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3 years ago

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Nuclear Size and Mass. The radii of most nuclei is given by the equation: R = R. 0. A1/3. R: radius of the nucleus, A: mass numb

er, R. 0. = 1.2x10-15 m. The mass of a nucleus

1 answer:

mina [271]3 years ago

3 0

Answer:

sssss

Explanation:

Hope this helps you

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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

Nikitich [7]

There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"

a) the maximum speed of the glider

The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span> $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
<span>where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
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:
</span> $E=K_{max} = \frac{1}{2}mv_{max}^2$
<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:
</span> $E=U_{max}= \frac{1}{2}k A^2$
<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
</span> $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$
<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.
The total mechanical energy is:
</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
</span> $E=U+K$
<span>we find the kinetic energy when the glider is at this position:
</span> $K=E-U=0.36 J - 0.05 J = 0.31 J$
<span>And then we can find the corresponding velocity:
</span> $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$
<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
$a_{max} = \omega^2 A$
where $\omega= \sqrt{ \frac{k}{m} }$ is the angular frequency, and A is the amplitude.
The angular frequency is:
$\omega = \sqrt{ \frac{450 N/m}{0.500 kg} }=30 rad/s$
and so the maximum acceleration is
$a_{max} = \omega^2 A = (30 rad/s)^2 (0.040 m) =36 m/s^2$

d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m

For a simple harmonic motion, the acceleration is given by
</span> $a(t)=\omega^2 x(t)$
<span>where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
</span> $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$
<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> $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

8 0

3 years ago

An extreme skier, starting from rest, coasts down a mountain slope that makes an angle of 25.0 with the horizontal. The coeffici

uranmaximum [27]

Answer:

V = 10.88 m/s

Explanation:

V_i =initial velocity = 0m/s

a= acceleration= gsinθ- $\mu_k$ cosθ

putting values we get

a= 9.8sin25-0.2cos25= 2.4 m/s^2

v_f= final velocity and d= displacement along the inclined plane = 10.4 m

using the equation

$v^2_f=v^2_i-2as$

$v^2_f=0^2-2(2.4)(10.4)$

v_f= 7.04 m/s

let the speed just before she lands be "V"

using conservation of energy

KE + PE at the edge of cliff = KE at bottom of cliff

(0.5) m V_f^2 + mgh = (0.5) m V^2

V^2 = V_f^2 + 2gh

V^2 = 7.04^2 + 2 x 9.8 x 3.5

V = 10.88 m/s

6 0

3 years ago

Why does the sky change colors at sunset?

KiRa [710]

Atmospheric refraction is the deviation of light or other electromagnetic wave from a straight line as it passes through the atmosphere due to the variation in air density as a function of height. ... Refraction not only affects visible light rays, but all electromagnetic radiation, although in varying degrees.

So in short, the answer is D.

(My answer got deleted because it didnt explain which is dumb)

8 0

3 years ago

Read 2 more answers

What do understand by the efficiency of a machine? By using a block and tackel a man can raise a load of 720 N by an effort of 1

motikmotik

Answer:

Efficiency of a machine is how well the machine works and what the machine is capable of doing.

Mechanical advantage=Load/Effort.

720/180=4

6 0

3 years ago

goldfiish [28.3K]

Explanation:

...domains are aligned which creates a magnetic field.

... are not always aligned so the resulting element had no apparent magentism

5 0

3 years ago