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3 years ago

Electricity costs 10p per unit. How much will it cost Sam to use a 85W laptop and a 60W lamp for an hour?

1 answer:

6 0

The standard unit is KW/hr, = 1,000W/hr.
(85 + 60) = 145W.
You need to find its fraction of 1,000W., so (145/1000) = 0.145 KWH.
(0.145 x 10p) = 1.45p. per hr.

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How many atoms are in one mole of potassium

Svetach [21]

Answer : 6.022• 10^23 atoms of potassium

5 0

3 years ago

Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place

Semmy [17]

Answer:

Ans= 9

See attached picture for clearer solution.

Explanation:

The net electrostatic force acting on charge A = 2/ 2 + 2 /(2) 2 − 2 /(3) 2 = 2 / 2 (1 + 1/4 – 1/9 ) = 41/36 2/2 .

The net electrostatic force acting on charge B = 2/2 + 2/(2)2 − 2/2 = 1/4 2/d2 .

The net electrostatic force acting on charge C = 2/2 + 2/(2)2 + 2/2 = 2/2 (1 + 1 4 + 1) = 9/4 2/2 .

The net electrostatic force acting on charge D = 2/2+ 2 /(2)2 + 2/(3)2 = 2 /2 (1 + 1/4 + 1/9 ) = 49/36 2/ 2 .

The ratio of the largest to the smallest net force = 9/4*2/2 / 1/4 2/2 . = 9

5 0

3 years ago

A positive charge of 8.0 × 10-4 C is in an electric field that exerts a force of 3.5 × 10-4 N on it. What is the strength of the

Gennadij [26K]

Answer:

E = 0.437 N/C

Explanation:

Given that,

Charge, $q=8\times 10^{-4}\ C$

Electric force, $F=3.5\times 10^{-4}\ N$

Let the strength of the electric field is E. We know that, the electric force is given by :

F = qE

Where

E is the electric field strength

$E=\dfrac{F}{q}\\\\E=\dfrac{3.5\times 10^{-4}}{8\times 10^{-4}}\\E=0.437\ N/C$

So, the strength of the electric field is equal to 0.437 N/C.

6 0

3 years ago

A 1.50-m string of weight 0.0125 N is tied to the ceil- ing at its upper end, and the lower end supports a weight W. Ignore the

Elena L [17]

The wave equation is missing and it is y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Answer:

A) 0.0534 seconds

B) 0.67N

C) 41

D) (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

Explanation:

we are given weight of string = 0.0125N

Thus, since weight = mg

Then, mass of string = 0.0125/9.8

Mass of string = 1.275 x 10⁻³ kg

Length of string; L= 1.5 m .

mass per unit length; μ = (1.275 x 10⁻³)/1.5

μ = 0.85 x 10⁻³ kg/m

We are given the wave equation: y(x,t) = (8.50 mm)cos(172 rad/m x − 4830 rad/s t)

Now if we compare it to the general equation of motion of standing wave on a string which is:

y(x,t) = Acos(Kx − ω t)

We can deduce that

angular velocity;ω = 4830 rad/s

Wave number;k = 172 rad/m

A) Velocity is given by the formula;

V = ω/k

Thus, V = 4830/172 m/s

V = 28.08 m /s

Thus time taken to go up the string = 1.5/28.08 = 0.0534 seconds

B) We know that in strings,

V² = F/μ

Where μ is mass per unit length and V is velocity.

Thus, F = V²*μ =28.08² x 0.85 x 10⁻³

F = 0.67N

C) Formula for wave length is given as; wave length;λ = 2π /k

λ = 2 x π/ 172

λ = 0.0365 m

Thus, number of wave lengths over whole length of string

= 1.5/0.0365 = 41

D) The equation for waves traveling down the string

= (8.50 mm)cos(172 rad/m x + 4830 rad/s t)

8 0

3 years ago

For a science project, you would like to horizontally suspend an 8.5 by 11 inch sheet of black paper in a vertical beam of light

liubo4ka [24]

Answer:

I = 3.9 x 10⁷ W/m²

Explanation:

given,

Sheet of black paper dimension = 8.5 x 11 inch

Area of sheet = 8.5 x 11 = 93.5 inch^2

1 inch =0.0254 m

Area = 0.06032 m²

mass of sheet = 0.80 g

Force = m g = 0.8 x 9.8 x 10⁻³ N

= 7.84 x 10⁻³ N

speed of light = c = 3 x 10⁸ m/s

Using equation

$F = \dfrac{IA}{c}$

where I is the intensity of light

$7.84 \times 10^{-3} = \dfrac{I\times 0.06032}{3 \times 10^8}$

$2.352 \times 10^{6} = I\times 0.06032$

$I = \dfrac{2.352 \times 10^{6}}{0.06032}$

I = 3.9 x 10⁷ W/m²

Intensity of the light is equal to I = 3.9 x 10⁷ W/m²

4 0

3 years ago