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3 years ago

How much voltage is required to run 0.42 A of current through a 150

Physics

2 answers:

IRINA_888 [86]3 years ago

8 0

Answer:

like the other person said, the answer is 6.3

Explanation:

igomit [66]3 years ago

4 0

The voltage in the resistor is 63 V

Explanation:

We can solve the problem by applying Ohm's law, which states the relationship between voltage, current and resistance in a resistor:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

For the resistor in this problem, we have:

I = 0.42 A is the current

$R=150 \Omega$ is the resistance

Substituting into the equation, we find the voltage needed:

$V=(0.42)(150)=63 V$

Learn more about voltage and current:

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Aleks [24]

Well sorry but this is the wrong language.

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3 years ago

Read 2 more answers

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

3 years ago

The work-energy theorem states that the work done on an object is equal to a change in which quantity?

Fynjy0 [20]

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

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2 years ago

What is the definition of physical change

jeka94

They are used to separate mixtures into their component compounds but can no usually be used to deprecate compounds into chemical elements or simpler compounds

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3 years ago

Whenever two apollo astronauts were on the surface of the moon, a third astronaut orbited the moon. assume the orbit to be circu

almond37 [142]

Missing question:
"Determine (a) the astronaut’s orbital speed v and (b) the period of the orbit"

Solution

part a) The center of the orbit of the third astronaut is located at the center of the moon. This means that the radius of the orbit is the sum of the Moon's radius r0 and the altitude ( $h=430 km=4.3 \cdot 10^5 m$ ) of the orbit:
$r= r_0 + h=1.7 \cdot 10^6 m + 4.3 \cdot 10^5 m=2.13 \cdot 10^6 m$
This is a circular motion, where the centripetal acceleration is equal to the gravitational acceleration g at this altitude. The problem says that at this altitude, $g=1.08 m/s^2$ . So we can write
$g=a_c= \frac{v^2}{r}$
where $a_c$ is the centripetal acceleration and v is the speed of the astronaut. Re-arranging it we can find v:
$v= \sqrt{g r}= \sqrt{(1.08 m/s^2)(2.13 \cdot 10^6 m)}=1517 m/s = 1.52 km/s$

part b) The orbit has a circumference of $2 \pi r$ , and the astronaut is covering it at a speed equal to v. Therefore, the period of the orbit is
$T= \frac{2 \pi r}{v} = \frac{2\pi (2.13 \cdot 10^6 m)}{1517 m/s} =8818 s = 2.45 h$
So, the period of the orbit is 2.45 hours.

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3 years ago