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Xelga [282]
1 year ago
7

Which of the following devices would you expect to consume the most energy for each hour that it operates? a portable tape recor

der an electric space heater a desk lamp an electric can openerwhich of the following devices would you expect to consume the most energy for each hour that it operates? the answer is an electric space heater?
Physics
2 answers:
Alisiya [41]1 year ago
8 0

Answer: electric space heater

Explanation: The consumption of electrical energy per unit hour  is given by the power rating on the electrical appliance. The power rating depends on the electric current it withdraws to operate and the resistance of the appliance.

Portable tape recorder, a desk lamp and an electric can opener are small devices as compared to an electric space heater and with draw less current. thus, the power rating of the space heater is higher than other given appliances and it consumes more energy per hour.

Charra [1.4K]1 year ago
3 0
You are correct, the answer is an electric space heater.
The reason for this is because the heating element generates heat by resisting the flow of current. For this reason, other similar devices such as irons take up a large amount of energy as well.
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Three disks are spinning independently on the same axle without friction. Their respective rotational inertias and angular speed
ololo11 [35]

Answer:

3/7 ω

Explanation:

Initial momentum = final momentum

I(-ω) + (2I)(3ω) + (4I)(-ω/2) = (I + 2I + 4I) ωnet

-Iω + 6Iω - 2Iω = 7I ωnet

3Iω = 7I ωnet

ωnet = 3/7 ω

The final angular velocity will be 3/7 ω counterclockwise.

7 0
2 years ago
Read 2 more answers
dopasuj wartości pracy z ramki do przedstawionych sytuacji ,a nastepnie wyraż te prace w dżulach uwaga jedna wartośc pracy nie b
Eduardwww [97]

Answer:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Explanation:

Całą część pytania można znaleźć na poniższym schemacie.

Z diagramu załączonego poniżej; mamy

A - samolot lotniczy

B - dźwig

C - podnoszenie ciężarów

Wszyscy to wiemy ;

1kJ = 1000 J

1MJ = 1000000 J

Mamy cztery opcje; i.e 200 MJ, 800 kJ, 1.6 kJ  and 250 mJ

Z czterech opcji można wykluczyć 250 mJ, ponieważ jest to 0,25 J, co przedstawia bardzo niską energię w porównaniu z trzema warunkami pokazanymi na schemacie.

Więc:

A (samolot) - 200 MJ = 200000000 J

B (dźwig) - 800 kJ = 800000 J

C (podnośnik)-1.6 kJ = 1600 J

Największą pracę wykona samolot. Jest tak, ponieważ ma bardzo dużą masę i bardzo dużą prędkość. W związku z tym istnieje potrzeba wytworzenia ogromnej ilości ciepła i energii.

Z drugiej strony żuraw może podnieść ładunek o wiele większy i przewyższa ciężar ciężaru, więc praca wykonywana przez dźwig musi być zdecydowanie większa niż praca ciężarka.

3 0
1 year ago
A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
adell [148]

Answer:

r = 0.114 m

Explanation:

To find the speed of the proton, from conservation of energy, we know that

KE = PE

Thus, we have;

(1/2)mv² = qV

Where;

V is potential difference = 1kv = 1000V

q is charge on proton which has a value of 1.6 x 10^(-19) C

m is mass of proton with a constant value of 1.67 x 10^(-27) kg

Let's make the velocity v the subject;

v =√(2qV/m)

v = √(2•1.6 x 10^(-19)•1000)/(1.67 x 10^(-27))

v = 4.377 x 10^(5) m/s

Now to calculate the radius of the circular motion of charge we know that;

F = mv²/r = qvB

Thus, mv²/r = qvB

Divide both sides by v;

mv/r = qB

Thus, r = mv/qB

Value of B from question is 0.04T

Thus,

r = (1.67 x 10^(-27) x 4.377 x 10^(5))/(1.6 x 10^(-19) x 0.04)

r = 0.114 m

r = 8.76 m

8 0
1 year ago
An engineer is designing a process for a new transistor. She uses a vacuum chamber to bombard a thin layer of silicon with ions
adelina 88 [10]

Answer:

E=5.7\times 10^{-3}\ V/m

Explanation:

Given that

m_p=5.18\times 10^{-26}\ kg

re= 46 cm

Vp= 180 m/s

We know that

E=\dfrac{\Delta V}{r}

\Delta V=\dfrac{1}{4}\dfrac{m_pv_p^2}{e}

So

E=\dfrac{1}{4}\dfrac{m_pv_p^2}{e.r_e}

Now by putting the all given values in the questions

E=\dfrac{1}{4}\times \dfrac{5.18\times 10^{-26}\times 180^2}{1.6\times 10^{-19}\times 0.46}

E=5.7\times 10^{-3}\ V/m

So the average electric field is E=5.7\times 10^{-3}\ V/m.

6 0
2 years ago
Calculate the current through a 10.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V
Kipish [7]

Answer:

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

Explanation:

Given:

Length = l = 10 meter

Radius = r = 0.321\ mm =0.321\times 10^{-3}\ meter

Resistivity=\rho=1.00\times 10^{-6}\ ohm\ meter

V = 12 Volt

To Find:

Current, I =?

Solution:

Resistance for 0.0-m long 22-gauge nichrome wire with a radius of 0.321 mm if it is connected across a 12.0-V battery given as

R=\dfrac{\rho\times l}{A}

Where,

R = Resistance

l = length

A = Area of cross section = πr²

\rho=Resistivity=1.00\times 10^{-6}\ ohm\ meter

Substituting the values we get

R=\dfrac{1\times 10^{-6}\times 10}{3.14\times (0.321\times 10^{-3})^{2}}

R=\dfrac{1\times 10^{-5}}{3.23\times 10^{-7}}

R=\dfrac{1\times 10^{2}}{3.23}

R=30.95\ ohm

Now by Ohm's Law,

V= I\times R

Substituting the values we get

I=\dfrac{V}{R}=\dfrac{12}{30.95}=0.3876\ Ampere

Therefore,

Current through Nichrome wire is 0.3879 Ampere.

4 0
1 year ago
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