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seraphim [82]
3 years ago
13

***PLEASE HELP WITH ANSWER AND EXPLANATION: Imagine the current in a current-carrying wire is flowing into the screen. What is t

he direction of the magnetic field around this current?
The magnetic field is clockwise.
The magnetic field is counterclockwise.
The magnetic field is moving out of the screen.
The magnetic field is moving into the screen.
Physics
2 answers:
DerKrebs [107]3 years ago
7 0

Answer:

the magnetic field is clockwise

Explanation:

aalyn [17]3 years ago
5 0

Magnetic field direction is given by right hand thumb rule.

If we put our thumb in the direction of current then curl of fingers will show the magnetic field direction around the wire.

Now here since current is going into the screen so we will put our thumb into the screen and then the curl of fingers is clockwise around it.

The magnetic field is clockwise.

So this would be the direction of magnetic field

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Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

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