Ask question

Ask question

All categories

3 years ago

5

Electric current is a measure of

2 answers:

raketka [301]3 years ago

8 0

Answer:

The answer is B which is the amount of electric charge that passes within a certain amount of time in my opinion.

Tpy6a [65]3 years ago

7 0

Answer:

b. the amount of electric charge that passes within a certain amount of time

You might be interested in

Explain the following defects of a simple electric cell:

eimsori [14]

Answer:

Explanation:

The two major defects of simple electric cells causes current supplied to be for short time. These defects are: polarization and local action.

a. Polarization: This is a defect caused by an accumulation of hydrogen bubbles at the positive electrode of the cell. It can be prevented by the use of vent, using a hydrogen absorbing material or the use of a depolarizer.

b. Local Action: This is the gradual wearing away of the electrode due to impurities in the zinc plate. It can be controlled by the amalgamation of the zinc plate before it is used.

4 0

3 years ago

you get a flat tire on her car while driving. You use a jack to change the tire. It exerts a force of 5,000 N to lift the car 0.

ch4aika [34]

Answer:

b is the anwser

Explanation:

you are super

6 0

3 years ago

Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ

Tomtit [17]

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

$\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}$

Put the value into the formula

$525=\sqrt{\dfrac{c+v}{c-v}}\times950$

$\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2$

$\dfrac{c+v}{c-v}=0.305$

$c+v=0.305\times(c-v)$

$v(1+0.305)=c(0.305-1)$

$v=\dfrac{0.305-1}{1+0.305}c$

$v=−0.532c$

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

7 0

3 years ago

A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the <em>x</em>-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its <em>y</em>-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

3 years ago

5. What is the amount of force required to accelerate a 20 kg object at a rate of 5 m/sz?

GenaCL600 [577]

Force required is 100 N

<u>Given that;</u>

Rate of acceleration = 5 m/s²

Mass of object = 20kg

<u>Find:</u>

Force required

<u>Computation:</u>

Force = Mass × Acceleration

Force required = Rate of acceleration × Mass of object

Force required = 20 × 5

Force required = 100 N

Learn more:

brainly.com/question/17506203?referrer=searchResults

3 0

2 years ago