The focal length of a concave lens 13.3 cm.Calculate its power.

Physics

1 answer:

cricket20 [7]3 years ago

4 0

Explanation:

Formula for finding power : 1/f

Substitution of values:

1 /13.3 = 0.07518 cm

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Please help! Average speed. Show work!

KiRa [710]

Answer:

3.78 m/s

Explanation:

Recall that the formula for average speed is given by

Speed = Distance ÷ Time taken

Where,

Speed = we are asked to find this

Distance = given as 340m

Time taken = 1.5 min = 1.5 x 60 = 90 seconds

Substituting the values into the equation:

Speed = Distance ÷ Time taken

= 340 meters ÷ 90 seconds

= 3.777777 m/s

= 3.78 m/s (round to nearest hundredth)

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3 years ago

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Which language style would be most appropriate for the given situation?

polet [3.4K]

Answer:

informal language

Explanation:

you do not need to be formal! you are not at a business conference. technical is not needed either since you are not discussing the intricacies of your job or some computer language.

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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,

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