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3 years ago

What is the definition of power.

Physics

2 answers:

alekssr [168]3 years ago

7 0

Answer:

Hey buddy you are in college and I am in 8th standard so, There must be a difference between your level and mine. (:

Explanation:

Power is energy to exert force on objects, machinery use etc.

Sorry if not helpful ):

Mandarinka [93]3 years ago

6 0

Answer:

What is the definition of power.

Explanation:

ability to do or act; capability of doing or accomplishing something.

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3 years ago

Depict the following path: You drive 14 blocks East, 7 blocks North, and 2 blocks West. Use the vertex of the graph (coordinates

horrorfan [7]

Answer:

Displacement = $12\hat{i} +7\hat{j}$

Magnitude of displacement = 13.89 units

Angle of displacement = 30.26°

Explanation:

Mark the co-ordinates of each of the end points of the vectors.

The marked co-ordinates are shown below.

Displacement is the shortest distance between two points. Here, displacement between the starting and end points is the vector $\overrightarrow{OA}$ .

Displacement vector for a point A $(x,y)$ from origin O is given as:

$\overrightarrow{OA}=(x-0)\hat{i}+(y-0)\hat{j}$

Magnitude is given as:

$|\overrightarrow{OA}|=\sqrt{x^{2}+y^{2}}$

Direction of the vector is given as:

$\theta=tan^{-1}\frac{y}{x}$

Therefore, displacement vector is:

$\overrightarrow{OA}=(12-0)\hat{i}+(7-0)\hat{j}\\ \overrightarrow{OA}=12\hat{i}+7\hat{j}$

Magnitude of displacement is given as:

$|\overrightarrow{OA}|=\sqrt{x^{2}+y^{2}}\\ |\overrightarrow{OA}|=\sqrt{12^{2}+7^{2}}\\ |\overrightarrow{OA}|=\sqrt{144+49}\\\\ |\overrightarrow{OA}|=13.89$

Angle of displacement is:

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$\theta=30.26$ °

8 0

4 years ago

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30° angle. The block’s initial speed is 10 m/s. What vert

Romashka [77]

Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, $Vi$ and $Ui$ are its initial vertical and horizontal components

$R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi = 10*sin(30) = 5 m/s\\Ui = 10*cos(30) = 8.66 m/s$

To find:

Max Height $h$ achieved

Calculation:

1) Using the $3^{rd}$ equation of motion, we know

$2*a*s = Vf^{2} - Vi^{2}$

2) In terms of gravity $g$ height $h$ and the vertical component of Velocity $Vf , Vi$ .

3) As $Vf = 0$ as at maximum height the vertical component of velocity is zero maximum height achieved

$2*g*h = Vf^{2} -Vi^{2}$

putting values

4) $h = 0.784 m/s$

5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)

3 0

4 years ago