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3 years ago

PLEASE ANSWER FAST I WILL GIVE BRAINLIEST

Physics

2 answers:

IrinaVladis [17]3 years ago

5 0

Answer:

Explanation:

got right on edg

TiliK225 [7]3 years ago

3 0

Explanation:

The answer is:

A squirrel runs up the trunk of a tree.

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Energy can come from electricity, but it can also come from water false true

Hitman42 [59]

Answer:

the answer is truth

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3 years ago

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The combined focal length of two thin lens is 24 cm and the focal length of one converging lens is 8

qwelly [4]

Answer: f = -12 cm

Explanation: Combined lenses is an array of simple lenses with a common axis. The combination is useful for correction of optical aberrations which cannot be corrected by simple lenses.

When two lenses are in contact and are thin, focal lengths are related as:

$\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}$

If there is a distance between the lenses, the focal length will be:

$\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}} -\frac{d}{f_{1}f_{2}}$

Since the lenses in the question above are thin and in contact, the focal length of one of them will be:

$\frac{1}{F} =\frac{1}{f_{1}} +\frac{1}{f_{2}}$

$\frac{1}{f_{2}} =\frac{1}{f_{1}} -\frac{1}{F}$

$\frac{1}{f_{2}} =\frac{1}{8} -\frac{1}{24}$

$\frac{1}{f_{2}} =\frac{-2}{24}$

$f_{2}=$ -12

The focal length of the other lens is -12 cm, with the negative sign meaning it's a converging lens.

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3 years ago

What is the momentum of an 18-kg object moving at 0.5 m/s?

Arada [10]

Answer:

9kgm per second

Explanation:

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2 years ago

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22.13 .. Two very long uniform lines of charge are parallel and are separated by 0.300 m. Each line of charge has charge per uni

zimovet [89]

Answer:

Two very long uniform lines of charge are parallel and are separated by 0.300 m. Each line of charge has charge per unit length.

Explanation:

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3 years ago

A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud

IceJOKER [234]

Answer: 1.51 km

Explanation:

Coulomb's Law: The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or, $\vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}$

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

Given:

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

So,

$\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514 \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}$

Therefore, the separation between the two charges (r) = 1.51 km

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2 years ago