Ask question

Ask question

All categories

jek_recluse [69]

2 years ago

13

A block is attached to a horizontal spring and oscillates back and forth on a frictionless horizontal surface at a frequency of

4.42 Hz. The amplitude of the motion is 8.26 x 10-2 m. At the point where the block has its maximum speed, it suddenly splits into two identical parts, only one part remaining attached to the spring.
(a) What is the amplitude and
(b) the frequency of the simple harmonic motion that exists after the block splits?

1 answer:

Nastasia [14]2 years ago

8 0

Answer:

Amplitude = 0.058m

Frequency = 6.25Hz

Explanation:

Given

Amplitude (A) = 8.26 x 10-2 m

Frequency (f) = 4.42Hz

Conversation of energy before split

½mv² = ½KA²

Make A the subject of formula

A = $v\sqrt{\frac{m}{k} }$

Conversation of energy after split

½(m/2)V'² = ½(m/2)V² = ½KA'²

½(m/2)V² = ½KA'²

Make A the subject of formula

First divide both sides by ½

(m/2)V² = KA'²

Divide both sides by K

$\frac{m}{2K}$ V² = A'²

$v\sqrt{\frac{m}{2k} }$ = A'

Substitute $v\sqrt{\frac{m}{k} }$ for A in the above equation

A' = A/√2

A' = 8.26 x 10^-2/√2

A' = 0.05840702012600882

Amplitude after split = 0.058 (Approximated)

Frequency (f') = f√2

f' = 4.42√2

f' = 6.25082394568908011

Frequency after split = 6.25Hz (approximated)

You might be interested in

Calculate A, E, μ, cv and S for 1 mole of Kr at 298 K and 1 atm (assuming ideal behavior)

vaieri [72.5K]

Answer:

internal energy E 3716.35 j

cv = 12.47 J/K

S = 12.47 J/K

A = 0.29 J

$\mu =3716.35 J/mole$

Explanation:

given data:

Kr atomic number = 36

degree of freedom = 3

1) internal energy E = $\frac{f}{2} n R T$

= $\frac{3}{2} *1*8.314*298 = 3716.35 j$

2) $cv = \frac{E}{T}$

$= \frac{3716.35}{298} = 12.47 J/K$

3) $S = cv =\frac{E}{T} = 12.47 J/K$

4) A, Halmholtz free energy = E -TS = 37146.35 - 12.47*298 = 0.29 J

5) $chemcial potential \mu = \frac{energy}{mole} = 3716.35 J/mole$

6 0

2 years ago

Select the correct answer.

Sergeeva-Olga [200]

Answer:

All of the above.

Explanation:

All of these steps are useful towards getting a job.

8 0

1 year ago

Why is temperature scalar?

shepuryov [24]

Temperature is just a measure of how HOT or COLD a substance is, which can be easily defined by a magnitude using a numerical value say “300 K” or “27°C”. Hence we can say it is a scalar quantity.

But the energy which transfer by virtue of a temperature difference is a vector quantity, as it has both magnitude and direction of motion (from High temperature to low temperature region).

3 0

2 years ago

The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is th

Darya [45]

Answer:

1.92 x 10⁻¹²J

Explanation:

The magnetic force from the magnetic field gives the circulating protons gives the particle the necessary centripetal acceleration to keep it orbiting round the circular path. And from Newton's second law of motion, the force(F) is equal to the product of the mass(m) of the proton and the centripetal acceleration(a). i.e

F = ma

Where;

a = $\frac{v^2}{r}$ [v = linear velocity, r = radius of circular path]

=> F = m $\frac{v^2}{r}$ ------------(i)

We also know that the magnitude of this magnetic force experienced by the moving charge (proton) in a magnetic field is given by;

F = q v B sin θ ----------(ii)

Where;

q = charge of the particle

v = velocity of the particle

B = magnetic field

θ = the angle between the velocity and the magnetic field.

Combining equations (i) and (ii) gives

m $\frac{v^2}{r}$ = q v B sin θ [θ = 90° since the proton is orbiting at the maximum orbital radius]

=> m $\frac{v^2}{r}$ = q v B sin 90°

=> m $\frac{v^2}{r}$ = q v B

Divide both side by v;

=> m $\frac{v}{r}$ = qB

Make v subject of the formula

v = $\frac{qBr}{m}$

From the question;

B = 1.25T

m = mass of proton = 1.67 x 10⁻²⁷kg

r = 0.40m

q = charge of a proton = 1.6 x 10⁻¹⁹C

Substitute these values into equation(iii) as follows;

v = $\frac{(1.6*10^{-19})(1.25)(0.4)}{(1.67*10^{-27})}$

v = 4.79 x 10⁷m/s

Now, the kinetic energy, K, is given by;

K = $\frac{1}{2}$ mv²

m = mass of proton

v = velocity of the proton as calculated above

K = $\frac{1}{2}(1.67*10^{-27} * (4.79 * 10^7)^2 )$

K = 1.92 x 10⁻¹²J

The kinetic energy is 1.92 x 10⁻¹²J

8 0

2 years ago

A pyramid was built with approximately 2.3 million stone blocks, each weighing 2.4 tons (1 ton = 2,000 lbm). Find the mass of th

pogonyaev

Total weight in tons

$\\ \sf\longmapsto 2.3\times 10^6\times 2.4$

$\\ \sf\longmapsto 5.52\times 10^6$

$\\ \sf\longmapsto 5520000ton$

1ton=1000kg

$\\ \sf\longmapsto 5520000\times 1000$

$\\ \sf\longmapsto 5520000000kg$

$\\ \sf\longmapsto 5.52\times 10^9Kg$

4 0

2 years ago

Read 2 more answers