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chubhunter [2.5K]
3 years ago
14

What kind of image is formed by a convex lens when the object is less than a focal length away from the object?

Physics
1 answer:
maw [93]3 years ago
3 0
Bigger image and can get clearer
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What is a partial and total lunar eclipse?
Minchanka [31]

Partial Lunar Eclipse:

A partial lunar eclipse is when the earth gets between the Sun and Moon. However, all three bodies are not in alignment meaning we are able to see some more like part of the moon's surface as it moves in route of the Earth's shadow.

Total Lunar Eclipse:

The three celestial bodies are perfectly aligned which allows for the earth to completely block the sun's rays from hitting/reaching the moon. The sun is positions is in back of the Earth which then causes the shadow of the earth to be cast on the Moon covering the moon completely. When that happens you get the phenomenon called a total lunar eclipse.

Hopefully this helped and good luck.



7 0
3 years ago
A water molecule consists of an oxygen atom with two hydrogen atoms bound to it. The angle between the two bonds is 106◦ . If ea
balu736 [363]

Answer:

easy 16

Explanation:

because i love jesus

3 0
3 years ago
Can you respond this two questions, please? :
Andrews [41]

Where's the diagram for question 1?

6 0
3 years ago
2 Ten identical lengths of wire are laid closely side-by-
Mice21 [21]

Answer:

(a) 0.71 mm

(b) 0.158 cubic cm

Explanation:

The width of one wire is the diameter of the wire.

(a) Let the diameter of each wire is d.

So, 10 d = 14.2 mm

d = 1.42 mm

radius of each wire, r = d/2 = 1.42/2 = 0.71 mm

(b) Length, L = 10 cm

The volume of the single wire is given by

V =\pi\times r^2\times h \\\\V =3.14\times 0.071^2\times 10\\\\V =0.158 cm^3

7 0
2 years ago
An automobile starter motor has an equivalent resistance of 0.055 Ï and is supplied by a 12.0 v battery which has a 0.0305 Ï int
LiRa [457]
12 V is the f.e.m. \epsilon of the battery. The potential difference that is applied to the motor is actually the fem minus the voltage drop on the internal resistance r:
\epsilon - Ir
this is equal to the voltage drop on the resistance of the motor R:
RI
so we can write:
\epsilon - Ir = RI
and using r=0.0305~\Omega and R=0.055~\Omega we can find the current I:
I= \frac{\epsilon}{R+r}=140~A
8 0
3 years ago
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