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2 years ago

Define the term potential energy of a body? How is it measured

Physics

2 answers:

kondaur [170]2 years ago

7 0

Answer:

The energy possessed by a body as a result of its position or condition rather than its motion.

Simplified, this formula can be written as: Potential Energy = mgh, where m is the mass, measured in kilograms; g is the acceleration due to gravity (9.8 m/s^2 at the surface of the Earth); and h is the height, measured in meters.

DENIUS [597]2 years ago

4 0

Answer:

potential energy. noun. the energy of a body or system as a result of its position in an electric, magnetic, or gravitational field. It is measured in joules (SI units), electronvolts, ergs, etcSymbol: E p, V, U, φ Abbreviation: PE.

Explanation:

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Which of the following has the fewest calence electrons?

dlinn [17]

Answer:

both magnesium and strontium have 2 valence electrons

Explanation:

potassium has 8 valence electrons, and nitrogen has 5.

8 0

2 years ago

A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal

Sati [7]

(a) The ball has a final velocity vector

$\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j$

with horizontal and vertical components, respectively,

$v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}$

$v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}$

The horizontal component of the ball's velocity is constant throughout its trajectory, so $v_{x,i}=v_{x,f}$ , and the horizontal distance x that it covers after time t is

$x=v_{x,i}t=v_{x,f}t$

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

$103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s$

The vertical component of the ball's velocity at time t is

$v_{y,f}=v_{y,i}-gt$

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

$-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}$

So, the initial velocity vector is

$\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j$

which carries an initial speed of

$\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}$

and direction θ such that

$\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}$

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height y at time t is

$y=v_{y,i}t-\dfrac12gt^2$

so that when it lands in the seats at t ≈ 6.38 s, it has a height of

$y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}$