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GarryVolchara [31]

3 years ago

13

An oil bath maintained at 50.5°C loses heat to its surroundings at the rate of 4.68 kJ/min. Its temperature is maintained by an

electrically heated coil with a resistance of 60 operated from a 110 V line. A thermoregulator switches the current on and off. What fraction of the time will the current be turned on?

1 answer:

elena-14-01-66 [18.8K]3 years ago

3 0

Answer:

The fraction of time for turn on is 0.3852

Solution:

As per the question:

Temperature at which oil bath is maintained, $T_{o} = 50.5^{\circ}$

Heat loss at rate, q = 4.68 kJ/min

Resistance, R = $60\Omega$

Operating Voltage, $V_{o} = 110 V$

Now,

Power that the resistor releases, $P_{R} = \frac{V_{o}^{2}}{R}$

$P_{R} = \frac{110^{2}}{60} = 201.67 W = 12.148 J/min$

The fraction of time for the current to be turned on:

$P_{R} = \frac{q}{t}$

$12.148 = \frac{4.68}{t}$

t = 0.3852

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Dispositivo que muestra tanto la dirección como la magnitud de la corriente eléctrica.

slamgirl [31]

Answer:

<em>Galvanómetro</em>

Explanation:

Un galvanómetro es un dispositivo eléctrico utilizado para <em>detectar la presencia de corriente y voltaje pequeños o para medir su magnitud.</em> Los galvanómetros se utilizan principalmente en puentes y potenciómetros.

5 0

3 years ago

Two particles are fixed to an x axis: particle 1 of charge −1.50 ✕ 10−7 c at x = 6.00 cm, and particle 2 of charge +1.50 ✕ 10−7

sleet_krkn [62]

Answer : $\underset{E_{R}}{\rightarrow} =-2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Explanation :

Given that,

Charge of particle 1 = $-1.50\times10^{-7} c$

Distance x = 6 cm

Charge of particle 2 = $1.50\times10^{-7} c$

Distance x = 27 cm

Total distance = $\dfrac{6+27}{2}$

$r = 16.5\ cm$

Particle 1 is at (6,0) and particle 2 is at (27,0) .

Therefore, midway (16.5, 0)

Now, $r = \dfrac{|6-16.5|}{2} = \dfrac{|27-16.5|}{2} = 10.5\ cm$

Formula of electric field

$E = \dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q}{r^{2}}$

Now, the the electric field due to particle 1

$\underset{E}{\rightarrow}\ = -\dfrac{9\times10^{9}\times1.50\times10^{-7 }}{10.5}\ \widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = \dfrac{13.5\times10^{2}}{(10.5\times10^{-2})^{2}}\widehat{i} \dfrac{N}{C}$

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Similarly, the electric field due to particle 2

$\underset{E}{\rightarrow} = -1.22\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Resultant Electric field

$\underset{E_{R}}{\rightarrow} = \underset{E_{1}}{\rightarrow} + \underset{E_{2}}{\rightarrow}$

$\underset{E_{R}}{\rightarrow} = -2.44\times10^{5}\ \widehat{i} \ \dfrac{N}{C}$

Hence, this is the required answer.

3 0

3 years ago

The current in the wires of a circuit is 60.0 milliAmps. If the voltage impressed across the ends of the circuit were doubled (w

Pavel [41]

Answer:

120 mA

Explanation:

The relationship between voltage, current and resistance in a circuit is given by Ohm's law

$I=\frac{V}{R}$

where

I is the current

V is the voltage

R is the resistance

In this problem, we have an initial current of

I = 60 mA

Then the voltage is doubled:

V' = 2 V

while the resistance is kept the same:

R' = R

So the new current is

$I' = \frac{V'}{R'}=\frac{2V}{R}=2\frac{V}{R}=2 I$

so, the current has doubled. Since I = 60 mA, the new current is

$I' = 2(60 mA)=120 mA$

3 0

3 years ago

Please help me with my physics! I do not understand and please provide work.

Anuta_ua [19.1K]

Yes I am here to answer

3 0

3 years ago

The intensity of sunlight reaching the earth is 1360 w/m2. Assuming all the sunligh is absorbed, what is the radiation pressure

krok68 [10]

(a) $1.15\cdot 10^9 N$

For an electromagnetic wave incident on a surface, the radiation pressure is given by (assuming all the radiation is absorbed)

$p=\frac{I}{c}$

where

I is the intensity

c is the speed of light

In this problem, $I=1360 W/m^2$ ; substituting this value, we find the radiation pressure:

$p=\frac{1360 W/m^2}{3\cdot 10^8 m/s}=4.5\cdot 10^{-6} Pa$

the force exerted on the Earth depends on the surface considered. Assuming that the sunlight hits half of the Earth's surface (the half illuminated by the Sun), we have to consider the area of a hemisphere, which is

$A=2pi R^2$

where

$R=6.37\cdot 10^6 m$

is the Earth's radius. Substituting,

$A=2\pi (6.37\cdot 10^6 m)^2=2.55\cdot 10^{14}m^2$

And so the force exerted by the sunlight is

$F=pA=(4.5\cdot 10^{-6} Pa)(2.55\cdot 10^{14} m^2)=1.15\cdot 10^9 N$

(b) $3.2\cdot 10^{-14}$

The gravitational force exerted by the Sun on the Earth is

$F=G\frac{Mm}{r^2}$

where

G is the gravitational constant

$M=1.99\cdot 10^{30}kg$ is the Sun's mass

$m=5.98\cdot 10^{24}kg$ is the Earth's mass

$r=1.49\cdot 10^{11} m$ is the distance between the Sun and the Earth

Substituting,

$F=(6.67\cdot 10^{-11} )\frac{(1.99\cdot 10^{30}kg)(5.98\cdot 10^{24} kg)}{(1.49\cdot 10^{11} m)^2}=3.58\cdot 10^{22}N$

And so, the radiation pressure force on Earth as a fraction of the sun's gravitational force on Earth is

$\frac{1.15\cdot 10^9 N}{3.58\cdot 10^{22}N}=3.2\cdot 10^{-14}$

6 0

4 years ago