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Usimov [2.4K]
3 years ago
12

A magnifying glass uses a convex lens of focal length 6.25 cm. When it is held 5.20 cm in front of an object, what is the image

distance?
(Mind your minus signs)
(Unit=cm)
Physics
1 answer:
mars1129 [50]3 years ago
7 0

Answer:

The answer is either 5 or I'm learning something different and I just can't read

Explanation:

I hope this helped...

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What do you call the curved upper surface of the water that comely appears when using a graduated cylinder
algol [13]
The curved surface of water is called the meniscus
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3 years ago
Read 2 more answers
A parallel-plate capacitor is made from two aluminum-foil sheets, each 5.9 cm wide and 5.6 m long. Between the sheets is a Teflo
Zarrin [17]

Answer:

1.98 x 10⁻⁷ F

Explanation:

w = width of the sheet = 5.9 cm = 0.059 m

L = length of the sheet = 5.6 m

Area of the sheet is given as

A = L w = (5.6) (0.059) = 0.3304 m²

d = distance between the sheets = 3.1 x 10⁻⁵ m

k = dielectric constant of teflon = 2.1

Capacitance is given as

C = \frac{k\epsilon _{o}A}{d}

C = \frac{(2.1)(8.85\times 10^{-12})(0.3304)}{3.1\times 10^{-5}}

C = 1.98 x 10⁻⁷ F

5 0
3 years ago
When a wave passes from one medium to another, its _________ remains constant.
Mashcka [7]
The frequency will not change
3 0
3 years ago
Water flows through a horizontal pipe. The diameter of the pipe at point b is larger than the diameter of the pipe at point a. W
Natalija [7]

The speed of the water is the greatest at point B

5 0
4 years ago
A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric
MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

J = \frac{I}{\pi R^2}

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}

This enclosed current is now to be used in Ampere’s Law.

\mu_o I_{enc} = \int {B} \, dl

Here, \int \, dl represents the circular path of radius r. So we can replace the integral with the circumference of the path, 2\pi r.

As a result, the magnetic field is

B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}

5 0
3 years ago
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