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3 years ago

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A 45 kg boy runs at a rate of 2.5 m/s and jumps on top of a stationary skateboard that has a mass of 4 kg. After jumping onto th

e board, the boy and the board travel off together. Assuming that momentum is conserved, what is the final speed of the boy and the board?

2 answers:

Natalija [7]3 years ago

7 0

Answer:

v = 2.29 m/s

Explanation:

As we know that the external force on the system of mass of boy + board is ZERO

So here we can use momentum conservation

now we have

$m_1v_1 + m_2v_2 = (m_1 + m_2) v$

now we have

$45 (2.5) + 4(0) = (45 + 4) v$

now we have

$v = \frac{45}{49} (2.5)$

$v = 2.29 m/s$

igomit [66]3 years ago

4 0

Answer:

V = 2.29 m/s

Explanation:

Given that,

Mass of the boy, $m_1=45\ kg$

Mass of the skateboard, $m_2=4\ kg$

Initial speed of the boy, v = 2.5 m/s

Let V is the final velocity of the boy and the board. The net momentum of the system remains constant. Using the conservation of linear momentum to find it as :

$45\times 2.5=(45+4)V$

$V=\dfrac{45\times 2.5}{(45+4)}$

V = 2.29 m/s

So, the velocity of the boat after Batman lands in it 2.29 m/s. Hence, this is the required solution.

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Read 2 more answers

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Papessa [141]

Answer:

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Explanation:

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The rate at which it decreases is $\frac{dB}{dt} = 24.5G/s = 24.5*10^{-4} T/s$

The distance from the center of field is $r = 1.5 cm = \frac{1.5}{100} = 0.015m$

According to Faraday's law

$\epsilon = - \frac{d \o}{dt}$

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Where the magnetic flux $\o = B* A$

E is the electric field

dl is a unit length

So

$\int\limits {E} \, dl = - \frac{d}{dt} (B*A)$

${E} l = - \frac{d}{dt} (B*A)$

Now $l$ is the circumference of the circular loop formed by the magnetic field and it mathematically represented as $l = 2\pi r$

A is the area of the circular loop formed by the magnetic field and it mathematically represented as $A= \pi r^2$

So

${E} (2 \pi r)= - \pi r^2 \frac{dB}{dt}$

$E = \frac{r}{2} [ - \frac{db}{dt} ]$

Substituting values

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Also $F_p = q E$

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Answer:

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Explanation:

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