What wavelength ? should the astrophysicist look for to detect a transition of an electron from the n=7 to the n=3 level?

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 k=450N/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:
 $E=U+K= \frac{1}{2}kx^2 + \frac{1}{2} mv^2$
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:
 $E=K_{max} = \frac{1}{2}mv_{max}^2$
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:
 $E=U_{max}= \frac{1}{2}k A^2$

Since the mechanical energy must be conserved, we can write
 $\frac{1}{2}mv_{max}^2 = \frac{1}{2}kA^2$
from which we find the maximum speed
 $v_{max}= \sqrt{ \frac{kA^2}{m} }= \sqrt{ \frac{(450 N/m)(0.040 m)^2}{0.500 kg} }= 1.2 m/s$

b) the speed of the glider when it is at x= -0.015m

We can still use the conservation of energy to solve this part.
The total mechanical energy is:
 $E=K_{max}= \frac{1}{2}mv_{max}^2= 0.36 J$

At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
 $U= \frac{1}{2}kx^2 = \frac{1}{2}(450 N/m)(-0.015 m)^2=0.05 J$
And since
 $E=U+K$
we find the kinetic energy when the glider is at this position:
 $K=E-U=0.36 J - 0.05 J = 0.31 J$
And then we can find the corresponding velocity:
 $K= \frac{1}{2}mv^2$
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.31 J}{0.500 kg} }=1.11 m/s$

c) the magnitude of the maximum acceleration of the glider;

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) the acceleration of the glider at x= -0.015m

For a simple harmonic motion, the acceleration is given by
 $a(t)=\omega^2 x(t)$
where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
 $a=(30 rad/s)^2 (-0.015 m)=-13.5 m/s^2$

e) the total mechanical energy of the glider at any point in its motion. 

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

8 0

3 years ago

An electroscope is a fork-shaped device commonly used to detect the presence of charge. The tin leaves of an electroscope will s

velikii [3]

Answer:

All these is caused by the repulsion force.

Explanation:

The electroscope produces a series of electric charges that produce a repulsion force when is putted in contact with a electric charged object.

As the physics law mentions, two different forces are repealed, the electrocospe is charged negatively and the object positively, causing a repulsion force that avoids that both objects touch the other.

7 0

3 years ago

How many neutrons does element X have if its atomic number is 23 and its mass number is 90?

givi [52]

Answer:

67

Explanation:

- The atomic number (Z) of an atom is equal to the number of protons in the nucleus

- The mass number (A) of an atom is equal to the sum of protons and neutrons in the nucleus

Therefore, calling p the number of protons and n the number of neutrons, for element X we have:

Z = p = 23

A = p + n = 90

Substituting p=23 into the second equation, we find the number of neutrons:

n = 90 - p = 90 - 23 = 67

4 0

3 years ago

Read 2 more answers

The gnaphosid spider Drassodes cupreus has evolved a pair of lensless eyes for detecting polarized light. Each eye is sensitive

BartSMP [9]

Answer:

Part A

The intensity is $I = 618 W/m^2$

Part B

The intensity is $I_1= 81.884 W/m^2$

Explanation:

From the question we are told that

The intensity of the light detected by first eye is $I = 700 W/m^2$

Now at initial state according the question the light ray is perpendicular to the eye so it means that it is at 90° the eye

Now the first question is to obtain the intensity the first eye (the first in this case is the one focused on the light )would detect when the head is rotated by 20° its previous orientation

This is mathematically evaluated as

$I = I_i cos^2 ( 20^o)$

$I = 700\ cos^2 (20)$

$I = 618 W/m^2$

Now the second question is to obtain the intensity the first eye (the first eye in this case is the one that is not focused on the light )would detect when the head is rotated by 20° its previous orientation

Now in this case the angle between the eye and the light is 90-20 = 70°

So

$I_1 = 700 \ cos^2 (70)$

$I_1= 81.884 W/m^2$

5 0

3 years ago

Which element's atoms have the greatest average number of neutrons ?

Mkey [24]

I am pretty sure this is uranium. it has 140 neutrons.

6 0

3 years ago

Read 2 more answers