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2 years ago

If a person wanted to engineer a new type of communication device, what field of study would be the most useful?

Physics

1 answer:

Fiesta28 [93]2 years ago

6 0

Answer: Option A: electrical engineering.

Explanation: The field of electrical engineering, sometimes called electronics engineering, mostly outside the US, is a field of engineering that covers a number of subtopics that include large scale power systems, electronics, control systems, signal processing and telecommunications. In this case the engineer would specifically have to know about electronics (computer chips), signal processing (cell signals) and telecommunications.

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3 years ago

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Answer:

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3 years ago

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3 years ago

A balloon of mass M is floating motionless in the air. A person of mass less than M is on a rope ladder hanging from the balloon

Tasya [4]

Let the mass of the person be m. Total momentum is conserved (because the exterior forces on the system are balanced), especially the component in the vertical direction.

Given that,

Mass of gallon is M

Let man mass be m

Velocity of man is v

Let velocity if ballot be Vb

When the person begin to move we have

Conservation of momentum

mv + MVb=0

MVb=-mv

Vb= -(m/M) v

Given that the mass of man is less than mass of balloon. i.e. m<M

So, if m<M, then, m/M <1

Therefore, .

Vb= -(m/M) v

Vb< -v

This implies that the velocity of balloon is less than the velocity of man and if is also moving in opposite direction

So the man is moving upward, then the balloon is moving downward and it's velocity is less than the velocity of man,

The answer is C

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3 years ago

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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.

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3 years ago