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3 years ago

A heating element on an electric range operating at 200. v has a resistance of 32.0 ohms. the current drawn by the element is

Physics

1 answer:

LenaWriter [7]3 years ago

4 0

Answer:

6.25 A

Explanation:

We cant find the current drawn by the element by using Ohm's law:

$V=RI$

where

V is the potential difference across the resistance of the element

R is the resistance

I is the current

In this problem, we know

V = 200 V

$R=32.0 \Omega$

Therefore we can re-arrange the equation and solve for the current:

$I=\frac{V}{R}=\frac{200}{32}=6.25 A$

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In the standing waves experiment, the string has a mass of 31.2 g and a length of 0.7 m. The string is connected to a mechanical

mestny [16]

Answer:

linear density of the string = 4.46 × 10⁻⁴ kg/m

Explanation:

given,

mass of the string = 31.2 g

length of string = 0.7 m

linear density of the string = $\dfrac{mass\ of\ string}{length}$

linear density of the string = $\dfrac{31.2\times 10^{-3}\ kg}{0.7\ m}$

linear density of the string = 44.57 × 10⁻³ kg/m

linear density of the string = 4.46 × 10⁻⁴ kg/m

7 0

3 years ago

What instrument will be used to measure the volume of a box

Anettt [7]

Answer:

A ruler.

Explanation:

Just measure height, length and width and multiply each.

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2 years ago

Two objects carry initial charges that are q1 and q2, respectively, where |q2| > |q1|. They are located 0.160 m apart and beh

mart [117]

Answer:

$\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force between two static point charges $\rm q_1$ and $\rm q_1$ , separated by a distance $\rm r$ , is given by

$\rm F = \dfrac{kq_1q_2}{r^2}.$

where k is the Coulomb's constant.

Initially,

$\rm r = 0.160\ m\\F_i = -1.30\ N.\\\\and \ \ |q_2|>|q_1|.$

The negative sign is taken with force F because the force is attractive.

Therefore, the initial electrostatic force between the charges is given by

$\rm F_i = \dfrac{kq_1q_2}{r^2}.\\-1.30=\dfrac{kq_1q_2}{0.160^2}\\\rm\Rightarrow q_2 = \dfrac{-1.30\times 0.160^2}{q_1k}\ \ \ ..............\ (1).$

Now, the objects are then brought into contact, so the net charge is shared equally, and then they are returned to their initial positions.

The force is now repulsive, therefore, $\rm F_f = +1.30\ N.$

The new charges on the two objects are

$\rm q_1'=q_2' = \dfrac{q_1+q_2}{2}.$

The new force is given by

$\rm F_f = \dfrac{kq_1'q_2'}{r^2}\\+1.30=\dfrac{k\left (\dfrac{q_1+q_2}{2}\right )\left (\dfrac{q_1+q_2}{2}\right )}{0.160^2}\\\Rightarrow \left (\dfrac{q_1+q_2}{2}\right )^2=\dfrac{+1.30\times 0.160^2}{k}\\(q_1+q_2)^2=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+q_2^2+2q_1q_2=\dfrac{4\times 1.30\times 0.160^2}{k}\\\\$

Using (1),

$\rm q_1^2+\left ( \dfrac{-1.30\times 0.160^2}{q_1k}\right )^2+2\left (\dfrac{-1.30\times 0.160^2}{k} \right )=\dfrac{4\times 1.30\times 0.160^2}{k}\\q_1^2+\dfrac 1{q_1^2}\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0$

$\rm q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{k}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{k} \right )=0\\q_1^4+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )=0\\q_1^4-q_1^2\left (\dfrac{6\times 1.30\times 0.160^2}{9\times 10^9} \right )+\left ( \dfrac{-1.30\times 0.160^2}{9\times 10^9}\right )^2=0$

$\rm q_1^4-q_1^2\left (2.22\times 10^{-11} \right )+\left ( 1.37\times 10^{-23}\right ) =0\\\Rightarrow q_1^2 = \dfrac{-(-2.22\times 10^{-11})\pm \sqrt{(-2.22\times 10^{-11})^2-4\cdot (1)\cdot (1.37\times 10^{-23})}}{2}\\=1.11\times 10^{-11}\pm 1.046\times 10^{-11}.\\=6.4\times 10^{-13}\ \ \ or\ \ \ 2.156\times 10^{-11}\\\Rightarrow q_1 = \pm 8.00\times 10^{-7}\ C\ \ \ or\ \ \ \pm 4.64\times 10^{-6}\ C.$

Using (1),

When $\rm q_1 = \pm 8.00\times 10^{-7}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 8.00\times 10^{-7}\times 9\times 10^9}=\mp4.6\times 10^{-6}\ C.$

When $\rm q_1=\pm 4.6\times 10^{-6}\ C$ ,

$\rm q_2=\dfrac{-1.30\times 0.160^2}{\pm 4.64\times 10^{-6}\times 9\times 10^9}=\mp7.97\times 10^{-7}\ C\approx 8.0\times 10^{-7}\ C.$

Since, $\rm |q_2|>|q_1|$

Therefore, $\rm |q_1|=8.0\times 10^{-7}\ C,\ \ \ |q_2| = 4.6\times 10^{-6}\ C.$

7 0

2 years ago

Which point on the graph shows a temperature of 40°C and the time of 25 minutes?

const2013 [10]

Answer:

'A' is the the point on the graph that shows a temperature of 40°C and the time of 25 minutes

8 0

3 years ago

Read 2 more answers

A small metal bar whose initial temperature was 20 degrees Celsius is dropped into a large container of boiling water. How long

Vesnalui [34]

Below is the answer. I hope it help.
T ( t ) = C e k t + T m where Tm is the temperature of the surroundings
T ( t ) = C e k t + T m
T ( 0 ) = 20
T ( 1 ) = T ( 0 ) + 2 = 22
C + T m = 20 C+Tm=20
C e k + T m = 22

4 0

2 years ago