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AleksAgata [21]

2 years ago

7

A block attached to a spring undergoes simple harmonic motion on a horizontal frictionless surface. Its total energy is 50 J. Wh

en the displacement is half the amplitude, the kinetic energy is

1 answer:

Nina [5.8K]2 years ago

3 0

Answer:

The kinetic energy at a displacement of half the amplitude is 37.5 J

Explanation:

Given;

total energy on the spring, E = 50 J

When the displacement is half the amplitude, the total energy in the spring is sum of the kinetic energy and elastic potential energy.

E = K + U

Where;

K is the kinetic energy

U is the elastic potential energy

K = E - U

K = E - ¹/₂KA²

When the displacement is half = ¹/₂(A) = A/₂

K = E - ¹/₂K(A/₂)²

K = E - ¹/₂K(A²/₄)

K = E - ¹₄(¹/₂KA²)

Recall, E = ¹/₂KA²

K = ¹/₂KA² - ¹₄(¹/₂KA²) (recall from simple arithmetic, 1 - ¹/₄ = ³/₄)

K = 1(¹/₂KA²) - ¹₄(¹/₂KA²) = ³/₄(¹/₂KA²)

K = ³/₄(¹/₂KA²)

But E = ¹/₂KA² = 50J

K = ³/₄ (50J)

K = 37.5 J

Therefore, the kinetic energy at a displacement of half the amplitude is 37.5 J

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A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b

daser333 [38]

Answer:

<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b) The bullet travels horizontally 110.6 m</h2>

Explanation:

a) Consider the vertical motion of bullet

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 1.5 m

Substituting

s = ut + 0.5 at²

1.5 = 0 x t + 0.5 x 9.81 xt²

t = 0.553 s

Time elapsed before the bullet hits the ground is 0.553 seconds.

b) Consider the horizontal motion of bullet

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 200 m/s

Acceleration, a = 0 m/s²

Time, t = 0.553 s

Substituting

s = ut + 0.5 at²

s = 200 x 0.553 + 0.5 x 0 x 0.553²

s = 110.6 m

The bullet travels horizontally 110.6 m

6 0

2 years ago

A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.

dalvyx [7]

Answer:

a. Acceleration, a = 1.88 m/s²

b. Time, t = 7.87 seconds.

Explanation:

Given the following data;

Initial velocity, U = 14.5m/s

Final velocity, V = 29.3m/s

Distance, S = 172m

a. To find the acceleration of the speedboat;

We would use the third equation of motion;

V² = U² + 2aS

Substituting into the formula

29.3² = 14.5² + 2a*172

858.49 = 210.25 + 344a

344a = 858.49 - 210.25

344a = 648.24

a = 648.24/344

Acceleration, a = 1.88 m/s²

b. To find the time;

We would use the first equation of motion;

V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

6 0

3 years ago

Explain the difference between what Isaac Newton and Louis de Broglie would have to say about the momentum of a particle that is

pishuonlain [190]

Answer:

Explained

Explanation:

Newton would resort to the classical mechanics and say that the momentum of the particle that is moving with a constant velocity will be given by: momentum = mass x velocity

this approach will highlight the particle nature and will not be relativistic.

De-Broglie will say that the momentum of the particle is related to its associated matter wave and the relation between them is given by:

$p = \frac{h}{\lambda}$

where \lambda = wavelength of the matter wave associated to the particle, h = planck's constant

and $p = \gamma\times mv$

thus, this highlights the wave nature of the particle and is also relativistic.

6 0

3 years ago

A 80 W light bulb (normally run at 120 V) is attached to a transformer. The voltage source in the transformer is 65 V and Np = 3

Marina CMI [18]

67.8 turns needed by the secondary coil to run the bulb.

<u>Explanation</u>:

We know that,

$\text { Electric power }(p)=\frac{V^{2}}{R}$

$\text { Hence, } \frac{P_{1}}{P_{2}}=\frac{V_{1}^{2} / R}{V_{2}^{2} / R}$

$\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

For calculating number of turns

$\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

Given that,

$80 \mathrm{W}\left(P_{1}\right) \text { bulb with voltage } 120 \mathrm{V}\left(V_{1}\right) \text { is connected to a transformer. }$

$\text { The source voltage of a transformer is }\left(V_{P}\right) \text { is } 65 \mathrm{V}$

$\text { The number of turns in primary winding of transformer is }\left(N_{P}\right) \text { is } 30 .$

We need to find the number of turns in the secondary winding $\left(N_{S}\right)$ to run the bulb at 120W $\left(P_{2}\right)$

Firstly find the secondary voltage in the transformer use, $\frac{P_{1}}{P_{2}}=\frac{V_{1}^{2}}{V_{2}^{2}}$

$\frac{80}{120}=\frac{120^{2}}{V_{2}^{2}}$

$V_{2}^{2}=\frac{120^{2} \times 120}{80}$

$V_{2}^{2}=\frac{1728000}{80}$

$V_{2}^{2}=21600$

$V_{2}=\sqrt{21600}$

$V_{2}=146.9 \mathrm{V}=V_{S}$

Now, finding the number of turns in secondary coil. Use, $\frac{N_{P}}{N_{S}}=\frac{V_{P}}{V_{S}}$

$\frac{30}{N_{S}}=\frac{65}{146.9}$

$N_{S}=\frac{30 \times 146.9}{65}$

$N_{S}=\frac{4407}{65}$ $N_{S}=67.8$

The number of turns in the secondary winding are 67.8 turns.

6 0

3 years ago

sleet_krkn [62]

Answer:

yes

Explanation:

8 0

3 years ago

Read 2 more answers