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lisabon 2012 [21]
3 years ago
15

The period of sound wave coming from an instrument is o.oo2. What is the frequency of the sound

Physics
1 answer:
Maslowich3 years ago
5 0

The frequency of the sound is 500

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A woman (mass= 50.5 kg) jumps off of the ground, and comes back down to the ground at a velocity of -8.4 m/s.
Blizzard [7]

Answer:

Approximately 1.6\times 10^{3}\; \rm N.

Explanation:

By the Impulse-Momentum Theorem, the change in this woman's momentum  will be equal to the impulse that is applied to her.

The momentum p of an object is equal to the product of its mass m and velocity v. That is: p = m \cdot v.

Let v(\text{before}) and v(\text{after}) represent the velocity of the woman before and after the landing. Let m represent the woman's mass.

  • The woman's momentum before the landing would be m \cdot v(\text{before}).
  • The woman's momentum after the landing would be m \cdot v(\text{after}).

Therefore, the change in this woman's momentum would be:

\begin{aligned}& \Delta p \\ & = p(\text{after}) - p(\text{before}) \\ &= m \cdot (v(\text{after})- v(\text{before}))\end{aligned}.

On the other hand, impulse is equal to force multiplied by the duration of the force. Let F represent the average force on the woman. The impulse on her during the landing would be F \cdot t.

Apply the Impulse-Momentum Theorem.

  • Impulse: F\cdot t.
  • Change in momentum: m \cdot (v(\text{after})- v(\text{before})).

Impulse is equal to the change in momentum:

F \cdot t = m \cdot (v(\text{after})- v(\text{before})).

After landing, the woman comes to a stop. Her velocity would become zero. Therefore, v(\text{after}) = 0\; \rm m \cdot s^{-1}.

\begin{aligned}F &= \displaystyle \frac{m \cdot (v(\text{after})- v(\text{before}))}{t} \\ &= \frac{50.5\; \text{kg} \times \left(0 \; \mathrm{m \cdot s^{-1}}- 8.4\; \mathrm{m \cdot s^{-1}}\right)}{0.27\; \rm s} \\ &\approx 1.6 \times 10^{3}\; \rm N\end{aligned}.

3 0
4 years ago
Which element(s) is/are not balanced in this equation?
Shkiper50 [21]
All elements are balanced. There are 1 Mg, 1 O, 2 Li's and 2 Cl's.
3 0
3 years ago
Does passing a magnet through a coil of wire break off it’s electric current
hichkok12 [17]
A magnetic field is actually generated by a moving current (or moving electric charge specifically). The magnetic field generated by a moving current can be found by using the right hand rule, point your right thumb in the direction of current flow, then the wrap of your fingers will tell you what direction the magnetic field is. In the case of current traveling up a wire, the magnetic field generated will encircle the wire. Similarly electromagnets work by having a wire coil, and causing current to spin in a circle, generating a magnetic field perpendicular to the current flow (again right hand rule).

So if you were to take a permenant magnet and cut a hole in it then string a straight wire through it... my guess is nothing too interesting would happen. The two different magnetic fields might ineteract in a peculiar way, but nothing too fascinating, perhaps if you give me more context as to what you might think would happen or what made you come up with this question I could help.

Source: Bachelor's degree in Physics.
7 0
3 years ago
A car traveling at 45 miles per hour is brought to a stop, at constant deceleration, 132 feet from where the brakes are applied.
Nataliya [291]

Answer:

c

Explanation:

4 0
3 years ago
The momentum of an object is 2.5 kg•m/s, and it is travelling at a speed of 100 m/s.
goblinko [34]

Answer:

Refer to the attachment!~

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