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just olya [345]
3 years ago
7

A single point charge q is located at the center of both an imaginary cube and an imaginary sphere. How does the electric flux t

hrough the surface of the cube compare to that through the surface of the sphere?
Physics
1 answer:
saveliy_v [14]3 years ago
7 0

Answer:

Explanation:

Electric flux is defined as the flow of electric field intensity through a given surface.

Mathematically:

\phi=E.A.cos\ \theta

where:

E = electric field

A = area

θ = angle between the area vector and the electric field

Electric flux through the surface of a sphere will be uniform throughout the surface area due to a charge at the center of the sphere. The distance of the surface from the center is always at a constant distance of radius of the sphere.

Electric flux through the surface of a cube will be varying as the surface area is at a varying distance from the center of the cube. The distance of the surface from the center is not at a uniform distance from the center of the cube and so the projection of solid angle changes.

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pychu [463]

Answer:

The time for the cake to cool off to room temperature is

approximately 30 minutes.

Let T_{0} = 70^{0}F be the temperature and T that of the body

Explanation:

 Our Tm = 70, the initial-value problem is

\frac{DT}{dt} = <em>k</em>(T − 70), T(0) = 300

Solving the equation, we get

\frac{DT}{t-70} = <em>kdt</em>

In [T-70]= <em>kt </em>+C_{1}

    T   =  70  + C_{2} e^{kt}

Finding he value for C_{2} using the initial value of T (0)= 300, therefore we get:

300=70+C_{2}

C_{2} = 230 therefore

T= 70+ 230 e^{kt}

Finding the value for <em>k </em>using T (3)  = 200, therefore we get

T (3) = 200

e^{3k} = \frac{13}{23}

<em>K </em>= \frac{1}{3} in \frac{13}{23}

= -0.19018

Therefore

T(t) = 70+230e^{-0.19018}

4 0
3 years ago
A person wishes to heat pot of fresh water from 20°C to 100°C in order to boil water for pasta. They calculate that their pot ho
nasty-shy [4]
Given:
Water, 2 kilograms
T1 = 20 degrees Celsius, T2 = 100 degrees Celsius.  

Required:
Heat produced  

Solution:
Q (heat) = nRT = nR(T2 = T1)
Q (heat) = 2 kilograms (4.184 kiloJoules per kilogram Celsius) (100 degrees Celsius – 20 degrees Celsius)
<u>Q (heat) = 669.42 Joules
</u>This is the amount of heat produced in boiling 2 kg of water.
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A solid disk has a mass of 162 kg and a radius of 1.30m. This
RideAnS [48]

Answer:

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Explanation:

The moment of inertia of the solid disk:

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I = 162*1.3^2/2 = 136.89 kgm^2

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let g = 9.8 m/s2

E_p = m_bgh = 22176.18 J

where m_b = 3 kg is the mass of block

3*9.8h = 22176.18

h = \frac{22176.18}{3*9.8} = 754.3 m

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