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3 years ago

Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 43.0 cm?

1 answer:

6 0

Using lens equation;

1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)

Substituting;

1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm

Therefore, the object should be place 99.23 cm from the lens.

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Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm

lina2011 [118]

Answer:

Difference in height = 7.5 cm

Explanation:

We are given;.

Height of ethyl alcohol;h2 = 20 cm = 0.2 m

Density of glycerin: ρ1 = 1260 kg/m³

Density of ethyl alcohol; ρ2 = 790 kg/m³

To get the difference in height, the pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring.

Thus;

P1 = P2

Formula for pressure is; P = ρgh

Thus;

ρ1 × g × h1 = ρ2 × g × h2

g will cancel out to give;

ρ1 × h1 = ρ2× h2

Making h1 the subject, we have;

h1 = (ρ2× h2)/ρ1

h1 = (790 × 0.2)/1260

h1 = 0.125 m

Difference in height will be;

Δh = h2 - h1

Δh = 0.2 - 0.125

Δh = 0.075 m = 7.5 cm

4 0

3 years ago

Calculate the acceleration of a car if it's velocity increases from 15 m/s to 75m/s in 5 seconds.

Andreas93 [3]

initial velocity=u=15m/s
Final velocity=v=75m/s
Time=t=5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{75-15}{5}$

$\\ \sf\longmapsto Acceleration=\dfrac{60}{5}$

$\\ \sf\longmapsto Acceleration=12m/s^2$

7 0

3 years ago

What is the primary atomic characteristic that Mendeleev used to organize the periodic table?

Fudgin [204]

Atomic number is the answer

7 0

3 years ago

Read 2 more answers

Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

wariber [46]

Answer:

For left = 0 N/C

For right = 0 N/C

At middle = $-7.6836 * 10^{-11} \vec{i}$ N/C

Explanation:

Given data :-

б = $6.8 * 10^{-22}$ C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

$E = \frac{\sigma }{2\varepsilon }$

1) To the left of the plate

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C

5 0

3 years ago

What is principal focus and focus length with examples

Shkiper50 [21]

Explanation:

Principle focus is the point on the axis of a convex lens, where the parallel rays of light from one side of the lens. meet on other side after refraction. Distance between optical centre to principle focus point is the focal length.

7 0

3 years ago