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3 years ago

A device is rated at 1.3kW when connected to a 120 V source. The equivalent resistance of this device in ohm is:

Physics

1 answer:

alexdok [17]3 years ago

7 0

Answer:

correct me if im wrong

brainlest plsss<333

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E) the time it takes the car to reach 20.0ms-1

Galina-37 [17]

4.2 thats the answer

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2 years ago

A 100-W lightbulb is placed in a cylinder equipped with a moveable piston. The lightbulb is turned on for 0.010 hour, and the as

Taya2010 [7]

Answer:

$w = - 508.53$ joules

$q = - 3091.47$ joules

Explanation:

Let us convert the time in hours into seconds

$0.010* 3600\\= 36$

Change in internal energy

$\delta E = p * \delta t$

where E is the internal energy in Joules

p is the power in watts

and t is the time in seconds

$\delta E = - 100 * 36\\$

$\delta E = - 3600$ Joules

Amount of work done by the system

$w = - P * \delta V$

where P is the pressure and V is the volume

Substituting the given values in above equation, we get -

$w = - 1 * ( 5.92 -0.90)\\$

$w = -5.02$ liter-atmospheres

Work done in Joules

$- 5.02 * 101.3\\$ $= 508.53$ Joules

$q = \delta E - w\\$

Substituting the given values we get -

$q = - 3600 - (-508.53)\\q = - 3091.47$

Thus

$w = - 508.53$ joules

$q = - 3091.47$ joules

7 0

3 years ago

The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr

Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

$I(t)=0.88e^{-t/6\times3600s}$

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

$q=\int\limits {I} \, dt$

Here the limits of integration is from 0 to infinite. So,

$q=\int\limits {0.88e^{-t/6\times3600s}}\, dt$

$q=0.88\times(-6\times3600)(0-1)$

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

$N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}$

N = 1.2 x 10²³

8 0

3 years ago

In which material does the light wave have the larger wavelength?

pochemuha

Answer: in the smaller/ thinner pieces.

8 0

2 years ago

Calculate the percentage increase in length of a wire of diameter 2.2 mm stretched by a load of

vesna_86 [32]

Answer:

0.21%

Explanation:

We are given;

Mass; m = 100 kg

Diameter; d = 2.2 mm = 2.2 × 10^(-3) m

Young's modulus; E = 12.5 x 10^(10) N/m².

Formula for area is;

A = πd²/4

A = (π/4) x (2.2 x 10^(-3))²

A = 3.8 x 10^(-6) m²

Force; F = mg

g is acceleration due to gravity and has a constant value of 9.8 m/s²

F = 100 × 9.8

F = 980 N

Formula for young's modulus is;

E = Stress/strain

Formula for stress = F/A

Formula for strain = ΔL/L

Thus;

E = (F/A)/(ΔL/L)

Making ΔL/L the subject, we have;

ΔL/L = (F/A)/E

Plugging in the relevant values;

ΔL/L = 980/(3.8 x 10^(-6) × 12.5 × 10^(10))

ΔL/L = 0.0021

Then percentage increase in length of a wire = 0.0021 × 100% = 0.21%

8 0

3 years ago