Ask question

Ask question

All categories

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.

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.

You might be interested in

Which simple machines make up a wheel barrow

7nadin3 [17]

Answer:

a lever and a wheel and axle (i guess)

Explanation:

don't have any

4 0

3 years ago

How do you convert from picoseconds to milliseconds using the conversion factors?

andrew11 [14]

There are 10⁹ picoseconds in 1 Ms

1 picosecond= 10¹² s

1 Ms = 10⁻³ s

so the number of picoseconds in one Ms=(10⁻³ s/1 Ms) * (10¹² Ps/ 1 s)=10⁹

Thus there are 10⁹ picoseconds in 1 Ms

8 0

3 years ago

A 100g block is initially compressing a spring 5.0 cm. The spring launches the block 50cm horizontally along the ground with a f

Setler [38]

Answer:

7200 N/m

Explanation:

Metric unit conversion

100g = 0.1 kg

5 cm = 0.05 m

50 cm = 0.5 m

As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground

The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

$W_f = F_fs = 15*0.5 = 7.5 J$

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

$E_p = mgh = 0.1*10*1.5 = 1.5 J$

Therefore, as elastic energy is converted to potential energy and work of friction:

$E_e = W_f + E_p$

$kx^2/2 = 7.5 + 1.5 = 9 J$

$k = 9*2/x^2 = 18/0.05^2 = 7200 N/m$

6 0

3 years ago

Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made o

Colt1911 [192]

Answer:

There is a loss of fluid in the container of 0.475L

Explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

$\Delta V = \beta V_0 \Delta T$

Where,

$\Delta V =$ Change in volume

$V_0 =$ Initial Volume

$\Delta T =$ Change in temperature

$\beta =$ coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

$\Delta V_T = \Delta V_l - \Delta V_c$

Where,

$\Delta V_l$ = Change in the volume of liquid

$\Delta V_c$ = Change in the volume of copper

Then replacing with the previous equation we have:

$\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T$

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

Our values are given as,

Thermal expansion coefficient for copper and the liquid to 20°C is

$\beta_c = 51*10^{-6}/\°C$

$\beta_l = 400*10^{-6}/\°C$

$V_0 = 16L$

$\Delta T = (95\°C-10\°C)$

Replacing we have that,

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

$\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)$

$\Delta V = 0.475L$

Therefore there is a loss of fluid in the container of 0.475L

6 0

3 years ago

When two positive charges are brought close together, what happens to the

telo118 [61]

Answer:

they will move away from each other

7 0

3 years ago