Ask question

Ask question

All categories

3 years ago

5

In our lab experiment on Ohm's law, the power supply is connected to a circuit containing one resistor, and a direct current of

1.5 A is drawn. If you connect a second identical resistor to the power supply in parallel with the first, (similar to what we did in the lab) how much current is drawn from the battery?
(A) 3.0 A
(B) 0.75 A
(C) 1.5 A
(D) 6.0 A

1 answer:

lapo4ka [179]3 years ago

3 0

Answer: (A) 3.0=A

Explanation: In order to explain this problem we have to use the OHM law, given by: V=R*I

Besides, we have to consider the resitance equivalent for a parallel connection. This is given by:

1/Re=1/R1+1/R2

If we connect the same resistance, the equivalent resistance is R/2.

Initlally the current is 1.5 A when one resistance is connected to the batttery. When a second resistance with the same value is connected in parallel to the battery, we have V=Re*Ifinal= (R/2)*Ifinal

also we know that V=R*Iinitial so Iinitial=V/R

then Ifinal= 2*V/R=2*Iinitial

You might be interested in

If two objects are moving at the same speed, and Object 1 has four times as much mass as Object 2, how much momentum does Object

Margaret [11]

The answer is number 1

7 0

3 years ago

During the motion of the slinky in a transverse wave, what do the particles of the slinky coil do?

Mazyrski [523]

Answer:

C.) The slinky particles move up and down

Explanation:

<u>Transverse Wave</u>-

<em>A wave that has a disturbance perpendicular to the wave motion</em>

<em></em>

<em>Hello! This is the correct answer! Have a blessed day! :)</em>

<em>If you are in K12, please review the lesson! :) It will give you some very helpful definitions! I hope this helped!</em>

<u />

5 0

2 years ago

Why are satellites placed into orbit at least 150 km above Earth’s surface?

dybincka [34]

Because its just enough to where its not out f the gravitational pull and not close enough to be pulled back to earth. Hope it helps<span />

8 0

2 years ago

Read 2 more answers

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago

The distance from Earth to the North Star is about 430 light-years. Which of the following statements describes why scientists u

vampirchik [111]

Answer:

D

Explanation:

D

7 0

2 years ago