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Schach [20]
3 years ago
11

The sun is 150,000,000 km from earth; its diameter is 1,400,000 km. A student uses a 5.2-cm-diameter lens with f = 10 cm to cast

an image of the sun on a piece of paper. Where should the paper be placed relative to the lens to get a sharp image?
Physics
1 answer:
Elenna [48]3 years ago
6 0

Answer:

The paper is placed on the focus of the lens then we get the sharp image.

Explanation:

Given that,

Object distance = 150000000 km

Diameter = 1400000 km

Diameter of lens = 5.2 cm

Focal length = 10 cm

We need to calculate the image distance

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{10}+\dfrac{1}{15\times10^{7}}

\dfrac{1}{v}=\dfrac{1}{1.0\times10^{-4}}+\dfrac{1}{15\times10^{7}}

Here,

v=0.0001\ km

v=1.0\times 10^{-4}\ km

Hence, The paper is placed on the focus of the lens then we get the sharp image.

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Please help me, I'm not good in physics.
expeople1 [14]

Answer:

Upward direction.

The friction is kinetic friction

3 0
3 years ago
Compute the resistance in ohms of a copper block 5.0 cm long and 0.10 cm2 in cross-sectional area. (ρ = 1.77 × 10-6 ohm-cm)
soldier1979 [14.2K]
R=ρ l/a
l=5 cm
a=0.10 cm²
ρ=1.77×10∧-6 Ωcm
R=(1.77×10∧-6) ×5/0.10
  =8.85 ×10∧-5 Ω

resistance is directly proportional to length and inversly proportional to area of cross section is given by above equation 
4 0
3 years ago
If you calculate W, the amount of work it took to assemble this charge configuration if the point charges were initially infinit
sp2606 [1]

Answer:

W = 0×(kq2L)

Explanation:

We know that the work to assemble a charge configuration of two charges a distance r from each other is simply W = kq2/r

If we want to assemble three charges A, B, and C. It's necessary to consider the distances between them

WABC = kq2/(rAB + rAC + rBC)

So, to assemble four charges A, B, C, & D....

WABCD = kq2/(rAB + rAC + rAD + rBC + rBD + rCD)

 

Considering a square charge configuration with sides L, such as in figure attached A, B, & C are positive & D is negative

rAB = L

rAC = L√2

rAD = L (-)

rBC = L

rBD = L√2 (-)

rCD = L (-)

⇒ W = kq2/(L + L√2 + (-L) + L + (-L√2) + (-L)

⇒ ∴ W = 0 × (kq2/L)

This way, working through each option...  

(a)

The positive charges are equidistant from each other at a distance of L.

rAB = L

rAC = L

rAD = ½L⋅sin(60) (-)

rBC = L

rBD = ½L⋅sin(60) (-)

rCD = ½L⋅sin(60) (-)

Wa = kq2/(3L - (3/2)L⋅(0.866))

⇒ ∴ Wa = (1/1.7) × (kq2/L) = (0.5879)× (kq2/L)

(b)

rAB = L

rAC = 2L

rAD = 3L (-)

rBC = L

rBD = 2L (-)

rCD = L (-)

Wb = kq2/(4L - 6L)

⇒ ∴ Wb = (-1/2) × (kq2/L) = (-0.5)× (kq2/L)

(c)

The factor doesn't matter, so Wc = 0 × (kq2/L)

In this case, the greater work is actually the less work. Therefore, the positive work represents the amount of work the system actually exhibits, that we don't have to do. If there is negative work, we have to make up that work in order to place the charges as desired.  

This way, charge configuration (a) requires the least amount of work.

5 0
3 years ago
The name (blank) means "all land"
mina [271]
I would say the answer is Pangaea.  I got this answer but google searching "all land" means what...  and the 3rd or 4th one down it says Answers.com
But i worn you answers.com is a very slow sight...
Hope it helps
and please mark me as brainiest

4 0
4 years ago
An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons ar
sleet_krkn [62]

Answer:

0.5 V

Explanation:

The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.

6 0
3 years ago
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