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Serga [27]
3 years ago
13

What is the resistance of a light bulb if a potential difference of 120 V will produce a current of 0.5 A in the bulb?

Physics
2 answers:
Yakvenalex [24]3 years ago
6 0

Explanation:

Ohm's law:

V = IR

120 V = (0.5 A) R

R = 240 Ω

Damm [24]3 years ago
3 0

Answer:

The resistance of alight bulb is 240 ohms.

Explanation:

Given that,

Potential difference of a light bulb, V = 120 V

Current produced in the light bulb, I = 0.5 A

We need to find the resistance of the light bulb. It can be calculated using Ohm's law. It is given by :

V = IR

R=\dfrac{V}{I}

R=\dfrac{120\ V}{0.5\ A}

R = 240 ohms

So, the resistance of alight bulb is 240 ohms. Hence, this is the required solution.

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An airplane flies 120 km at a constant altitude in a direction 30.0° north of east. a wind is blowing that results in a net hori
astraxan [27]
 Best Answer:<span>  </span><span>The net angle between the direction of flight of the aircraft and the wind in the opposing direction is 20. THe component opposing the aircraft is 2.4*cos 20 KN 
= 2400 * 0.9397 = 2255.2622 N. The distance covered is 120 Km 
The work done by the aircraft overcoming the wind is 
= 2255.2622 * 120000 = 270631464 = 2.71 x 10^8 N 
As the question is trickily worded as : the work done on the plane by the air (wind) the answer is -2.71 x 10^8 J . (fourth option)</span>
6 0
3 years ago
Read 2 more answers
In an isolated system, Bicycle 1 and Bicycle 2, each with a mass of 10 kg,
dimulka [17.4K]

Answer: 20 kgm/s

Explanation:

Given that M1 = M2 = 10kg

V1 = 5 m/s , V2 = 3 m/s

Since momentum is a vector quantity, the direction of the two object will be taken into consideration.

The magnitude of their combined

momentum before the crash will be:

M1V1 - M2V2

Substitute all the parameters into the formula

10 × 5 - 10 × 3

50 - 30

20 kgm/s

Therefore, the magnitude of their combined momentum before the crash will be 20 kgm/s

7 0
3 years ago
draw a velocity graph with Vi = 4 m/s and decreasing uniformly so that velocity at 2 seconds is 2 m/s and remaining constant fro
Shkiper50 [21]

For the velocity graph: start at 0s and 4m/s and draw a straight line to 2s and 2 m/s. Then draw a straight horizontal line to 4s and 2m/s

For the acceleration graph: start with a horizontal line from 0s and 2m/s/s to 2s and 2m/s/s. The draw another line from 2 s and 0m/s/s to 4 s and 0m/s/s

6 0
3 years ago
Sketch a reaction progress curve for a reaction that has an activation energy of 22 kj and the total energy change is -103kj.
crimeas [40]

Answer:

Do find the answer in the attachment herein.

Explanation:

From the attached diagram:

I. Activation energy = Activated complex - ∆H(reactants)

Activation energy = 162-140 = 22Kj.

II. ∆H(reaction) = ∆H(products) - ∆H(reactants)

∆H(reaction) = 37 - 140 = -103Kj.

8 0
3 years ago
A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha
Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

x(t) = A\cdot \cos (\omega \cdot t + \phi)

Where:

x(t) - Position of the mass with respect to the equilibrium position, measured in centimeters.

A - Amplitude of the mass-spring system, measured in centimeters.

\omega - Angular frequency, measured in radians per second.

t - Time, measured in seconds.

\phi - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)

Where the maximum acceleration of the system is represented by \omega^{2}\cdot A.

The natural frequency of the mass-spring system is:

\omega = \sqrt{\frac{k}{m} }

Where:

k - Spring constant, measured in newtons per meter.

m - Mass, measured in kilograms.

If k = 12\,\frac{N}{m} and m = 0.40\,kg, the natural frequency is:

\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }

\omega \approx 5.477\,\frac{rad}{s}

Lastly, the maximum acceleration of the system is:

a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)

a_{max} = 359.970\,\frac{cm}{s^{2}}

The maximum acceleration of the system is 359.970 centimeters per square second.

7 0
3 years ago
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