All categories

2 years ago

Explain how magnets have an attractive force and repulsive force

1 answer:

7 0

The force between two parallel wires carrying currents in the same direction is attractive. It is repulsive if the currents are in opposite directions.

You might be interested in

Which of the following paraphrases Hubble Law?Select one:A. The greater the distance to a galaxy, the greater its redshift. B. T

olya-2409 [2.1K]

The correct answer is:

~A. The greater the distance to a galaxy, the greater its redshift.

Hope this helps!!!

4 0

3 years ago

All ball is thrown up with a vertical velocity of 54 m/s and a horizontal velocity of 39 m/s. Calculate how many seconds it will

VikaD [51]

5.5 s

Explanation:

The time it takes for the ball to reach its maximum height can be calculated using

$v_y = v_{0y} - gt \Rightarrow t = \dfrac{v_{0y}}{g}$

since $v_y = 0$ at the top of its trajectory. Plugging in the numbers,

$t = \dfrac{(54\:\text{m/s})}{(9.8\:\text{m/s}^2)} = 5.5\:\text{s}$

8 0

2 years ago

A student performs a lab measuring the velocities of toy cars of different masses

stiv31 [10]

The answer is Car 1 and Car 2.

5 0

3 years ago

Read 2 more answers

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 x-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 y-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 x- and y-components of the copter's position vector after t = 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

Astar is 10 light years away from the earth. Suppose it brightens up suddenly today, after how long can we see this change?

Blababa [14]

The phrase "light year" is a distance ... it's the distance that light travels through vacuum in one year.

When you look at an object located 1 light year away from you, you see it as it was 1 year ago.

If a star located 10 light years away from us suddenly brightens, or dims, or explodes, we see the event 10 years later.

7 0

3 years ago