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Bingel [31]
3 years ago
5

Six members of a synchronized swim team wear earplugs to protect themselves against water pressure at depths, but they can still

hear the music and perform the combinations in the water perfectly. One day, they were asked to leave the pool so the dive team could practice a few dives, and they tried to practice on a mat, but seemed to have a lot more difficulty. Why might this be
Physics
1 answer:
tangare [24]3 years ago
3 0

Answer:

The sound travels differently in different medium according the density of the medium.

Explanation:

The sound travels faster in dense medium and can be heard by the vibration of the bone present in the ear. The ear plugs reduce the sound intensity in both medium water and on land (air).

In air the sound is not heard properly due to the earplugs that stops the as the vibration are not able to produce as sound is not able to reach to middle ear, but Navy researchers have discovered that sound under water is heard by the bone present behind the ear, vibrations mastoid.

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What happened to the kinetic energy when the velocy of a moving body is doubled?​
dem82 [27]

Answer:

If velocity is doubled, Kinetic Energy increases by 4 times. Kinetic energy of a body is the energy possessed by it, by virtue of its motion, i.e. if the body is moving it will always have kinetic energy.Nov 1, 2018

Explanation:

8 0
2 years ago
Read 2 more answers
A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the
olasank [31]

Explanation:

The given data is as follows.

   Spring constant (k) = 78 N/m,     \theta = 30^{o}

 Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

                mgh sin \theta = \frac{1}{2}kx^{2}

       h = \frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}

          = \frac{78 \times 0.04}{2 \times 0.5 \times 9.8 \times 0.5}

          = 0.64 m

Thus, we can conclude that the distance traveled by the block is 0.64 m.

4 0
2 years ago
The mass of the earth is 5.98 1024 kg, the mass of the moon is 7.36 1022 kg, and the distance between the centers of the earth a
Setler [38]

Answer:

 x_{cm} = 4.644 10⁶ m

Explanation:

The center of mass is given by the equation

         x_{cm} = 1 / M_{total}  ∑ x_{i}  m_{i}

Where M_{total} is the total masses of the system, x_{i} is the distance between the particles and m_{i} is the masses of each body

Let's apply this equation to our problem

        M = Me + m

        M = 5.98 10²⁴ + 7.36 10²²

        M = 605.36 10²² kg

Let's locate a reference system located in the center of the Earth

Let's calculate

       x_{cm} = 1 / 605.36 10²²   [Me 0 + 7.36 10²² 3.82 10⁸]

       x_{cm} = 4.644 10⁶ m

4 0
2 years ago
An apparatus like the one Cavendish used to find G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.
Ber [7]

Answer:

The magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N.

Explanation:

Given that,

Mass of first lead ball, m_1=5.2\ kg

Mass of the other lead ball, m_2=0.046\ kg

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N

So, the magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N. Hence, this is the required solution.

5 0
2 years ago
A wheel initially has an angular velocity of 18 rad/s, but it is slowing at a constant rate of 2 rad/s 2 . By the time it stops,
ZanzabumX [31]

Answer:

5) 13 revolutions (approximately)

Explanation:

We apply the equations of circular motion uniformly accelerated :

ωf²= ω₀² + 2α*θ Formula (1)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed  ( rad/s)

Data:

ω₀ = 18 rad/s

ωf = 0

α = -2 rad/s²  ; (-) indicates that the wheel is slowing

Revolutions calculation that turns the wheel until it stops

We apply the formula (1)

ωf²= ω₀² + 2α*θ

0 = (18)² + 2( -2)*θ

4*θ =  (18)²

θ =  (18)²/4 = 81 rad

1 revolution = 2π rad

θ = 81 rad * 1 revolution / 2πrad

θ =  13 revolutions approximately

8 0
3 years ago
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