A 10 g dime and a 200 g gold coin are dropped from the Empire State Building in

Physics

1 answer:

Lyrx [107]2 years ago

5 0

Answer:

Explanation:

The both hit the ground at the same time. There is no air resistance and that includes the factors brought up in B.

C and D are both eliminated. If you dropped an 800 kg honda Fit it would hit the ground at the same time as the two coins. When you look at the formulas used to derive time, they look like

d = vi * t + 1/2 a* t^2

vi = 0

the distance to fell is the same for all three masses.

a is the same for all 3 masses.

So when you solve for t, there is no mention of the mass of the object falling. Therefore t is dependent only on the height.

You might be interested in

What are the three places where ribisomes occur in a cell

irina [24]

Answer: Rough endoplasmic reticulum, cytoplasm, inside mitochondria

Explanation:

4 0

3 years ago

What time will the lunar eclipse happen central time?.

alexandr402 [8]

It was about 9:30 p.m. sorry if the answer is wrong

7 0

2 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

Assoli18 [71]

Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$