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DENIUS [597]
3 years ago
7

The cheetah is one of the fastest-accelerating animals, because it can go from rest to 16.2 m/s (about 36 mi/h) in 2.4 s. If its

mass is 102 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in the following units.
Physics
1 answer:
Scilla [17]3 years ago
7 0
<h2>Power of cheetah is 5576.85 W = 7.48 hp</h2>

Explanation:

Power is the ratio of energy to time.

Here we need to consider kinetic energy,

Mass, m = 102 kg

Initial velocity = 0 m/s

Final velocity = 16.2 m/s

Time, t = 2.4 s

Initial kinetic energy = 0.5 x Mass x Initial velocity² = 0.5 x 102 x 0² = 0 J

Final kinetic energy = 0.5 x Mass x Final velocity² = 0.5 x 102 x 16.2² = 13384.44 J

Change in energy = Final kinetic energy - Initial kinetic energy

Change in energy = 13384.44 - 0

Change in energy = 13384.44 J

Power = 13384.44  ÷ 2.4 = 5576.85 W = 7.48 hp

Power of cheetah is 5576.85 W = 7.48 hp

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2. Tomas is hanging from a tree limb, that is inclined at a 65° angle. The force
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Answer:

57 N

Explanation:

Were are told that the force

of gravity on Tomas is 57 N.

And it acts at an inclined angle of 65°

Thus;

The vertical component of the velocity is; F_y = 57 sin 65

While the horizontal component is;

F_x = 57 cos 65

Thus;

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The net force will be;

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In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

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\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

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\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

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Answer:

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