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Luden [163]
3 years ago
10

A 10 g bullet moving horizontally with a speed of 2000 m/s strikes and passes through a 4.0 kg block moving with a speed of 4.2

m/s in the opposite direction on a horizontal frictionless surface. If the block is brought to rest by the collision, what is the kinetic energy of the bullet as it emerges from the block. Round off your answer to zero decimal places.
Physics
1 answer:
Masteriza [31]3 years ago
3 0

Answer:

The answer is "512 J".

Explanation:

bullet mass m_1 = 10 g= 10^{-2} \ kg\\\\

initial speed u_1 = 2\ \frac{Km}{s}= 2000\ \frac{m}{s}\\\\

block mass m_2 = 4\ Kg

initial speed v_2 =-4.2 \frac{m}{s}

final speed v_2= 0

Let v_1 will be the bullet speed after collision:

throughout the consevation the linear moemuntum

\to M_1V_1+m_2v_2=M_1U_1+m_2u_2\\\\\to (10^{-2} kg) V_1 +0 = (10^{-2} kg)(2000 \frac{m}{s}) + (4 \ kg)(-4.2 \frac{m}{s}) \\\\\ \to 10^{-2} v_1 = 20 -16.8\\\\

                = 320 \frac{m}{s}

The kinetic energy of the bullet in its emerges from the block

k=\frac{1}{2} m_1 v_1^2

   =\frac{1}{2} \times 10^{-2} \times 320\\\\=512 \ J

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Vsevolod [243]

Answer:

v = 4.2 \ m/s

Explanation:

Given data:

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Kinetic energy, K = 0.013 \ \rm J

Let the velocity of the paper clip when it is thrown be <em>v</em>.

Thus,

K = \frac{1}{2}mv^{2}

0.013 = 0.5 \times 0.0015 \times v^{2}

\Rightarrow \ v = 4.16 \ m/s

v = 4.2 \ m/s.  (rounding to nearest tenth)

3 0
3 years ago
A force off 700 newtons is applied to a 600 kg bowling ball. What is the acceleration of the bowling ball once the force is appl
Readme [11.4K]

Answer:

<h2>1.17 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{700}{600}  =  \frac{7}{6}  \\  = 1.1666666...

We have the final answer as

<h3>1.17 m/s²</h3>

Hope this helps you

6 0
3 years ago
Block B has mass 6.00 kg and sits at rest on a horizontal, frictionless surface. Block A has mass 2.50 kg and sits at rest on to
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Answer:

Explanation:

Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 )  friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .

Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A

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net force on block A = P - 9.8

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for common acceleration

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( P - 9.8 ) / 2.5 = 1.63333

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Answer:

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8 0
3 years ago
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Answer:

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