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Veseljchak [2.6K]
3 years ago
11

Why are coal,nuclear,oil,and natural gas determined to be non-renewable types of energy resources ​

Physics
1 answer:
IrinaVladis [17]3 years ago
7 0

Answer:

Once used as a energy source they cannot be charged/used again.

Explanation:

You might be interested in
What affect does a tripling of the net force have upon the acceleration of the object ? Be quantitative
vovikov84 [41]
The formula of net Force is:
F = ma
where m is the mass of the object
a is the acceleration of the object

so if we triple the net force applied to the object:
3F = ma
a = 3F / m

so the acceleration will also be tripled. because from the equation, the force is directly proportional to the acceleration
5 0
3 years ago
Which is an example of convection currents
Viefleur [7K]

Answer:

A radiator emitting warm air and drawing in cool air is an example of convection current

Explanation:

I am guessing you meant "A radiator emitting warm air and drawing in cool air'

8 0
3 years ago
Read 2 more answers
A 10-cm-long thin glass rod uniformly charged to 8.00 nCnC and a 10-cm-long thin plastic rod uniformly charged to - 8.00 nCnC ar
ch4aika [34]

Complete Question

A 10-cm-long thin glass rod uniformly charged to 8.00 nC and a 10-cm

long thin plastic rod uniformly charged to -8.00 nC are placed side by

side, 4.20 cm apart. What are the electric field strengths E_1 to E_3 at

distances 1.0 cm, 2.0 cm, and 3.0 cm from the glass rod along the line

connecting the midpoints of the two rods

a.) Specify the electric field strength E1

b.) Specify the electric field strength E2

c.) Specify the electric field strength E3

Answer:

              E_1=7.13*10^5 N/C

             E_2= 2.95*10^{5} N/C

              E_3= 3.84*10^5 N/C

Explanation:

  From the question we are told that

          The length of the thin glass is  L = 10 cm

          The  charge on the glass rod is  q_g = 8.00nC = 8* 10^{-9} C

           The length of the plastic rod is  L_p = 10cm

             The charge on the  plastic rod is q_p =- 8.00nC = -8.0*10^{-9}C

           The distance between the materials  is d = 4.20cm = \frac{4.2}{100} =0.042m

          The various distances to obtain electric field of are r_1 = 1.0cm

                                                                                                r_2 = 2.0cm

                                                                                                 r_3 = 3.0cm

The objective of the solution is to obtain the electric field E_1 , E_2 \ and E_3 at distance d_1 , d_2 \ and \ d_3  from the glass rod  along the line connecting its mid point  

   Generally electric field of a charge rod at a distance of r the line dividing the rod  into half  is mathematically represented as

                              E = k \frac{2Q}{r\sqrt{L^2 + 4r^2} }

For the  r_2 = 1.0cm = \frac{1}{100} = 0.01m

The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

                               E_l =  k \frac{2Q }{r \sqrt{L^2 + 4r^2_1} }

The electric filed by the positively charge glass rod on the right  side of the dividing line is mathematically represented as  

                            E_r =  k \frac{2Q }{(0.044 - r_1) \sqrt{L^2 + 4r^2_1} }

The net electric field is,

            E_{net} =E_1= E_l + E_r

                    = k \frac{2Q}{r_1\sqrt{L^2 + 4 r^2_1 } } + k \frac{2Q}{(0.04-r_1) \sqrt{L^2 + 4 (0.044 -r_1)^2} }

Where k is  know as the coulomb's constant  with a constant value of

                  k = 9*10^9 \ kgm^3 s^{-4} A^{-2}

           =(9*10^9) \frac{(2) (8*10^{-9})}{(0.01)\sqrt{(0.01^2 + 4(0.01)^2)} }  + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.01)\sqrt{(0.01)^2 + (4) (0.042 - 0.01)^2} }

                           = 6.44*10^5 + 6.9*10^4

                           E_1=7.13*10^5 N/C

For the  r_2 = 2.0cm = \frac{2}{100} = 0.02m

           The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

                               E_l =  k \frac{2Q }{r_2 \sqrt{L^2 + 4r^2_2} }

The electric filed by the positively charge glass rod on the right  side of the dividing line is mathematically represented as  

                            E_r =  k \frac{2Q }{(0.044 - r_2) \sqrt{L^2 + 4r^2_2} }

The net electric field is,

            E_{net} =E_2= E_l + E_r

                    = k \frac{2Q}{r_2\sqrt{L^2 + 4 r^2_2 } } + k \frac{2Q}{(0.04-r_2) \sqrt{L^2 + 4 (0.044 -r_2)^2} }

Where k is  know as the coulomb's constant  with a constant value of

                  k = 9*10^9 \ kgm^3 s^{-4} A^{-2}

           =(9*10^9) \frac{(2) (8*10^{-9})}{(0.02)\sqrt{(0.02^2 + 4(0.02)^2)} }  + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.02)\sqrt{(0.02)^2 + (4) (0.042 - 0.02)^2} }

            = 1.6*10^{5}+ 1.3*10^{5}

             E_2= 2.95*10^{5} N/C

For the  r_3 = 3.0cm = \frac{3}{100} = 0.03m

           The electric filed by the positively charge glass rod on the left side of the dividing line is mathematically represented as

                               E_l =  k \frac{2Q }{r_3 \sqrt{L^2 + 4r^2_3} }

The electric filed by the positively charge glass rod on the right  side of the dividing line is mathematically represented as  

                            E_r =  k \frac{2Q }{(0.044 - r_3) \sqrt{L^2 + 4r^2_3} }

The net electric field is,

            E_{net} =E_3= E_l + E_r

                    = k \frac{2Q}{r_3\sqrt{L^2 + 4 r^2_3 } } + k \frac{2Q}{(0.04-r_3) \sqrt{L^2 + 4 (0.044 -r_3)^2} }

Where k is  know as the coulomb's constant  with a constant value of

                  k = 9*10^9 \ kgm^3 s^{-4} A^{-2}

           =(9*10^9) \frac{(2) (8*10^{-9})}{(0.03)\sqrt{(0.03^2 + 4(0.03)^2)} }  + (9* 10^9 ) \frac{(2)(8*10^{-9})}{(0.0420 - 0.03)\sqrt{(0.03)^2 + (4) (0.042 - 0.03)^2} }

        = 7.2 *10^{4} + 3.1*10^5

      E_3= 3.84*10^5 N/C                

8 0
3 years ago
A go-kart and rider have a mass of 14 kg. If the cart accelerates at 6 m/s² during a 40 m sprint in 100 seconds, how much power
Nutka1998 [239]

Answer:

<em>P=33.6 \ W</em>

Explanation:

<u>Mechanical Work and Power</u>

Work is the amount of energy transferred by a force. It's a scalar quantity, with SI units of joules.

Being \vec F the force vector and \vec s the displacement vector, the work is calculated as:

W=\vec F\cdot \vec s

If both the force and displacement are parallel, then we can use the equivalent scalar formula:

W=F.s

Mechanical Power is the amount of energy transferred or converted per unit of time. The SI unit of power is the watt, equal to one joule per second.

The power can be calculated as:

\displaystyle P=\frac {W}{t}

Where W is the work and t is the time.

If an object of mass m has an acceleration a, the net force is:

F=m.a

The go-kart and rider have a mass of m=14 kg and accelerate at 6 m/s^2, thus the net force applied is:

F=14\cdot 6 = 84\ N

The work done by the cart when traveling d= 40 m is:

W=84\cdot 40

W=3,360\ J

Finally, the power for t= 100 seconds is:

\displaystyle P=\frac {3,360}{100}

P=33.6 \ W

5 0
3 years ago
A jet aircraft is traveling at 262 m/s in horizontal flight. The engine takes in air at a rate of 85.9 kg/s and burns fuel at a
Fittoniya [83]

Answer:

F_T=60132.52N

P=15814852.76W

Explanation:

From the question we are told that

Velocity of aircraft  V=263m/s

Engine air intake rate \triangle M_a=85.9kg/s

Fuel burn rate  \triangle M_f =3.92kg/s

Velocity of exhaust gas V_e =921m/s

Generally the Mass change rate of Rocket is mathematically given by

 \triangle M = \triangle M_a+\triangle M_f

 \triangle M= 85.9+3.92

 \triangle M=89.82kg/s

Generally the Trust of the rocket is given mathematically by

 F_T=(\triangle M *V_e)-(dM_a/dt)*(V)

 F_T=(89.82 *921)-(85.9)*(263)

 F_T=60132.52N

Generally the Rocket's delivered power is mathematically given by

Delivered power P

 P=V*F_T

 P=263*60132.52N

 P=15814852.76W.

8 0
3 years ago
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