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Viktor [21]
3 years ago
7

What is a disadvantage and advantage for Electrical Energy??

Physics
1 answer:
Keith_Richards [23]3 years ago
8 0
There are many advances done to our society and civilization by electricity, which rely on electrical energy, electricity is produce either by modern coal powerplant, solar power, hydro electric, and nuclear power plant. Life is much easier to communicate to far away places by means of satellite communications. Computers, cars, all industry uses electricity. The only disadvantages of electricity is, if the source of energy to power our society suddenly disrupted or destroyed, this will result to total break down of society that rely on electricity.
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What is the gravity force between two stars with mass of 5,000,000 kg and 1,000,000 kg if the distance between them is 100 m
Luda [366]

Answer:

0.0334N

Explanation:

Given parameters:

M1  = 5 x 10⁶kg

M2  = 1 x 10⁶kg

Distance  = 100m

Unknown:

Gravitational force  = ?

Solution:

To solve this problem, we use the Newton's law of universal gravitation.

     Fg  = \frac{G m1 m2}{r^{2} }  

G is the universal gravitation constant

m is the mass

r is the distance

         Fg  = \frac{6.67 x 10^{-11} x 5 x 10^{6}  x 1 x 10^{6} }{100^{2} }   = 0.0334N

8 0
3 years ago
Two beams of coherent light travel different paths, arriving at point P. If the destructive interference is to occur at point P,
swat32

Answer:

Explanation:

When two coherent light beams travel different paths and arrive at a point , there will be difference in the length of path travelled by them . If this difference is zero then both will reinforce each other and their brightness will add up . Hence there will be constructive interference .

If their path difference is not zero but it is equal to odd multiple of their half wavelength like λ / 2 , 3 λ / 2 , 5 λ /2 , 7 λ /2 etc , then instead of reinforcing each other , they will destroy each other . This is called destructive interference . As a result of it , darkness will prevail at the point where they meet or interfere.

6 0
3 years ago
One way to cool a gas is to let it expand. When a certain gas under a pressure of 5.00 x 10^6 Pa at 25.°C is allowed to expand t
hichkok12 [17]

Answer:

The final temperature is T2= 5.35°C

Explanation:

Apply the Gay-lussacs's law we have

\frac{P1}{T1} = \frac{P2}{T2}

P1, initial pressure= 5.00 x 10^6 Pa

T1, initiation temperature= 25.°C

P2, final pressure= 1.07 x 10^6 Pa

T2, final temperature= ?

\frac{5.00 * 10^6}{25} =\frac{1.07 *10^6}{T2} \\

Cross multiplying and making T2 subject of formula we have

T2 =\frac{1.07 *10^6*25}{5.00 * 10^6} \\\\T2= \frac{26.75}{5} \\T2= 5.35

T2= 5.35°C

6 0
4 years ago
Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of th
Yuki888 [10]

Complete Question

Diagram for this  shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is I{xx}= 0.119 kg \cdot m^2

Explanation:

From the question we are told that

        The density `of wooden handle is  \rho_w = 860 kg/m^3

        The density `of soft-metal head  is \rho_s =8000kg/m^3

Generally the mass of the wooden can be mathematically obtained with this formula

          m_w = \rho_w A_w l_w

Where A_w is mass of wooden handle which is  mathematically obtain with the formula

             A_w = \frac{\pi}{4} d^2_w

Where d_w is the diameter  of the wooden handle which from the diagram is

       27mm = \frac{27}{1000} = 0.027m

So  A_w = \frac{\pi}{4} * 0.027^2

      l_w is the length of the the wooden handle which is given in the diagram as   l_w = 315mm = \frac{315}{1000} = 0.315m

Substituting these value into the formula for mass

      m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315

            = 0.155kg

Generally the mass of the soft-metal head can be mathematically obtained with this formula

           m_s = \rho_s A_s l_s

Where A_s is mass of soft-metal head which is  mathematically obtain with the formula

            A_s = \frac{\pi}{4} d^2_s

Where d_s is the diameter  of the soft-metal head which from the diagram is            

       36mm = \frac{36}{1000} = 0.036m

So  A_s = \frac{\pi}{4} * 0.036^2

 l_s is the length of the the soft-metal head which is given in the diagram

     as   l_s = 90mm = \frac{90}{1000} = 0.090m

Substituting these value into the formula for mass  

                  m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090

                       =0.733kg

Generally the mass moment of inertia about x-axis for the wooden handle is

                  (I_{xx})_w  =    [\frac{1}{3}m_w + l_w^2 ]  

Substituting values

                   (I_{xx})_w  =    [\frac{1}{3}*0.155 + 0.315^2 ]

                              =5.12*10^{-3}kg \cdot m^2  

Generally the mass moment of inertia about x-axis for the soft-metal head is

    (I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2]

Where b is the distance from the centroid to the axis of the head which is mathematically given as

                   b=l_w +\frac{d_s}{2}

Substituting values

                 b = 0.315 + \frac{0.036}{2}

                    = 0.336m

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

                            (I_{xx})_s = [\frac{1}{12} *0.733*  0.090^2 + 0.336^2]

                                      =0.113 kg \cdot m^2

Generally the mass moment of inertia about x-axis is mathematically represented as

         I_{xx} = (I_{xx})_w + (I_{xx})_s

                = [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2]

Substituting values

        I_{xx} = 5.12*10^{-3} +0.113

               I{xx}= 0.119 kg \cdot m^2

             

             

8 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
Tasya [4]

Answer:

I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

I(x)  = k ({r}^{6}/6)^(5)_0 ×  {θ/2 - sin2θ/4}^{\pi /2}_0

I(x)  = k × ({6}^{6}/6) × (  {\pi /4} - sin\pi /4)

I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

and we can say I(x) = I(y)   by the symmetry rule

and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

3 0
3 years ago
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