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Scilla [17]
3 years ago
5

A 210-kg woorden raft floats on a lake. When a 75-kg man stands on the raft, it sinks 5.0 cm deeper into the water. When he step

s off, the raft vibraties for a while. A. What is the frequency of vibration? B. What is the total energy of vibration? Hint: ignore damping.
Physics
1 answer:
GaryK [48]3 years ago
7 0

Answer:

A: The frequency of the vibration is 1.3329 Hz

B: The total energy of the vibration is  18.39375 J

Explanation:

The force of the man his weight causes the raft to sink, and that causes the water to put a larger upward force on the raft.  This extra force is a restoring force, because it is in the opposite direction of the force put on the raft by the man.  Then when the man steps off, the restoring force pushes upward on the raft, and thus the raft – water system acts like a spring, with a spring constant found as follows:

k= F/x = ((75 kg) * (9.81 m/s²))/(5*10^-2 m) = 14715 N/m

The frequency of the vibration is determined by the spring constant (k) and the mass of the raft (210kg).

fn = 1/2π * √(k/m)  = 1/2π * √(14715 / 210) = <u>1.3329 Hz</u>

<u>The frequency of the vibration is 1.3329 Hz</u>

<u />

<u>b) </u>

Since the gravitational potential energy can be ignored, the total energy will be :

Etot = 1/2 k* A² = 1/2 * (14715 )*(0.05)² = 18.39375 J

<u>The total energy of the vibration is  18.39375 J</u>

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A 10.0 kg weather rocket generates a thrust of 230 NN . The rocket, pointing upward, is clamped to the top of a vertical spring.
blondinia [14]

Answer: 0.2m

Explanation: Firstly only the Rocket's Weight Compress the spring which can be found by

F_r=M_r*g\\F_r=10*9.81\\F_r=98.1N

According to Hooks Law

F_r=k*x\\x=F_r/k\\x=98.1/480\\x=0.2m

The part b and c of this question is done in the attachment

7 0
3 years ago
65. The weight of a body when totally immersed in a liquid is 4.2N if he weight of the liquid displaced is 2.5N. Find the weight
Anna35 [415]

Answer:

Given, Apparent weight(W₂)=4.2N

          Weight of liquid displaced (u)=2.5N

          Let weight of body in air = W₁

Solution,

             U=W₁-W₂

              W₁=4.2=2.5=6.7N

∴Weight of body in air is 6.7N

5 0
2 years ago
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai
taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

v=f_1\lambda_1

Wavelength for f = 45 Hz is,

\lambda_1=\dfrac{v}{f_1}

\lambda_1=\dfrac{343}{45}=7.62\ m

Wavelength for f = 375 Hz is,

\lambda_2=\dfrac{v}{f_2}

\lambda_2=\dfrac{343}{375}=0.914\ m/s

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0
3 years ago
The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discover
givi [52]

Answer:

Acceleration due to gravity, a=4.61\times 10^{-14}\ m/s^2

Explanation:

It is given that,

Mass of Pluto, m=1.4\times 10^{22}\ kg

Distance between Neptune and Pluto, r=4.5\times 10^{12}\ m

The force of gravity is balanced by the gravitational force between Neptune and Pluto. It is given by :

a=\dfrac{Gm}{r^2}

a=\dfrac{6.67\times 10^{-11}\times 1.4\times 10^{22}}{(4.5\times 10^{12})^2}

a=4.61\times 10^{-14}\ m/s^2

So, the acceleration due to gravity at Neptune due to Pluto is 4.61\times 10^{-14}\ m/s^2. Hence, this is the required solution.  

4 0
4 years ago
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