im a boy and does anyone want to be my gf im 13 and in 7th grade and if you dont i get and if you want to yell at me i get

The info below shows three kettles with their powers and the time they take to boil 500cm3 of water. If electricity costs 9p per

dimaraw [331]

The cost of boiling 500 cm³ of water using the 3 KW Kettle is 1.35p

<h3>What is power? </h3>

This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:

Power (P) = Energy (E) / time (t)

P = E / t

<h3>How to determine the energy</h3>

Power (P) = 3 KW
Time (t) = 3 mins = 3 / 60 = 0.05 h
Energy (E) =?

E = Pt

E = 3 × 0.05

E = 0.15 KWh

<h3>How to determine the cost</h3>

Energy (E) = 0.15 KWh
Cost per unit = 9p
Cost =?

Cost = Energy × cost per unit

Cost = 0.15 × 9

Cost = 1.35p

Learn more about electrical power:

brainly.com/question/64224

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8 0

2 years ago

Pls help it's due today

mario62 [17]

Answer:

Breh seriously. Ugh fine.

1.B

2.D

3.C

4.C

5.D,A and B

6.A,C and D

3 0

2 years ago

The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be

bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

$F=\frac{Gm_{1} m_{2} }{d^{2} }$ ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

$1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }$

$d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}$

$d=\sqrt{1.48\times10^{17}}$

d = 3.85 x 10⁸ m

3 0

3 years ago

The climate of the earth throughout history has always _____.

Vlad [161]

Answer:

The climate of the earth throughout history has always been <em><u>fluctuated between hot and cold periods. </u></em>

7 0

4 years ago

Read 2 more answers

A car is running at the speed of 45km/hr see a child 25 meter ahead and suddenly apllies a brakes. If the retradation of the car

kenny6666 [7]

Answer:

The car stops in 7.78s and does not spare the child.

Explanation:

In order to know if the car stops before the distance to the child, you take into account the following equation:

$x=x_o+v_ot-\frac{1}{2}at^2$ (1)

vo: initial speed of the car = 45km/h

a: deceleration of the car = 2 m/s^2

t: time

xo: initial distance to the child = 25m

x: final distance to the child = 0m

It is necessary that the solution of the equation (1) for time t are real.

You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:

$45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}$

$0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s$

You take the first value t1 because it has physical meaning.

The solution for t is real, then, the car stops in 7.78s and does not spare the child.

4 0

3 years ago