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3 years ago

Absolute magnitude and apparent magnitude are the same thing. True False

Physics

2 answers:

Furkat [3]3 years ago

6 0

False because absolute magnitude is the real magnitude but apparent is not

Digiron [165]3 years ago

3 0

False because those are not the same thing

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While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction b

Citrus2011 [14]

Answer:

$\theta = 66.7 degree$

Explanation:

since force is applied downwards at some angle with the horizontal

so here we will have

$F_n = mg + Fsin\theta$

now we know that the box will not move if applied force is balanced by frictional force on it

so we will have

$Fcos\theta = \mu F_n$

$F cos\theta = \mu (mg + F sin\theta)$

$F(cos\theta - \mu sin\theta) = \mu mg$

$F = \frac{\mu mg}{cos\theta - \mu sin\theta}$

so here we can say

$cos\theta - \mu sin\theta > 0$

$tan\theta = \frac{1}{\mu}$

$\theta = tan^{-1}\frac{1}{\mu}$

$\theta = tan^{-1}(\frac{1}{0.43})$

$\theta = 66.7 degree$

6 0

3 years ago

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.50 mm. A 25

torisob [31]

Answer:

The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

Explanation:

Given that,

Area =7.70 cm²

Distance = 1.50 mm

Potential difference = 25.0 V

Suppose we find the electric field between the plates, the surface charge density, the capacitance and the charge on each plates.

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{V}{d}$

Put the value into the formula

$E=\dfrac{25.0}{1.50\times10^{-3}}$

$E=16666.66\ V/m$

We need to calculate the charge density

Using formula of charge density

$\sigma=E\times\epsilon_{0}$

Put the value into the formula

$\sigma=16666.66\times8.85\times10^{-12}$

$\sigma=1.474\times10^{-7}\ C/m^2$

We need to calculate the capacitance

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.60\times10^{-4}}{1.50\times10^{-3}}$

$C=4.484\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

$q=CV$

Put the value into the formula

$q=4.484\times10^{-12}\times25.0$

$q=112.1\times10^{-12}\ C$

Hence, The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

8 0

3 years ago

Does a black hole have 2 sides that pull light together?

Ostrovityanka [42]

Check this Light doesn't have mass or gravity right?
So if it doesn't have mass or gravity so light can only affect objects with mass
Does that make sense?
The black hole has gravity and remember light doesn't have gravity so does it affect the light?
To answer that yes, and since light doesn't have gravity it gets "pulled" into the black hole

I hope this helps you

8 0

3 years ago

If youre going 70 miles an hour for 70 miles, how long would it take you to travel those 70 miles

Anna11 [10]

490 miles in total ur welcome

6 0

3 years ago

Monochromatic light of wavelength 687 nm is incident on a narrow slit. On a screen 1.65 m away, the distance between the second

Sophie [7]

Answer:

a ) 1.267 radian

b ) 1.084 10⁻³ mm

Explanation:

Distance of screen D = 1.65 m

Width of slit d = ?

Wave length of light λ = 687 nm.

Distance of second minimum fro centre y = 2.09 cm

Angle of diffraction = y / D

= 2.09 /1.65

= 1.267. radian

Angle of diffraction of second minimum

= 2 λ / d

so 2 λ / d = 1.267

d = 2 λ / 1.267 = (2 x 687 ) /1.267 nm

=1084.45 nm = 1.084 x 10⁻³ mm.

3 0

3 years ago