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2 years ago

POSSIBLE POINTS: 60

Physics

1 answer:

Dima020 [189]2 years ago

4 0

Answer:

B. 57.4 N horizontal (right)

<u>Explanation</u>:

using cos method:

$cos(35) = \frac{vertical force}{100}$

$vertical force = 57.4 N$

<em>As its vertical, the direction of this force should be up.</em>

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A6 cm object is 8 cm from a convex lens that has a focal length of 2.7 cm. The image is 4 cm from the lens. The height of the im

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7.8cm

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3 years ago

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A transformer consists of a 500 turn primary coil and a 2000-turnsecondary coil. If the current in the secondary is 3.0A, what i

Neporo4naja [7]

Answer:

The correct solution will be "12.0 A".

Explanation:

The given values are:

$N_p= 500 \ turn$

$N_s= 200 \ turn$

$I_s= 3.0 \ A$

By using the transformer formula, we get

⇒ $\frac{N_p}{N_s} =\frac{I_s}{I_p}$

⇒ $I_p = I_s\times \frac{N_s}{N_p}$

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⇒ $=3.0 \ A\times \frac{2000}{500}$

⇒ $=12.0 \ A$

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3 years ago

When sugar is poured from the box into the sugar bowl, the rubbing of sugar grains creates a static electric charge that repels

Afina-wow [57]

Answer:

2.6×10⁻³ N

Explanation:

From coulomb's law,

F = kq'q/r²................ Equation 1

Where F = Repulsive force, q' = charge on the first sugar grain, q = charge on the second sugar grain, r = distance of separation between the sugar grain, k = proportionality constant.

From the question,

since q' = q

Then,

F = kq²/r²..................... Equation 2

Given: q = 1.79×10⁻¹¹ C, r = 3.45×10⁻⁵ m,

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Substitute into equation 2

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F = 9×10⁹(3.2041×10⁻²²)/(11.9025×10⁻¹⁰)

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3 years ago

Coherent light of wavelength 525 nm passes through two thin slits that are 4.15Ã—10^(âˆ’2) mm apart and then falls on a screen 7

IRINA_888 [86]

A) 4.7 cm

The formula for the angular spread of the nth-maximum from the central bright fringe for a diffraction from two slits is

$sin \theta=\frac{n \lambda}{d}$

where

n is the order of the maximum

$\lambda$ is the wavelength

$a$ is the distance between the slits

In this problem,

n = 5

$\lambda=525 nm =5.25\cdot 10^{-7} m$

$a=4.15\cdot 10^{-2} mm=4.15\cdot 10^{-5} m$

So we find

$\theta=sin^{-1} (\frac{(5)(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=3.62^{\circ}$

And given the distance of the screen from the slits,

$D=75.0 cm = 0.75 m$

The distance of the 5th bright fringe from the central bright fringe will be given by

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B) 8.1 cm

The formula to find the nth-minimum (dark fringe) in a diffraction pattern from double slit is a bit differente from the previous one:

$sin \theta=\frac{(n+\frac{1}{2}) \lambda}{d}$

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$\theta=sin^{-1} (\frac{(8+\frac{1}{2})(5.25\cdot 10^{-7} m)}{4.15\cdot 10^{-5} m})=6.17^{\circ}$

And the distance of the 8th dark fringe from the central bright fringe will be given by

$y=D tan \theta = (0.75 m)tan 6.17^{\circ}=0.081 m = 8.1 cm$

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Answer:

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