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makvit [3.9K]
3 years ago
10

Why does gravitational energy increase away from earth

Physics
1 answer:
aliina [53]3 years ago
4 0
Because of the rule of mask.
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Hi, please just say the Letter (A,B,C,D,E)
bija089 [108]
The letter that answers this question correctly is E .
4 0
3 years ago
Legacy issues $570,000 of 8.5%, four-year bonds dated January 1, 2019, that pay interest semiannually on June 30 and December 31
choli [55]

Answer:

1) Determine the total bond interest expense to be recognized.

Total bond interest expense over life of bonds:

Amount repaid:    

8 payments of $24,225:           $193,800    

Par value at maturity:                 $570,000    

Total repaid:                                   $763800 (193,800 + 570,000)  

Less amount borrowed:         $508050    

Total bond interest expense: $255750 (763800 - 508,050)

2)Prepare a straight-line amortization table for the bonds' first two years.

Semiannual Interest Period­ End; Unamortized Discount; Carrying Value

01/01/2019                                      61,950                           508,050  

06/30/2019                                      54,206                          515,794  

12/31/2019                                       46,462                         523,538  

06/30/2020                                       38,718                        531,282  

12/31/2020                                         30,974                          539,026

3) Record the interest payment and amortization on June 30:

June 30            Bond interest expense, dr                         31969  

                       Discount on bonds payable, Cr     (61950/8)  7743.75

                                        Cash, Cr                     ( 570000*8.5%/2)  24225  

4) Record the interest payment and amortization on December 31:

Dec 31                 Bond interest expense, Dr               31969  

                           Discount on bonds payable, Cr  7744  

                                    Cash, Cr                                24225

6 0
3 years ago
What parts of an electric generator are connected?
Charra [1.4K]

Answer:

The field and winding

Explanation: they are both connected

6 0
2 years ago
....................................
IRISSAK [1]
No question........ ok then :)
3 0
2 years ago
A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 cm , giving it a ch
vladimir1956 [14]

A) The electric field inside the paint layer is zero

B) The electric field just outside the paint layer is 3.2\cdot 10^7 N/C (radially inward)

C) The electric field at 6.00 cm from the surface is 1.2\cdot 10^7 N/C (radially inward)

Explanation:

A)

We can solve the problem by applying Gauss Law, which states  that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:

\int EdS = \frac{q}{\epsilon_0}

where

E is the magnitude of the electric field

dS is the element of the surface

q is the charge contained within the surface

\epsilon_0 is the vacuum permittivity

By taking a sphere centered in the origin,

\int E dS = E \cdot 4\pi r^2

where 4\pi r^2 is the surface of the Gaussian sphere of radius r.

In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than

R=9.0 cm = 0.09 m (radius of the plastic sphere is half of the diameter)

Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:

q=0

And therefore,

E4\pi r^2 = 0\\\rightarrow E = 0

So, the electric field inside the plastic sphere is zero.

B)

Here we apply again Gauss Law:

E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}

In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so

r=R=0.18 m

The charge contained within the Gaussian sphere is therefore

q=-29.0 \mu C = -29.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C

And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is

E=3.2\cdot 10^7 N/C

C)

For this part again, we apply Gauss Law:

E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}

In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be

r = 9 cm + 6 cm = 15 cm = 0.15 m

While the charge contained within the sphere is again

q=-29.0 \mu C = -29.0\cdot 10^{-6}C

Therefore, the electric field in this case is

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C

And again, this is radially inward, so according to the sign convention asked in the problem,

E=1.2\cdot 10^7 N/C

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

3 0
3 years ago
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