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swat32
3 years ago
13

At time t0 = 0 the mass happens to be at y0 = 4.05 cm and moving upward at velocity v0 = +4.12 m/s. (Mind the units!) Calculate

the amplitude A of the oscillating mass. Answer in units of cm.
Physics
1 answer:
Katen [24]3 years ago
4 0

Answer:

A = 5.727 cm

Explanation:

at time t = 0s the displacement of mass is 4.05 cm and velocity 4.12 m/s

we know that velcoity for simple harmonic motion is given as

v_o = \omega \sqrt {A^2 - Y_O^2}

W KNOW THAT

\omega = \frac{v_0}{Y_o}

Therefore we  get

Y_O^2 = A^2 - Y_O^2

A =\sqrt 2 * Y_O

A =\sqrt 2 * 4.05\\

A = 5.727 cm

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You drive 200 miles in 3 hours before stopping for lunch and gas. After lunch you
Paraphin [41]

Given :

You drive 200 miles in 3 hours before stopping for lunch and gas. After lunch you  travel 250 miles in an hour and a half.

To Find :

Average speed.

Solution :

We know, average speed is given by :

v=\dfrac{\text{Total distance covered}}{\text{Total time taken}}\\\\v=\dfrac{200+250}{3+1.5}\ miles/hr\\\\v=\dfrac{450}{4.5}\ miles/hr\\\\v=100\ miles/hr

Therefore, average speed of the journey is 100 miles/hr.

Hence, this is the required solution.

4 0
4 years ago
Earth is slightly closer to the Sun in January than in July. How does the area swept out by Earth's orbit around the Sun during
suter [353]

Answer: Option <em>a.</em>

Explanation:

Kepler's 2nd law of planetary motion states:

<em>A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.</em>

It tells us that it doesn't matter how far Earth is from the Sun, at equal times, the area swept out by Earth's orbit it's always the same independently from the position in the orbit.

3 0
3 years ago
PLEASEEEE HEEEEELP!!!!!
8090 [49]

Answer:

Before:

p_{truck}=16400\ kg.m/s

p_{car}=10000\ kg.m/s

After:

p_{truck}=8000\ kg.m/s

p_{car}=8400\ kg.m/s

v_{fcar}=8.4\ m/s

F=9333.33 \ Nw

Explanation:

<u>Conservation of Momentum</u>

Two objects of masses m1 and m2 moving at speeds v1o and v2o respectively have a total momentum of

p_1=m_1v_{1o}+m_2v_{2o}

After the collision, they have speeds of v1f and v2f and the total momentum is

p_2=m_1v_{1f}+m_2v_{2f}

Impulse J is defined as

J=F.t

Where F is the average impact force and t is the time it lasted

Also, the impulse is equal to the change of momentum

J=\Delta p

As the total momentum is conserved:

p_1=p_2

m_1v_{1o}+m_2v_{2o}=m_1v_{1f}+m_2v_{2f}

We can compute the speed of the second object by solving the above equation for v2f

\displaystyle v_{2f}=\frac{ m_1v_{1o}+m_2v_{2o} -m_1v_{1f} }{  m_2 }

The given data is

m_1=2000\ kg

m_2=1000\ kg\\v_{1o}=8.2\ m/s\\v_{2o}=0\ m/s\\v_{1f}=4\ m/s

a) The impulse will be computed at the very end of the answer

b) Before the collision

p_{truck}=2000\cdot 8.2=16400\ kg.m/s

p_{car}=1000\cdot 0=0\ kg.m/s

c) After collision

p_{truck}=2000\cdot 4=8000\ kg.m/s

Compute the car's speed:

\displaystyle v_{2f}=\frac{ 16400+0 -8000 }{ 1000 }

v_{2f}=8.4\ m/s

And the car's momentum is

p_{car}=1000\cdot 8.4=8400\ kg.m/s

The Impulse J of the system is zero because the total momentum is conserved, i.e. \Delta p=0.

We can compute the impulse for each object

J_1=\Delta p_1=2000(4-8.2)=-8400 \N.s

The force can be computed as

\displaystyle F=\frac{J}{t}=-\frac{8400}{0.9}=-9333.33 \ Nw

The force on the car has the same magnitude and opposite sign

7 0
4 years ago
A car slows down from 27.7 m/s to 10.9 m/s in 2.37 s, what is the acceleration?
larisa [96]

Answer:

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Explanation:

4 0
3 years ago
A diver comes off a board with arms straight up and legs straight down, giving her a moment of inertia about her rotation axis o
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Answer:

θ₁ = 0.5 revolution

Explanation:

We will use the conservation of angular momentum as follows:

L_1=L_2\\I_1\omega_1=I_2\omega_2

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I₁ = initial moment of inertia = 18 kg.m²

I₂ = Final moment of inertia = 3.6 kg.m²

ω₁ = initial angular velocity = ?

ω₂ = Final Angular velocity = \frac{\theta_2}{t_2} = \frac{2\ rev}{1.2\ s} = 1.67 rev/s

Therefore,

(18\ kg.m^2)\omega_1 = (3.6\ kg.m^2)(1.67\ rev/s)\\\\\omega_1 = \frac{(3.6\ kg.m^2)(1.67\ rev/s)}{(18\ kg.m^2)}\\\\\omega_1 = \frac{\theta_1}{t_1} =  0.333\ rev/s\\\\\theta_1 = (0.333\ rev/s)t_1

where,

θ₁ = revolutions if she had not tucked at all = ?

t₁ = time = 1.5 s

Therefore,

\theta_1 = (0.333\ rev/s)(1.5\ s)\\

<u>θ₁ = 0.5 revolution</u>

4 0
3 years ago
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