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AURORKA [14]
3 years ago
8

This diagram of the first right-hand rule relates which two quantities?

Physics
2 answers:
Reptile [31]3 years ago
8 0

PLATO ANSWER!!!!!!

current (thumb) to magnetic field (fingers)

AURORKA [14]3 years ago
7 0

Option A, current (thumb) to magnetic field (fingers)

As per the First right-hand rule,

Using right hand, if we suppose that thumb points towards the electric current

fingers curl towards the magnetic field

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An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m
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Answer:

\displaystyle y_m=3.65m

Explanation:

<u>Motion in The Plane</u>

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of v_o and \theta\\ as the initial speed and angle, then we have

\displaystyle v_x=v_o\ cos\theta

\displaystyle v_y=v_o\ sin\theta-gt

\displaystyle x=v_o\ cos\theta\ t

\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}

If we want to know the maximum height reached by the object, we find the value of t when v_y becomes zero, because the object stops going up and starts going down

\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt

Solving for t

\displaystyle t=\frac{v_o\ sin\theta }{g}

Then we replace that value into y, to find the maximum height

\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2

Operating and simplifying

\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}

We have

\displaystyle v_o=20\ m/s,\ \theta=25^o

The maximum height is

\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}

\displaystyle y_m=3.65m

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3 years ago
300 Jules energy are used to push an object with a force of 75N. what is the maximum distance the object can move?
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Energy consumed in doing the work = 300 Joules

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Work done by the force is determined by the force applied and the displacement happened in the direction of the force applied.

Work done = Force x displacement

300 = 75 x d

d = \frac{300}{75}

d = 4 m

Hence, the maximum distance moved by the object = 4 meters

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