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3 years ago

13

Mass = 10Kg velocity = 5m/s what is the kinetic energy = ? Joules

2 answers:

nirvana33 [79]3 years ago

5 0

Answer:

Hello There!!

Explanation:

You use the formula:K.E. = 1/2 m v2.

kinetic energy=1 / 2 × 10 × 5 ²(the power of 2 means you multiply the number by itself)

half of 10 is 5 so you do 5×5².

kinetic energy=5×5²=(5×5=25) so 25×5=125 Joules.

hope this helps,have a great day!!

~Pinky~

alexdok [17]3 years ago

3 0

Answer:

kinetic energy = 1 / 2 mv²

kinetic energy = 1 / 2 × 10 × 5 ²

kinetic energy = 5 × 25

kinetic energy = 125 joules

..........

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3 years ago

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Answer:

Incomplete question: "Each block has a mass of 0.2 kg"

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It is neccesary calculate the force:

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Here, g = gravity = 9.8 m/s²

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3 years ago

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3 years ago

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr

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Answer:

a

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b

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c

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

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3 years ago