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Ber [7]
3 years ago
13

A virtual image produced by a lens is always A. located in front of the lens. B. located in the back of the lens. C. smaller tha

n the object. D. larger than the object.
Physics
2 answers:
Scorpion4ik [409]3 years ago
6 0
I'd go for D on this one. The wording of B is odd, because it suggests to me that back surface of the lens - that is the surface of the lens that is remote from where the object is. 
kirza4 [7]3 years ago
3 0
A virtual image is usually located in back of the lens and is smaller than the object.

A few other things to know about virtual images:

virtual image - image that cannot be project on a screen because it is formed by rays that do not converge.

The images that are formed by concave lenses or mirrors are always: virtual, erect and diminished.

hope this helps :)

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Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The
Brut [27]

Answer:

The resultant vector is \vec R = \vec A + \vec B = 7.196\,i + 6.464\,j.

Explanation:

First, each vector is determined in terms of absolute coordinates:

6-meter vector with direction: 30º north of east.

\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)

\vec A = 5.196\,i + 3\,j

4-meter vector with direction: 30º east of north.

\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)

\vec B = 2\,i + 3.464\,j

The resultant vector is obtaining by sum of components:

\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j

The resultant vector is \vec R = \vec A + \vec B = 7.196\,i + 6.464\,j.

5 0
4 years ago
Two people are talking through a thick wall, 3 meters high, between them. This fact can best be explained by the phenomenon of A
dezoksy [38]
The answer is Ai just took it
6 0
3 years ago
A fixed mass of an ideal gas is heated from 50°C to 80°C (a) at constant volume and (b) at constant pressure. For which case do
soldi70 [24.7K]

Answer:

Specific heat at constant pressure is =  1.005 kJ/kg.K

Specific heat at constant volume is =  0.718 kJ/kg.K

Explanation:

given data

temperature T1 =  50°C

temperature T2 = 80°C

solution

we know energy require to heat the air is express as

for constant pressure and volume

Q  = m ×  c × ΔT     ........................1

here m is mass of the gas and c is specific heat of the gas and Δ T is change in temperature of the gas

here both Mass and temperature difference is equal and energy required is dependent on specific heat of air.

and here at constant pressure Specific heat  is greater than the specific heat at constant volume,

so the amount of heat required to raise the temperature of one unit mass by one degree at constant pressure is

Specific heat at constant pressure is =  1.005 kJ/kg.K

and

Specific heat at constant volume is =  0.718 kJ/kg.K

3 0
4 years ago
A student lists a set of materials
alexdok [17]

Answer:

D. wood, hydrogen gas, milk, and water

Explanation:

Waves will transmit through wood, hydrogen gas, milk and water.

Waves are disturbances that propagates energy usually through a material medium.

Most mechanical waves pass through the given medium. Only electromagnetic waves do not really require a material medium for their propagation. They can be propagated through a vacuum.

7 0
3 years ago
Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelen
xxTIMURxx [149]

Answer:

A) first laser

B) 0.08m

C) 0.64m

Explanation:

To find the position of the maximum you use the following formula:

y=\frac{m\lambda D}{d}

m: order of the maximum

λ: wavelength

D: distance to the screen = 4.80m

d: distance between slits

A) for the first laser you use:

y_1=\frac{(1)(d/20)(4.80m)}{d}=0.24m\\

for the second laser:

y_2=\frac{(1)(d/15)(4.80m)}{d}=0.32m

hence, the first maximum of the first laser is closer to the central maximum.

B) The difference between the first maximum:

\Delta y=y_2-y_1=0.32m-0.24m=0.08m=8cm

hence, the distance between the first maximum is 0.08m

C) you calculate the second maximum of laser 1:

y_{m=2}=\frac{(2)(d/20)(4.80m)}{d}=0.48m

and for the third minimum of laser 2:

y_{minimum}=\frac{(m+\frac{1}{2})(\lambda)(D)}{d}\\\\y_{m=3}=\frac{(3+\frac{1}{2})(d/15)(4.80m)}{d}=1.12m

Finally, you take the difference:

1.12m-0.48m=0.64m

hence, the distance is 0.64m

3 0
3 years ago
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