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just olya [345]

3 years ago

8

An object moves in simple harmonic motion with amplitude 11m and period 4 minutes. At time t = 0 minutes, its displacement d fro

m rest is -11m, and initially, it moves in a positive direction. Give the equation modeling the displacement d as a function of time t.d = ?

1 answer:

Sergio039 [100]3 years ago

8 0

Answer:

Explanation:

Angular velocity ω = 2π / T = 2 x 3 . 14 / (4 x 60 ) = .026 rad / s.

d = D Sin ( ωt + θ ) where D = amplitude = 11 m θ = initial phase.

-11 = 11 sin ( 0 + θ ) = sin θ = -1 , θ = - π /2

So, d = D Sin ( ωt - θ )

d = 11 Sin ( .026 t - π /2 ) is the required equation.

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If a train (travelling from Albuquerque to Atlanta) is moving 33.9 meters per second when it accelerates at -2.02 m/s^2 for 1.01

maxonik [38]

Answer:

The final velocity is 31.86 m/s.

Explanation:

The final velocity can be found using the following equation:

$V_{f} = V_{0} + at$

Where:

$V_{f}$ is the final velocity =?

$V_{0}$ is the initial velocity = 33.9 m/s

a is the acceleration = -2.02 m/s²

t is the time = 1.01 s

$V_{f} = V_{0} + at = 33.9 m/s + (-2.02 m/s^{2})*1.01 s = 31.86 m/s$

Therefore, the final velocity is 31.86 m/s.

I hope it helps you!

5 0

2 years ago

A satellite in a circular orbit of radius R around planet X has an orbital period T. If Planet X had one-fourth as much mass, th

Iteru [2.4K]

<h2>Answer: 2T</h2>

According to the Third Kepler’s Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $R$ of its orbit.

This Law is originally expressed as follows (in the case of planet X and assuming we have a circular orbit):

$T^{2}=\frac{4\pi^{2}}{GM}R^{3}$ (1)

Where:

$G$ is the Gravitational Constant

$M=1.9(10)^{27}kg$ is the mass of planet X

$R$ is the radius of the orbit of the satellite around planet X

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=2\pi\sqrt{\frac{R^{3}}{GM}}$ (2)

Now, we are asked to find the period when tha mass of the planet is $\frac{1}{4}M$ . In order to do this, we have to rewrite equation (2) with this new value:

$T=2\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (3)

Solving:

$T=4\pi\sqrt{\frac{R^{3}}{G(\frac{1}{4}M)}}$ (4)

On the other hand, if we multiply both sides of equation (2) by 2, we have:

$2T=4\pi\sqrt{\frac{R^{3}}{GM}}$ (5)

As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.

Hence, the answer is:

If Planet X had <u>one-fourth </u>as much mass, the <u>orbital period</u> of this satellite in an orbit of the same radius would be <u>2T.</u>

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Masteriza [31]

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2 years ago

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Norma-Jean [14]

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3 years ago

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A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is

gregori [183]

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is $d = 0.313 \ m$

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is $\theta = 25 ^o$

$sin \theta = \frac{h}{d}$

Where $h = h_b - h_a$

So $d sin \theta = h_b - h_a$

From the question we are told that

The mass of the block is $m = 20 \ kg$

The mass of the pendulum is $m_p = 2 \ kg$

The velocity of the pendulum at the bottom of swing is $v_p = 15 m/s$

The coefficient of restitution is $e =0.7$

The coefficient of kinetic friction is $\mu _k = 0.5$

The velocity of the block after the impact is mathematically represented as

$v_2 f = \frac{m_b - em_p}{m_b + m_p} * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p$

Where $v_2 i$ is the velocity of the block before collision which is 0

$= \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20} * 15$

Substituting value

$v_2 f = 2.310\ m/s$

According to conservation of energy principle

The energy at point a = energy at point b

So $PE_A + KE _A = PE_B + KE_B + E_F$

Where

$PE_A$ is the potential energy at A which is mathematically represented as

$PE_A = m_b gh_a$ = 0 at the bottom

$KE _A$ is the kinetic energy at A which is mathematically represented as

$K_A = \frac{1}{2} m_b * v_2f^2$

$PE_B$ is the potential energy at B which is mathematically represented as

$PE_B = m_b gh$

From the diagram $h = h_b -h_a$

$PE_B = m_b g(h_b - h_a)$

$KE _B$ is the kinetic energy at B which is 0 (at the top )

Where is $E_F$ is the workdone against velocity which from the diagram is

$\mu_k m_b g cos 25 *d$

So

$\frac{1}{2} m_b v_2 f^2 = m_b g h_b + \mu_k m_b g cos \25 * d$

Substituting values

$\frac{1}{2} * 20 * 2.310^2 = 20 * 9.8 * d sin(25) + 0.5* 20 * 9.8 * cos 25 * d$

So

$d = 0.313 \ m$

6 0

3 years ago