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Brut [27]
3 years ago
10

Elizabeth .........................................................................

Physics
2 answers:
Ksivusya [100]3 years ago
8 0

Answer:

What about it?

Explanation:

Umnica [9.8K]3 years ago
7 0

Answer:

who's elizabeth? haha

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What electric field strength would store 12.5 JJ of energy in every 6.00 mm3mm3 of space?
Tcecarenko [31]

To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as

The potential energy,  U = 13.0 J

The volume,  V = 6.00 mm^3 = 6.00*10^{-9}m^3

The potential energy per unit volume is defined as the energy density.

u = \frac{U}{V}

u= \frac{(13.0 J)}{(6.00*10^{-9} m^3)}

u= 2.167109 J/m^3

The energy density related with electric field is given by

u = \frac{1}{2} \epsilon_0 E^2

Here, the permitivity of the free space is

\epsilon_0 = 8.85*10^-{12} C^2/N \cdot m^2

Therefore, rerranging to find the electric field strength we have,

E = \sqrt{\frac{2u}{\epsilon_0}}

E = \sqrt{\frac{2(2.167109)}{8.85*10^{-12}}}

E = 2.211010 V/m

Therefore the electric field is 2.21V/m

8 0
3 years ago
What do active solar heat systems have that passive systems do not have?.
tamaranim1 [39]

Answer:

Passive solar heating uses building design to utilize sunlight, while active solar heating uses technology.

Explanation:

lol

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2 years ago
A billiard ball is dropped from a height of 64 feet. Use the position function s(t) = –16???? 2 + ????0???? + ????0 to answer th
Delicious77 [7]

Answer:

s(t) = -16*t^2 + 64

v(t) = -32*t

a(t) = -32 ft/s^2

v(t) = 64 ft/s ... At impact

Explanation:

Given:-

- The height of the billiard ball t = 0 , h = 64 ft.

- The position function of an object under gravity is given by:

                                    s(t) = -16*t^2 + v_o*t + s_o

Find:-

a. Determine the position function s(t),

b. the velocity function v(t),

c. the acceleration function a(t).

d. What is the velocity of the ball at impact?

Solution:-

- To determine the position function we must initialize our problem and use the given general equation.

- s(t) is the position of the billiard ball from the ground at time t. So when t = 0, then s(t) = h. Hence, we have:

                                  s(t) = s_o = h = 64 ft

- Similarly we know that v_o is the initial velocity of the ball. Since, the ball was dropped we say that the initial velocity v_o = 0. Hence, the position of the ball from ground is given by following expression:

                                  s(t) = -16*t^2 + 64  

- To find the velocity expression v(t) we will take the time derivative of the position expression s(t) as follows:

                                  v(t) = d s(t) / dt

                                  v(t) = -16*2*t + 0

                                  v(t) = -32*t ft/s

- Similarly, the expression for acceleration a(t) is given by the time derivative of the velocity expression v(t) as follows:

                                  a(t) = d v(t) / dt

                                  a(t) = -32*t

                                  a(t) = -32 ft/s^2

- The velocity of ball at impact can be determined by evaluating s(t) = 0 and find the value for time t. Then that time t can be substituted in the velocity expression v(t) for final velocity. Or we could use the following 3rd kinematic equation as follows:

                                 v(t)^2 - 0^2 = 2*a(t)*s_o

                                 v(t)^2 = 2*(32)*(64)

                                 v(t) = 64 ft/s

- The ball has a velocity of 64 ft/s at impact!

6 0
3 years ago
What determines the state of matter for any
noname [10]

Answer:

I think the answer is B. amount of energy present but I'm not 100% sure

Explanation:

7 0
2 years ago
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What prevents the pressure from increasing as a cloud contracts due to its gravity?
Vesnalui [34]

Thermal energy is converted to radiative energy via molecular collisions and released as photons.

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