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12345 [234]
3 years ago
12

the atlantic torpedo is a large electric fish capable of generating a voltage of 220V between its tail and its head. This drives

a pulse of current of 15A lasting for a time of 2.0ms. The fish produces 200 pulses per second. What is the average power output of the fish? a.33W b.1.3KW c.3.3KW d.6.6KW
Physics
1 answer:
Elza [17]3 years ago
7 0
The average power output:
P = V * I * t
V = 220 V, I = 15 A;
t = 2 ms * 200 = 400 ms = 0.4 s
P = 220 V * 15 A * 0.4 s
P = 1320 W ≈ 1.3 kW
Answer:
b. 1.3 kW
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Protons in an atomic nucleus are typically 10−15 m apart. what is the electric force (in n) of repulsion between nuclear protons
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<span>The electric force is given by: 
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3 years ago
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A source produces 20 crests and 20 troughs in 4 seconds. The second crest is 3 cm away from the first crest.Calculate :
sineoko [7]

Answer:

Solution given:

No of waves[N] =20crests & 20 troughs

=20waves

Time[T]=4seconds

distance[d]=3cm=0.03m

Now

<u>Wave</u><u> </u><u>length</u><u>=</u>3cm=3 × {10}^{ - 2}m

<u>Frequency</u>=\frac{No of waves}{time}

=\frac{20}{4}=5Hertz

and

Wave speed:wave length×frequency=3 × {10}^{ - 2}m×5=1.5 × {10}^{ - 1} \tt{ {ms}^{ - 1}}.

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When an object is heated, its molecules
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A

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Mary and her younger brother Alex decide to ride the 17-foot-diameter carousel at the State Fair. Mary sits on one of the horses
lorasvet [3.4K]

Answer:

\omege_A=\omega_M

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M denotes Mary

r = Distance from center

Mary and Alex will have the equal displacements in equal interval of time as they are in uniform circular motion. So,

\omega_A=\omega_M

Tangential speed speed is given by

\dfrac{v_M}{v_A}=\dfrac{r_M\omega_M}{r_A\omega_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{r_M}{r_A}\\\Rightarrow \dfrac{v_M}{v_A}=\dfrac{8}{5}\\\Rightarrow \dfrac{v_M}{v_A}=1.6\\\Rightarrow v_M=1.6v_A

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5 0
3 years ago
A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i
timurjin [86]

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}

\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}

\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s

4)Kinetic energy before:

E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J

Kinetic energy after:

E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

3 0
3 years ago
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