All categories

3 years ago

A TV is rated at 0.1 kW. This TV is used 3 hours per day. How much energy does the TV use per month (in kWh)?

Physics

1 answer:

exis [7]3 years ago

6 0

Answer:

A.9 kWh

Explanation:

Tv is rated 0.1 kW

in one day energy used will be 0.1kW ×3hr=0.3kwh

in 30 days energy used will be 0.3kwh×30=9kwh

You might be interested in

Steam enters a long, horizontal pipe with an inlet diameter of D1 = 16 cm at 2 MPa and 300°C with a velocity of 2.5 m/s. Farther

polet [3.4K]

Answer:

m = 0.4005 kg/s

Q_out = 45.1 KJ/s

Explanation:

Given

Pipe inlet diameter D1 = 16 cm

Steam inlet pressure P1 = 2 Mpa

Steam inlet temperature T1 = 300 °C

Pipe outlet diameter D2 = 14 cm

Steam inlet velocity V1 = 2.5 m/s

Steam outlet pressure P2 = 1.8 MPa

Steam outlet temperature T2 = 250 °C

Required

Determine

(a) The mass flow rate of steam.

(b) The rate of heat transfer.

Assumptions

Kinetic and potential energy changes are negligible.

This is a steady flow process.

There is no work interaction.

Solution

Part a From steam table (A-6) at P1 = 2 Mpa , T1 = 300 °C

vl = 0.12551 m^3/Kg

h1 = 3024.2 KJ/Kg

The mass flow rate of steam could be defined as the following

m = 1/v1*A1*V1

m = 0.4005 kg/s

Part b We take the pipe as our system.The energy balance could be defined as the following

E_in -E_out =ΔE_sys = 0

E_in = E_out

mh_1 = Q_out + mh_2

Q_out = m(h_1-h_2)

From steam table (A-6) at P2= 1.8 Mpa T2 = 250 °C h2= 2911.7 KJ/Kg The heat transfer could be defined as the following

Q_out = m(h_1-h_2)

Q_out = 0.4005*(3024.2 -2911.7) =45.1 KJ/s

6 0

3 years ago

What happens to the brightness of bulb a when the switch is closed and bulb b lights up?

eimsori [14]

The problem describes the relationship of "bulb a" and "bulb b" to be in connected in series. When the switch is open then no current can flow, on the other hand, when it is closed, current will pass through.

When only "bulb a" is connected to the battery then more current is flowing to "bulb a" causing it to be bright.

Closing the switch would mean that "bulb b" is already included in the circuit and the battery will push small current to flow around the whole circuit. The more bulbs are connected, the harder for the current to flow because the resistance will be very high.

So the light of "bulb a" will be dimmer.

8 0

4 years ago

You start at (8, 1). You move right 2 units. Where do you end?

ivann1987 [24]

I'm pretty sure should be shifting your transformation those two units to the right from the original function.

Moving right = subtracting

Moving left = adding

4 0

3 years ago

In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of

aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

L= r × p The raduis and momentum are perpendicular.

L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]

w = 158.6/208.96

w = 0.76rad/s

7 0

3 years ago

A tire is filled with air at 10 ∘C to a gauge pressure of 250 kPa. Part A If the tire reaches a temperature of 45 ∘C, what fract

Alex_Xolod [135]

Answer:

The air fraction to be removed is 0.11

Given:

Initial temperature, T = $10^{\circ}$ = 283 K

Pressure, P = 250 kPa

Finally its temperature increases, T' = $45^{\circ}$ = 318 K

Solution:

Using the ideal gas equation:

PV = mRT

where

P = Pressure

V = Volume

m = no. of moles of gas

R = Rydberg's Constant

T = Temperature

Now,

Considering the eqn at constant volume and pressure, we get:

mT = m'T'

Thus

$\frac{m}{m'} = \frac{T'}{T}$ (1)

Now, the fraction of the air to be removed for the maintenance of pressure at 250 kPa:

$y = \frac{m - m'}{m} = 1 - \frac{m'}{m}$

From eqn (1):

$y = 1 - \frac{T}{T'}$

$y = 1 - \frac{283}{318} = 0.11$

6 0

4 years ago