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Lubov Fominskaja [6]
3 years ago
11

A 250 watt electric bulb is lighted for 5 hours daily and four 6 watt bulbs are lighted for 4.5 hours daily. Calculate the energ

y consumed (in kWh) in the month of February.
Physics
1 answer:
densk [106]3 years ago
5 0

Answer:

38.024 KWh

Explanation

250 watts multiplied with 5 hours daily multiplied with 28 days monthly equals 35000 Watt hours. Divided by a 1000 to get the KWh equals 35 KWh.

= 250×5hrs× 28days

= 35000watts

= 35000/1000

= 35KWh

Four 6 watt lightbulbs equals 24 watts 4x6=24

Hence, 24×4.5hrs×30days

= 3024watts

= 3024/1000

= 3.024KWhr

The total amount of energy consumed in the month of February = 35 KWh + 3.024 KWh = 38.024 KWh

Note that I had to use 28days since we are considering the month of February.

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What happens when parallel rays of light hit a smooth mirror surface
Gnesinka [82]

Answer:it rotate

Explanation:

it just spin

8 0
3 years ago
5.
iris [78.8K]

Answer:

t = 1.659s

Explanation:

We can use the kinematics equations to solve this questions:

v = u + at

v^{2} = u^{2} +2as

where v = Final Velocity, u = initial velocity, a = acceleration, t = time, s = displacement

a) Given information from the question,

u = \frac{70km}{h} =\frac{(70*1000)m}{(1*3600)s} = 19.444m/s (Convert km/h to m/s first)

a = 2m/s^{2}

s = 35m

Now we can substitute these values into the 2nd kinematics equation to find v, final velocity.

v^{2} =(19.444)^{2} +2(2)(35)\\v=\sqrt{(19.444)^{2} +2(2)(35)} \\v= 22.761m/s (5.sf)\\

b) Now we have the final velocity, we can substitute the values into the first kinematics equation to find t , the time taken.

v = u + at

22.761 = 19.444 + 2t

2t = 22.761 - 19.444

t =\frac{22.761-19.444}{2}

t = 1.659s

7 0
2 years ago
What statement applies to the horizontal rows or periods in the periodic table? A. The properties are all the same. B. The eleme
Sliva [168]
The right answer for the question that is being asked and shown above is that: "C. <span>. The properties change going across each row. " the </span>statement that applies to the horizontal rows or periods in the periodic table is that t<span>he properties change going across each row. </span>
7 0
3 years ago
How many joules are in 10,000 Electron Volts
Semenov [28]
1.602177e-15 joules are in 10,000 electron volts

5 0
3 years ago
Read 2 more answers
Problem 12: A 20.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong w
avanturin [10]

Answer:

The deformation in the pole due to force is 0.70 mm.

Explanation:

Given that,

Height = 20.0 m

Diameter = 4.00 cm

Force = 1100 N

We need to calculate the  area

Using formula of area

A=\pi\times r^2

A=\pi\times(2.00\times10^{-2})^2

A=0.00125\ m^2

A=1.25\times10^{-3}\ m^2

We need to calculate the deformation

Using formula of deformation

\Delta x=\dfrac{1}{s}(\dfrac{F}{A}\times L)

Where, s = shear modulus

F = force

l = length

A = area

Put the value into the formula

\Delta x=\dfrac{1}{2.5\times10^{10}}\times(\dfrac{1100}{1.25\times10^{-3}}\times 20.0)

\Delta x=0.000704\ m

\Delta x=7.04\times10^{-4}\ m

\DElta x=0.70\ mm

Hence, The deformation in the pole due to force is 0.70 mm.

3 0
4 years ago
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