Stems tend to grow in Or against the direction of gravity

A 15.0 cm object is 12.0 cm from a convex mirror that has a focal length of -6.0 cm. What is the height of the image produced by

alisha [4.7K]

<h2>Answer: 5 cm</h2>

In convex mirrors the focus is virtual and the focal distance is negative. This is how the reflected rays diverge and only their extensions are cut at a point on the main axis, resulting in a virtual image of the real object .

The Mirror equation is:

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ (1)

Where:

$f=-6cm$ is the focal distance

$u=12cm$ is the distance between the object and the mirror

$v$ is the distance between the image and the mirror

We already know the values of $f$ and $u$ , let's find $v$ from (1):

$v=\frac{u.f}{u-f}$ (2)

$v=\frac{(12cm)(-6cm)}{12cm-(-6cm)}$

$v=-4cm$ (3)

On the other hand, the magnification $m$ of the image is given by the following equations:

$m=-\frac{v}{u}$ (4)

$m=\frac{h_{i}}{h_{o}}$ (5)

Where:

$h_{i}$ is the image height

$h_{o}=15cm$ is the object height

Now, if we want to find the image height, we firstlu have to find $m$ from (4), substitute it on (5) and find $h_{i}$ :

Substituting (3) in (4):

$m=-\frac{-4cm}{12cm}$

$m=\frac{1}{3}$ (6)

Substituting (6) in (5):

$\frac{1}{3}=\frac{h_{i}}{15cm}$

$h_{i}=\frac{15cm}{3}$

Finally we obtain the value of the height of the image produced by the mirror:

$h_{i}=5cm$

6 0

4 years ago

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A uniformly charged ring of radius 10.0 cm has a total charge of 71.0 μC. Find the electric field on the axis of the ring at the

oee [108]

Answer:

General Expression: E = kql/(l² + r²)^(3/2)

(a) 6.3 MN/C

(b) 22.8 MN/C

(c) 6.1 MN/C

(d) 0.63 MN/C

Explanation:

The general expression for electric field along axis of a uniformly charged ring is:

E = kqL/(L² + r²)^(3/2)

where,

E = Electric Field Strength = ?

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q = Total Charge = 71 μC = 71 x 10⁻⁶ C

L = Distance from center on axis

r = radius of ring = 10 cm = 0.1 m

(a)

L = 1 cm = 0.01 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.01 m)/[(0.01 m)² + (0.1 m)²]^(3/2)

E = (6390 N.m³/C)/(0.00101 m³)

E = 6.3 x 10⁶ N/C = 6.3 MN/C

(b)

L = 5 cm = 0.05 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.05 m)/[(0.05 m)² + (0.1 m)²]^(3/2)

E = (31950 N.m³/C)/(0.00139 m³)

E = 22.8 x 10⁶ N/C = 27.4 MN/C

(c)

L = 30 cm = 0.3 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(0.3 m)/[(0.3 m)² + (0.1 m)²]^(3/2)

E = (191700 N.m³/C)/(0.03162 m³)

E = 6.1 x 10⁶ N/C = 6.1 MN/C

(d)

L = 100 cm = 1 m

Therefore,

E = (9 x 10⁹ N.m²/C²)(71 x 10⁻⁶ C)(1 m)/[(1 m)² + (0.1 m)²]^(3/2)

E = (639000 N.m³/C)/(1.015 m³)

E = 0.63 x 10⁶ N/C = 0.63 MN/C

8 0

3 years ago

It usually takes more force to start an object sliding than it does to keep an object sliding because static friction is usually

labwork [276]

The answer is already in the blank for, its was greater

5 0

4 years ago

Stability is a term that means...

Aleksandr-060686 [28]

The correct answer is A

7 0

3 years ago

With a bar magnet where are the lines of force closest together

attashe74 [19]

At the tip of either of the magnets poles

8 0

3 years ago

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