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Sonbull [250]
3 years ago
11

A stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two

fragments, each 1.0 kg, move in directions that make 60o angle above and below the positive x-axis and their speeds are 60 m/s each. What is the velocity of the 4.0-kg fragment
Physics
1 answer:
natta225 [31]3 years ago
4 0

Answer:

-15 m/s

Explanation:

The computation of the velocity of the 4.0 kg fragment is shown below:

For this question, we use the correlation of the momentum along with horizontal x axis

Given that

Weight of stationary shell = 6 kg

Other two fragments each = 1.0 kg

Angle = 60

Speed = 60 m/s

Based on the above information, the velocity = v is

1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0

\frac{60}{2} + \frac{60}{2} - 4\ v = 0

v = \frac{60}{4}

= -15 m/s

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Answer:

8.91 J

Explanation:

mass, m = 8.20 kg

radius, r = 0.22 m

Moment of inertia of the shell, I = 2/3 mr^2

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n = 6 revolutions

Angular displacement, θ = 6 x 2 x π = 37.68 rad

angular acceleration, α = 0.890 rad/s^2

initial angular velocity, ωo = 0 rad/s

Let the final angular velocity is ω.

Use third equation of motion

ω² = ωo² + 2αθ

ω² = 0 + 2 x 0.890 x 37.68

ω = 8.2 rad/s

Kinetic energy,

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.265 x 8.2 x 8.2

K = 8.91 J

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3 years ago
What is the pressure at a depth of 15 cm brine of density 1.2/cm³? ​
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P = density × gravity acceleration × height

P = 1200 × 9.81 × 15/100

P = 1765.8

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3 years ago
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An object is pulled to the left by a force of 50 N. The same amount of force pulls it to the right. The object will ____.
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<h2><u>Required</u><u> </u><u>Answer</u><u>:</u></h2>

The body will <u>stay at rest </u>(Option D). It is because a force of magnitude 50 N is pulled towards left and another force is pulling it towards right with same magnitude 50 N. So, the direction of force is opposite and magnitude is same i.e. 50 N. So, they will cancel each other and net force is 0. Hence, there would be no acceleration.

  • Option A - Showing acceleration
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Hence, these options are incorrect because they are only possible when net external force is non-zero. Staying at rest i.e. Option D means there is no motion and hence no acceleration, this shows that net force is 0.

<u>━━━━━━━━━━━━━━━━━━━━</u>

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The second object, the one that had twice the force applied to it, would move twice as far, I believe.
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What type of circuit is illustrated?
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Answer:

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