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svetlana [45]
3 years ago
6

The light energy that falls on a square meter of ground over the course of a typical sunny day is about 20 MJ. The average rate

of electric energy consumption in one house is 1.0 kW
On average, how much energy does one house use during each 24 hr day?
Express your answer using two significant figures.
Physics
1 answer:
Free_Kalibri [48]3 years ago
7 0

Answer:

86MJ

Explanation:

THE TOPIC FROM WHICH THE QUESTION IS OBTAINED IS WORK, ENERGY AND POWER

  • Power is defined as the time rate of doing work.
  • Power =\frac{Energy used/ work done}{time}
  • If the energy use is in joules and the time is in seconds, then power is a measure of watts. 1watt = 1 joule per second.
  • According to the question, we need to find how much energy does one house use during each 24 hour day ( Already given time = 24hours) when the average rate of electric energy consumption in one house is 1.0 kW ( we have been given power = 1kW).
  • Power = 1kW = 1 × 10³W
  • Time = 24hours = 24 ×3600 = 86400seconds.
  • It is important we have our time in seconds so as to be consistent with unit and dimension

Power = \frac{Energy used/ work done}{time}

1 × 10³W =\frac{Energy used}{86400}

  • Energy used =1 × 10³W × 86400
  • Energy used = 86400000J = 86.4 × 10^{6}W  = 86.4MJ  

The question requested the answer in 2 significant figures and the answer is 86MJ

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I really need help with this
katen-ka-za [31]

<u>C</u> is the correct answer, because energy cannot be created neither destroy. The energy is changing from chemical to from electric to light, and from light to heat.  

8 0
3 years ago
Cart a having a mass of 150 kg initially moving to the right at a speed of 8 m/s collides with cart b with a mass of 150 kg, ini
kakasveta [241]

The speed of cart b is 6m/s while the total momentum of the systmen is 4200 kg m/s

<h3>Conservation of Linear Momentum</h3>

Given Data

  • Mass of cart one M1  = 150kg
  • Initial Velocity U1 = 8m/s
  • Final VelocityV1 = 5 m/s

Mass of cart two M2  = 150kg

Velocity U2 = 6m/s

Applying the principle of conservation of linear momentum we have

M1U1+M2U2 = M1V1+ M2V2

a. what is the speed of cart b after collision

substituting our given data we have

150*8+ 150*6 = 150*5+150*V2

1200 + 900 = 1200+ 150V2

2100 - 1200 = 150V2

900 = 150V2

Divide both sides by 150

V2 = 900/150

V2 = 6m/s

b. what is the total momentum of the system before and after collision

Total Momentum in the system is

Total momentum = Momentum before Impact+ Momentum after Impact

Total momentum = M1U1+M2U2 + M1V1+ M2V2

Total momentum = 1200 + 900 + 1200+ 900

Total momentum = 4200 kg m/s

Learn more about Conservation of Linear Momentum here:

brainly.com/question/7538238

6 0
2 years ago
Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th
timama [110]

Answer:

Explanation:

Given

length of window h=2.9\ m

time Frame for which rock can be seen is \Delta t=0.134\ s

Suppose h is height above which rock is dropped

Time taken to cover h+2.9 is t_1

so using equation of motion

y=ut+\frac{1}{2}at^2

where  y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h  is

h=0+0.5\times g\times (t_2)^2---2

Subtract 1 and 2 we get

2.9=0.5g(t_1^2-t_2^2)

5.8=g(t_1+t_2)(t_1-t_2))

and from equation t_1-t_2=0.134\ s

so t_1+t_2=\frac{5.8}{9.8\times 0.134}

t_1+t_2=4.416\ s

and t_1=t_2+\Delta t

so t_2+\Delta t+t_2=4.416

2t_2+0.134=4.416

t_2=0.5\times 4.282

t_2=2.141\ s

substitute the value of t_2 in equation 2

h=0.5\times 9.8\times (2.141)^2

h=22.46\ m

                                                     

8 0
4 years ago
In a 2-dimensional coordinate system, the x- and y-axes are typically _____.
dem82 [27]
<span>In a 2-dimensional coordinate system, the x- and y-axes
are typically perpendicular to each other. (C) </span>
7 0
4 years ago
A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g gla
miss Akunina [59]

Answer:

The ball traveled 0.827 m

Explanation:

Given;

distance between the metal plates of the room, d = 3.1 m

mass of the glass, m = 1.1g

charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

F = qV/d

|F| = (4.7 x 10⁻⁹ x 3 x  10⁶) / (3.1)

|F| = 4.548 x 10⁻³ N

F = - 4.548 x 10⁻³ N

The net horizontal force experienced by this ball is;

F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N

The work done between the ends of the plate is equal to product of the  magnitude of net force on the ball and the distance traveled by the ball.

W = F_{net} *h\\\\W = 15.328 *10^{-3} *  h

W = K.E

15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m

Therefore, the ball traveled 0.827 m

4 0
4 years ago
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