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yuradex [85]
1 year ago
9

The density of lead is 11.3 g/cm3. what mass of lead is required to make a 1.00 cm3 fishing sinker?

Physics
1 answer:
Elza [17]1 year ago
8 0

The mass of lead required to make a 1.00 cm3 fishing sinker is 11.3g.

What is mass?

Mass is a metric used in physics to express inertia, a fundamental characteristic of all matter. A mass of matter's resistance to altering its direction or speed in response to the application of a force is what it essentially is. The change that an applied force produces is smaller the more mass a body has.

Given :

Density of lead = 11.3 g/cm3

Volume of  sinker  =  1.00 cm3

One of a substance's attributes is density, which is calculated by dividing the mass by the volume. Mathematically:

Density : Mass / volume

therefore after putting the values,

mass= 11.3g

To learn more about density click on the link below:

brainly.com/question/18939565

#SPJ4

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A 0.500-kg glider, attached to the end of an ideal spring with force constant undergoes shm with an amplitude of 0.040 m. comput
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There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"

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</span>E=U+K=  \frac{1}{2}kx^2 + \frac{1}{2} mv^2
<span>where
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v is the speed of the glider at position x

When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
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<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
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<span>
Since the mechanical energy must be conserved, we can write
</span>\frac{1}{2}mv_{max}^2 =  \frac{1}{2}kA^2
<span>from which we find the maximum speed
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<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m

We can still use the conservation of energy to solve this part. 
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</span>E=K_{max}=  \frac{1}{2}mv_{max}^2= 0.36 J
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span>U= \frac{1}{2}kx^2 =  \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J
<span>And since 
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</span>K=E-U=0.36 J - 0.05 J = 0.31 J
<span>And then we can find the corresponding velocity:
</span>K= \frac{1}{2}mv^2
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<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
For a simple harmonic motion, the magnitude of the maximum acceleration is given by
a_{max} = \omega^2 A
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\omega =  \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s
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d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

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<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>

we have already calculated it at point b), and it is given by
</span>E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J
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Answer:

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Explanation:

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