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3 years ago

13

Electric eels generate electric pulses along their skin that can stun an enemy. In a test, an electric eel generated pulse of 50

0 V that produced a current of 80 mA for 10ms. Assuming a steady current, what was the power of the pulse?
A. 63 W
B. 16 W
C. 40 W
D.59 W

1 answer:

nydimaria [60]3 years ago

3 0

Answer:

The power of the pulse is 40 Watts

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Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its

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Answer:

a

The orbital speed is $v= 2.6*10^{3} m/s$

b

The escape velocity of the rocket is $v_e= 3.72 *10^3 m/s$

Explanation:

Generally angular velocity is mathematically represented as

$w = \frac{2 \pi}{T}$

Where T is the period which is given as 1.6 days = $1.6 *24 *60*60 = 138240 sec$

Substituting the value

$w = \frac{2 \pi}{138240}$

$= 4.54*10^ {-5} rad /sec$

At the point when the rocket is on a circular orbit

The gravitational force = centripetal force and this can be mathematically represented as

$\frac{GMm}{r^2} = mr w^2$

Where G is the universal gravitational constant with a value $G = 6.67*10^{-11}$

M is the mass of the earth with a constant value of $M = 5.98*10^{24}kg$

r is the distance between earth and circular orbit where the rocke is found

Making r the subject

$r = \sqrt[3]{\frac{GM}{w^2} }$

$= \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }$

$= 5.78 *10^7 m$

The orbital speed is represented mathematically as

$v=wr$

Substituting value

$v= (5.78*10^7)(4.54*10^{-5})$

$v= 2.6*10^{3} m/s$

The escape velocity is mathematically represented as

$v_e = \sqrt{\frac{2GM}{r} }$

Substituting values

$= \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }$

$v_e= 3.72 *10^3 m/s$

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Melanoma skin cancer

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Can a runner on a track be accelerating even if she is running at a constant speed why or why not

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4 years ago

The driver of a car wishes to pass a truck that is traveling at a constant speed of 20.0 m/s (about 45 mi/h). Initially, the car

Alex73 [517]

Answer:

15.8640053791 s

392.780107582 m

29.5184032275 m/s

Explanation:

0 denotes initial

x denotes displacement

c denotes car

t denotes truck

r denotes rear

$x_0_{cr}=-49.5\ m$

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$x_0_{t}=0$

$v_0_{c}=v_0_{t}$

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$x_c=x_0+v_0_{c}t+\dfrac{1}{2}at^2$

The displacement of the truck will be

$x_t=v_tt$

From the above two equations we get

$x_c-x_t=x_0+v_0_{c}+\dfrac{1}{2}at-v_{t}t=26\\\Rightarrow 26=x_0+\dfrac{1}{2}at^2\\\Rightarrow 26+49.5=\dfrac{1}{2}0.6t^2\\\Rightarrow t=\sqrt{\dfrac{2(26+49.5)}{0.6}}\\\Rightarrow t=15.8640053791\ s$

The time taken is 15.8640053791 s

$x-x_0=v_{0}_{c}t+\dfrac{1}{2}at^2\\\Rightarrow x-x_0=20\times 15.8640053791+\dfrac{1}{2}0.6\times 15.8640053791^2\\\Rightarrow x-x_0=392.780107582\ m$

The distance the car travels is 392.780107582 m

$v=v_0+at\\\Rightarrow v=20+0.6\times 15.8640053791\\\Rightarrow v=29.5184032275\ m/s$

The velocity of the car is 29.5184032275 m/s

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