Ask question

Ask question

All categories

MA_775_DIABLO [31]

3 years ago

12

A resistor of 5 Ω is placed in a circuit. The voltage drop across the resistor is 12 V. What is the current through the resistor

?

1 answer:

Kaylis [27]3 years ago

5 0

<span>Data:

</span>R = 5 Ω
U = 12 V
i = ?
<span>
Formula:

</span> $U = R*i$ → $i = \frac{U}{R}$ <span>

Solving:

</span> $i = \frac{U}{R}$
$i = \frac{12}{5}$
$\boxed{i = 2.4A}$

Answer:

<span>The current through the resistor is 2.4 Amperes</span><span>

</span>

You might be interested in

Subduction occurs at which of the following tectonic plate boundaries?

exis [7]

When two tectonic plates collide and form a converging plate boundry, normally one of the plates will slide underneath the other and that is when Subduction occurs.

5 0

3 years ago

The name used to describe the range of EM waves

Nimfa-mama [501]

Answer: the electromagnetic spectrum

Explanation:

I’m not sure but I think this is it!

8 0

3 years ago

A ball is droped from a height of 16m how much time will pass before the ball hits the ground

sergey [27]

Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

Hence we use free fall formula to calculate the time by the ball to hit the ground

$h= \frac{1}{2}g{t}^{2}$

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

Also, let take

$g = 9.8m{s}^{-2}$

By substitution we obtain,

$16= \frac{1}{2}\times 9.8{t}^{2}$

$\implies32=9.8{t}^{2}$

Diving through by 9.8

$\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}$

$\implies{t}^{2} =3.265$

square root both sides, we obtain

$\implies t= \sqrt{3.265}$

$t=1.8s$

4 0

3 years ago

An electron is projected with an initial speed of 5 3.2 10 / ⋅ m s directly toward a proton that is fixed in place. If the elect

Dvinal [7]

Answer:

The distance is $d =1.66*10^{-9}m$

Explanation:

From the question we are told that

The initial speed of the electron is $v_i = 3.2 *10^5 m/s$

The mass of electron is $m = 9.11*10^{-31}kg$

Let $d$ be the distance between the electron and the proton when the speed of the electron instantaneously equal to twice the initial value

Let $KE_i$ be the initial kinetic energy of the electron \

Let $KE_d$ be the kinetic energy of the electron at the distance $d$ from the proton

Considering that energy is conserved,

The energy at the initial position of the electron = The energy at the final position of the electron

i.e

$KE_i +U_1 = KE_d + U_2$

$U_1 \ and \ U_2$ are the potential energy at the initial position of the electron and at distance d of the electron to the proton

Here $U_1 = 0$

So the equation becomes

$\frac{1}{2} mv_i^2 = \frac{1}{2} mv_d^2 + \frac{kq_1 q_2}{d}$

Here $q_1 \ and \ q_2$ are the charge on the electron and the proton and their are the same since a charge on an electron is equal to charge on a proton

$k$ is electrostatic constant with value $8.99*10^9 N \cdot m^2 /C^2$

i.e $q = 1.602 *10^{-19}C$

$v_d$ is the velocity at distance d from the proton = 2 $v_i$

So the equation becomes

$\frac{1}{2}mv_i^2 = \frac{1}{2} m (2v_i)^2 -\frac{k(q)^2}{d}$

$\frac{1}{2} mv_i^2 = 4 [\frac{1}{2}mv_i^2 ]- \frac{k(q)^2}{d}$

$3[\frac{1}{2}mv_i^2 ] = \frac{k(q)^2}{d}$

Making d the subject of the formula

$d = \frac{2k(q)^2}{3mv_i^2}$

$= \frac{2* 8.99*10^9 *(1.602*10^{-19}^2)}{3 * 9.11*10^{-31} *(3.2*10^5)^2}$

$=1.66*10^{-9}m$

6 0

3 years ago

Here is a picture showing how two magnets will stick to one another when opposite poles align. When these magnets are pulled sli

7nadin3 [17]

Answer:

C

Explanation:

A magnetic field exerts its force beyond just direct touch.

7 0

3 years ago

Read 2 more answers