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2 years ago

17 points

Physics

1 answer:

const2013 [10]2 years ago

3 0

The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.

The question can be solved, using Newton's second law of motion.

Note: Momentum of the cannon = momentum of the cannonball.

<h3>Formula:</h3>

MV = mv................. Equation 1

<h3>Where:</h3>

M = mass of the cannon
m = mass of the cannonball
V = velocity of the cannon
v = velocity of the cannonball

Make v the subject of the equation.

v = MV/m................ Equation 2

From the question,

<h3>Given: </h3>

M = 500 kg
V = 3 m/s
m = 10 kg.

Substitute these values into equation 2.

v = (500×3)/10
v = 150 m/s.

Hence, The velocity of the cannonball is 150 m/s, the right option is B. 150 m/s.

Learn more about Newton's second law here: brainly.com/question/25545050

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Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),

yuradex [85]

Answer:

$v=3.564\ m.s^{-1}$

$\Delta v =2.16\ m.s^{-1}$

Explanation:

Given:

mass of John, $m_J=30\ kg$

mass of William, $m_W=30\ kg$

length of slide, $l=3\ m$

(A)

height between John and William, $h=1.8\ m$

<u>Using the equation of motion:</u>

$v_J^2=u_J^2+2 (g.sin\theta).l$

where:

v_J = final velocity of John at the end of the slide

u_J = initial velocity of John at the top of the slide = 0

Now putting respective :

$v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3$

$v_J=5.94\ m.s^{-1}$

<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

<em>Sum of initial momentum of kids before & after collision must be equal.</em>

$m_J.v_J+m_w.v_w=(m_J+m_w).v$

where: v = velocity with which they move together after collision

$30\times 5.94+0=(30+20)v$

$v=3.564\ m.s^{-1}$ is the velocity with which they leave the slide.

(B)

frictional force due to mud, $f=105\ N$

<u>Now we find the force along the slide due to the body weight:</u>

$F=m_J.g.sin\theta$

$F=30\times 9.8\times \frac{1.8}{3}$

$F=176.4\ N$

<em><u>Hence the net force along the slide:</u></em>

$F_R=71.4\ N$

<em>Now the acceleration of John:</em>

$a_j=\frac{F_R}{m_J}$

$a_j=\frac{71.4}{30}$

$a_j=2.38\ m.s^{-2}$

<u>Now the new velocity:</u>

$v_J_n^2=u_J^2+2.(a_j).l$

$v_J_n^2=0^2+2\times 2.38\times 3$

$v_J_n=3.78\ m.s^{-1}$

Hence the new velocity is slower by

$\Delta v =(v_J-v_J_n)$

$\Delta v =5.94-3.78= 2.16\ m.s^{-1}$

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A heavy anvil is suspended by a 0.75 m long steel wire that has a mass of 12 g. When the wire is plucked, it hums at its fundame

Dima020 [189]

Explanation:

It is given that,

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Mass of the wire, m = 12 g = 0.012 kg

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$f=\dfrac{v}{2l}$

v is the speed of the mass

Speed is given by :

$v=\sqrt{\dfrac{T}{\mu}}$

$\mu$ is the mass per unit length, $\mu=\dfrac{m}{l}$

$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$

T is the tension in the wire,

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T = 518.4 N

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$m'=\dfrac{T}{g}$

$m'=\dfrac{518.4}{9.8}$

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