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Olenka [21]
2 years ago
15

Identify the vibrating media in three different types of musical instruments.

Physics
1 answer:
mart [117]2 years ago
3 0

Explanation:

Hole. Hole. Different notes can be played on the flute by blocking holes. ...

Drum skin. Drum skin. Hitting the bongo drum makes its tight elastic skin vibrate.

String. String. ...

Sound. hole. ...

Bow. Bow.

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You just found a metallic object in a creek bed. you have found its mass to be 96.5 g, and it displaces 5mL of water. what is th
MaRussiya [10]
Judging by the density (19.3 g/ml) there's a very good chance that it could be gold.
4 0
2 years ago
Read 2 more answers
A 0.0450-kg golf ball initially at rest is given a speed of 25.0 m/s when a club strikes. If the club and ball are in contact fo
LuckyWell [14K]

Answer:

Average force, F = 562.5 N

Explanation:

Mass of the golf ball, m = 0.045 kg

Initially, it is at rest, u = 0

Final speed of the ball, v = 25 m/s

The club and the ball are in contact for, t=2\ ms=2\times 10^{-3}\ s

We need to find the average force acting on the ball. It can be calculated using the formula as :

F\times t=m(v-u)

F=m\dfrac{v-u}{t}

F=0.045\times \dfrac{25}{2\times 10^{-3}}  

F = 562.5 N

So, the average force acting on the ball is 562.5 N. Hence, this is the required solution.

5 0
3 years ago
25. How long does it take a child on a swing to complete one swing if her center of gravity is 4.00 m below the pivot?
Lina20 [59]
  • Length=l=4m

\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{l}{g}}

\\ \rm\rightarrowtail T=2\pi\sqrt{\dfrac{4}{9.8}}

\\ \rm\rightarrowtail T=2\pi(0.63887)

\\ \rm\rightarrowtail T=1.27774\pi

\\ \rm\rightarrowtail T=4.012s

4 0
2 years ago
List several examples of applied force, normal force, and friction that you’ve observed in your life.
grandymaker [24]

applied forces would be push for example.

normal forces would seem to be a force such as gravity.

friction for example when you try to slide on carpet but the fabric or whatever its made of stops you.

3 0
3 years ago
In an atomic clock there are approximately 9.193 × 109oscillations of the specified light emitted by cesium-133 atoms. The text
Aleks [24]

Explanation:

6000 years = 6000 x 365 x 24 x 60 x 60

= 1.892 x 10¹¹ second

 gain is 1 second

1 second is equivalent to 9.193 × 10⁹ oscillations .

In 1.892 x 10¹¹ second ,  change in oscillation is 9.193 × 10⁹ oscillation

in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹

=  4.859 x 10⁻² oscillations .

5 0
2 years ago
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