exactly 299,792,458 m/s. A beam of light has a wavelength of 651 nm in vacuum. This light propag

sukhopar [10]

Answer:

$v=2.58\times10^8m/s$

Explanation:

The index of refraction is equal to the speed of light c in vacuum divided by its speed v in a substance, or $n=\frac{c}{v}$ . For our case we want to use $v=\frac{c}{n}$ , which for our values is equal to:

$v=\frac{c}{n}=\frac{299792458m/s}{1.16}=258441774.138m/s$

Which we will express with 3 significant figures (since a product or quotient must contain the same number of significant figures as the measurement with the <em>least</em> number of significant figures):

$v=2.58\times10^8m/s$

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3 years ago

Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

Explanation:

Given that,

Positive charge = 24.00 μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

$\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})$

$\theta=\sin^{-1}(\dfrac{1}{4})$

$\theta=14.47^{\circ}$

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

$E=\dfrac{2k\lambda}{r}\times2\cos\theat$

Put the value into the formula

$E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}$

$E=408094.00\ N/C$

$E=4.08\times10^{5}\ N/C$

Hence, The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

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3 years ago

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lapo4ka [179]

Answer:

1 percent

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Instantaneous speed?

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The correct answer my friend would be C

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