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vladimir1956 [14]

3 years ago

12

An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o

f the apple.

1 answer:

jenyasd209 [6]3 years ago

5 0

Answer:

Image distance of apple=-6.7 cm

Magnification of apple=0.33

Explanation:

We are given that an apple is placed 20.cm in front of a diverging lens.

Object distance=u=-20 cm

Focal length=f=-10 cm

Because focal length of diverging lens is negative.

We have to find the image distance and magnification of the apple.

Lens formula

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Substitute the values then we get

$-\frac{1}{10}=\frac{1}{v}+\frac{1}{20}$

$\frac{1}{v}=-\frac{1}{10}-\frac{1}{20}$

$\frac{1}{v}=\frac{-2-1}{20}=-\frac{3}{20}$

$v=-\frac{20}{3}=-6.7 cm$

Image distance of apple=-6.7 cm

Magnification=m= $\frac{v}{u}=\frac{-\frac{20}{3}}{-20}$

Magnification of apple= $\frac{1}{3}=0.33$

Hence, the magnification of apple=0.33

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You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you

vekshin1

Answer:

Approximately $5.9\; {\rm m\cdot s^{-1}}$ (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

the antinode at the end of the spring that is being shaken, and
the antinode between the two ends of this spring.

Overall, the nodes and antinodes on this spring would be:

node at the end that is being held still,
antinode (as mentioned in the question),
node (inferred, not mentioned in the question), and
antinode at the end that is being shaken.

The distance between two adjacent nodes is equal to one-half (that is, $(1/2)$ ) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, $(1/4)$ ) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is $\lambda$ , the length of this spring would be:

$\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda$ .

The question states that the length of this coiled spring is $7.2\; {\rm m}$ . In other words, $(3/4) \, \lambda = 7.2\; {\rm m}$ . The wavelength of this wave would be $(7.2\; {\rm m}) / (3/4) = 9.6\; {\rm m}$ .

The frequency $f$ of this wave is the number of cycles in unit time:

$\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}$ .

Hence, the speed $v$ of this wave would be:

$\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}$ .

3 0

2 years ago

Coherent light of wavelength 540 nm passes through a pair of thin slits that are 3.4 × 10-5 m apart. At what angle away from the

Scrat [10]

Answer: $1.8\°$

Explanation:

The diffraction angles $\theta_{n}$ when we have a slit divided into $n$ parts are obtained by the following equation:

$dsin\theta_{n}=n\lambda$ (1)

Where:

$d=3.4(10)^{-5}m$ is the width of the slit

$\lambda=540 nm=540(10)^{-9}m$ is the wavelength of the light

$n$ is an integer different from zero.

Now, the second-order diffraction angle is given when $n=2$ , hence equation (1) becomes:

$dsin\theta_{2}=2\lambda$ (2)

Now we have to find the value of $\theta_{2}$ :

$sin\theta_{2}=\frac{2\lambda}{d}$ (3)

Then:

$\theta_{2}=arcsin(\frac{2\lambda}{d})$ (4)

$\theta_{2}=arcsin(\frac{2(540(10)^{-9}m)}{3.4(10)^{-5}m})$ (5)

Finally:

$\theta_{2}=1.8\°$ (6)

5 0

3 years ago

How did Orsted discover Electromagnetism?

Liula [17]

Answer:

Hans Christian Oersted began a new scientific epoch when he discovered that electricity and magnetism are linked. He showed by experiment that an electric current flowing through a wire could move a nearby magnet. The discovery of electromagnetism set the stage for the eventual development of our modern technology-based world.

Explanation:

7 0

3 years ago

blagie [28]

Assuming Adam is on earth g= 9.8 m/s and m= weight/ gravity = 667/9.8 = 68 kg

8 0

3 years ago

Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri

miv72 [106K]

Answer:

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

T = m_A a

Axis y

N- W_A = 0

Body B

Vertical axis

W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

T = m_A a

W_B –T = M_B a

We solve this system of equations

m_B g = (m_A + m_B) a

a = m_B / (m_A + m_B) g

In this initial case

m_A = M

m_B = M

a = M / (1 + 1) M g

a = ½ g

Let's find the tension

T = m_A a

T = M ½ g

T = ½ M g

Now we change the mass of the second block

m_B = 2M

a = 2M / (1 + 2) M g

a = 2/3 g

We seek tension for this case

T’= m_A a

T’= M 2/3 g

Let's look for the relationship between the tensions of the two cases

T’/ T = 2/3 M g / (½ M g)

T’/ T = 4/3

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

4 0

3 years ago