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4 years ago

An alarm clock draws 0.5 A of current when connected to a 120 volt circuit. Calculate its resistance.

1 answer:

8 0

<span>We need to calculate the resistance of a clock which draws current I=0.5A and is connected to a circuit of V=120V. We can find the resistance of a circuit from Ohm's law: I=U/R where I is the current, U is the voltage and R is the resistance. So lets manipulate Ohm's law to find the resistance: R=U/I. Now we input our numbers: R=120V/0.5A and we get R=240 Ohm. So the resistance is R=240 Ohm.</span>

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Water is poured into a bowl at a constant rate of 17.0 cm^3/s. The bowl has a circular cross section, but does not have a unifor

7nadin3 [17]

Answer:

10.29 cm/s

Explanation:

Discharge in to the bowl = 17.0 cm³/s

Diameter of the bowl, d₁ = 1.45 cm

Now,

Rate at which water level rise at its diameter = $\frac{\textup{Discharge}}{\textup{Area of cross-section}}$

also,

Area of cross-section = $\frac{\pi}{\textup{4}}\times1.45^2$

Area of cross-section = 1.651 cm²

Therefore,

Rate at which water level rise at its diameter = $\frac{\textup{17}}{\textup{1.651}}$

Rate at which water level rise at its diameter = 10.29 cm/s

8 0

4 years ago

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist

Sidana [21]

Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is $g = 167.2 \ m/s^2$

Explanation:

From the question we are told that

The length of the simple pendulum is $L = 1.081.08 \ m$

The number of cycles is $N = 101$

The time take is $t = 2.00 *10^{2 \ }s$

Generally the period of this oscillation is mathematically evaluated as

$T = \frac{N}{t }$

substituting values

$T = \frac{101}{2.0*10^2 }$

$T = 0.505 \ s$

The period of this oscillation is mathematically represented as

$T = 2 \pi \sqrt{\frac{l}{g} }$

making g the subject of the formula we have

$g = \frac{L}{[\frac{T}{2 \pi } ]^2 }$

$g = \frac{4 \pi ^2 L }{T^2 }$

Substituting values

$g = \frac{4 * 3.142 ^2 * 1.08 }{505.505^2 }$

$g = \frac{4 * 3.142 ^2 * 1.08 }{0.505^2 }$

$g = 167.2 \ m/s^2$

7 0

3 years ago

The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of

goldfiish [28.3K]

Answer:

$6.6*10^{27}e/m^3$

Explanation:

When we are dealing with Hall voltage, it is necessary to have the values of the current, the magnetic field, the length, the area and the number of carriers at hand. The Hall voltage equation is given by, $V_h = \frac{iB}{neL}$