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dlinn [17]
3 years ago
9

Stephen schwartz created the musical godspell as well as __________.

Physics
1 answer:
Alik [6]3 years ago
8 0

Answer:

Stephen Lawrence Schwartz (born March 6, 1948) is an American musical theater lyricist and composer. In a career spanning over five decades, Schwartz has written such hit musicals as Godspell (1971), Pippin (1972), and Wicked (2003).

Explanation:

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Two large metal plates of area 0.7 m2 face each other. They are 5.1 cm apart and have equal but opposite charges on their inner
german

Explanation:

The given data is as follows.

    Area (A) = 0.7 m^{2}.

  Electric field between the plates, (E) = 55 N/C

Since, electric field is related to surface charge density as follows.

          \sigma = \frac{Q}{A}

Also,    E = \frac{\sigma}{\epsilon_{o}}

or,        \sigma = \epsilon_{o} E

Therefore, charge will be calculated as follows.

          Q = \sigma \times A = \epsilon_{o} E \times A

             = 8.854 \times 10^{-12} \times 55 \times 0.7

             = 3.41 \times 10^{-10} C

             = 341 pC

Thus, we can conclude that magnitude of the charge on each plate is 341 pC.

8 0
4 years ago
Which human activity causes the most erosion?
Ksivusya [100]

Answer:

B

Explanation:

3 0
3 years ago
Does the elf owl's small size allow it to live in cacti
kaheart [24]
Habitat
-In the Sonoran Desert region, elf owls are found mainly in riparian habitats (places where there is water), or in areas where saguaro cactus are plentiful.

Range
-Elf Owls are found from the southwest USA to Central Mexico and Baja California. Northern populations winter in Central Mexico and on the Pacific slope north to Sinaloa, Mexico.

Wild Status
-The most important threat to the elf owl is habitat loss both of its riparian forest habitat and desert-scrub habitats. In Arizona, the elf owl is not uncommon, but it's numbers are decreasing in California and Texas.

-So I mean like ya it's possible that they could live in cacti. Sorry that I couldn't give you a straight answer.



couldlive in
3 0
3 years ago
A mass of 1.00×10−6 μkg is the same as
tankabanditka [31]
I'm not sure but I know u is 10^6
6 0
3 years ago
How many significant digits should the answer to the following problem have? (2.49303 g) * (2.59 g) / (7.492 g) =
Feliz [49]

The number of significant digits to the answer of the following problem is four.

<h3>What are the significant digits?</h3>

The number of digits rounded to the approximate integer values are called the significant digits.

The following problem is

(2.49303 g) * (2.59 g) / (7.492 g) =

On solving we get

= 0.86184566204

The answer is approximated to  0.86185

Thus, the significant digits must be four.

Learn more about significant digits.

brainly.com/question/1658998

#SPJ1

6 0
2 years ago
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