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3 years ago

What is the relationship between PE+KE and total energy? Provide details/data from the simulation.

Physics

1 answer:

kondaur [170]3 years ago

4 0

When the skateboarder is at the highest point of either side of the ramp, he has the highest potential energy. At the lowest point aka the middle point of the graph, it has the highest kinetic energy. When the skateboarder is on the section between the highest point and middle/lowest point on either side, it has an equal amount of kinetic energy compared to potential energy.

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A 4 kg object is moving at a speed of 5 m/sec. How much kinetic energy does the object have? A. 10 joules B. 20 joules C. 50 jou

sp2606 [1]

Answer:

A. 10 joules

Explanation:

Given parameters:

Mass of object = 4kg

Speed = 5m/s

Unknown:

Kinetic energy of the the object = ?

Solution:

Kinetic energy is the energy due to a moving body.

So;

mathematically;

K.E = $\frac{1}{2}$ x m v²

m is the mass

v is the velocity

K.E = $\frac{1}{2}$ x 4 x 5 = 10J

5 0

3 years ago

A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-

SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency $\omega_0$ at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

Where

L = Inductance

C = Capacitance

Our values are given as

$C = 15*10^{-6}\mu F$

$L = 0.280*10^{-3}mH$

Replacing we have,

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})$

$f= 2455.81Hz$

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

7 0

4 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

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