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Fiesta28 [93]
2 years ago
13

The frequency of a sound wave is 457 Hz. What is the period?

Physics
1 answer:
kondor19780726 [428]2 years ago
7 0
Period: 1/Frequency
1/457 = 0.002188 seconds
Round it to 0.0022 seconds

The answer is D
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A brick of mass 5 kg is released from rest at a height of 3 m. How fast is it going when it hits the ground? Acceleration due to
sineoko [7]

Taking into account the definition of kinetic, potencial and mechanical energy, when the brick hits the ground, it has a speed of 7,668 m/s.

<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

Kinetic energy is defined as the amount of work necessary to accelerate a body of a given mass and at rest, until it reaches a given speed. Once this point is reached, the amount of accumulated kinetic energy will remain the same unless there is a change in speed or the body returns to its state of rest by applying a force.

The kinetic energy is represented by the following expression:

Ec= ½ mv²

Where:

  • Ec is the kinetic energy, which is measured in Joules (J).
  • m is the mass measured in kilograms (kg).
  • v is the speed measured in meters over seconds (m/s).

<h3>Potential energy</h3>

On the other hand, potential energy is the energy that measures the ability of a system to perform work based on its position. In other words, this is the energy that a body has at a certain height above the ground.

Gravitational potential energy is the energy associated with the gravitational force. This will depend on the relative height of an object to some reference point, the mass, and the force of gravity.

So for an object with mass m, at height h, the expression applied to the gravitational energy of the object is:

Ep= m×g×h

Where:

  • Ep is the potential energy in joules (J).
  • m is the mass in kilograms (kg).
  • h is the height in meters (m).
  • g is the acceleration of fall in m/s².
<h3>Mechanical energy</h3>

Finally, mechanical energy is that which a body or a system obtains as a result of the speed of its movement or its specific position, and which is capable of producing mechanical work. Then:

Potential energy + kinetic energy = total mechanical energy

<h3>Principle of conservation of mechanical energy </h3>

The principle of conservation of mechanical energy indicates that the mechanical energy of a body remains constant when all the forces acting on it are conservative (a force is conservative when the work it does on a body depends only on the initial and final points and not the path taken to get from one to the other.)

Therefore, if the potential energy decreases, the kinetic energy will increase. In the same way, if the kinetics decreases, the potential energy will increase.

<h3>This case</h3>

A brick of mass 5 kg is released from rest at a height of 3 m. Then, at this height, the brick of mass has no speed, so the kinetic energy has a value of zero because it depends on the speed or moving bodies. But the potential energy is calculated as:

Ep= 5 kg× 9.8 \frac{m}{s^{2} }× 3 m

Solving:

<u><em>Ep= 147 J</em></u>

So, the mechanical energy is calculated as:

Potential energy + kinetic energy = total mechanical energy

147 J +  0 J= total mechanical energy

147 J= total mechanical energy

The principle of conservation of mechanical energy  can be applied in this case. Then, when the brick hits the ground, the mechanical energy is 147 J. In this case, considering that the height is 0 m, the potential energy is zero because this energy depends on the relative height of the object. But the object has speed, so it will have kinetic energy. Then:

Potential energy + kinetic energy = total mechanical energy

0 J +  kinetic energy= 147 J

kinetic energy= 147 J

Considering the definition of kinetic energy:

½  5 kg×v²= 147 J

v=\sqrt{\frac{2x147 J}{5 kg} }

v=7.668 m/s

Finally, when the brick hits the ground, it has a speed of 7,668 m/s.

Learn more about mechanical energy:

brainly.com/question/17809741

brainly.com/question/14567080

brainly.com/question/12784057

brainly.com/question/10188030

brainly.com/question/11962904

#SPJ1

6 0
2 years ago
a motorcycle is capable of accelerating at 5.1 m/s starting from rest how far can it travell in 1.5 sec
Katarina [22]
The answer is 21m because the motion is in one dimension with constant acceleration.

The initial velocity is 0, because it started from rest, the acceleration <span>ax</span> is <span>4.7<span>m<span>s2</span></span></span>, and the time t is <span>3.0s</span>

Plugging in our known values, we have

<span>Δx=<span>(0)</span><span>(3.0s)</span>+<span>12</span><span>(4.7<span>m<span>s2</span></span>)</span><span><span>(3.0s)</span>2</span>=<span>21<span>m</span></span></span>

5 0
3 years ago
Please help on this one
PolarNik [594]

If the mass of both of the objects is doubled, then the force of gravity between them is quadrupled; and so on. Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces

8 0
3 years ago
Read 2 more answers
Electricity and magnetism
stellarik [79]

Answer:

Electricity is which is possessed by the flow of electrons.

Magnitism is the process in which the magnet attract the things which consist magnetic substance.

6 0
2 years ago
How do you solve this problem?
Katarina [22]

The particle has acceleration vector

\vec a=\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\vec\jmath

We're told that it starts off at the origin, so that its position vector at t=0 is

\vec r_0=\vec0

and that it has an initial velocity of 12 m/s in the positive x direction, or equivalently its initial velocity vector is

\vec v_0=\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)\,\vec\imath

To find the velocity vector for the particle at time t, we integrate the acceleration vector:

\vec v=\vec v_0+\displaystyle\int_0^t\vec a\,\mathrm d\tau

\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\displaystyle\int_0^t\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)\,\mathrm d\tau\right]\,\vec\jmath

\vec v=\left[12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right]\,\vec\imath+\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\vec\jmath

Then we integrate this to find the position vector at time t:

\vec r=\vec r_0+\displaystyle\int_0^t\vec v\,\mathrm d\tau

\vec r=\left[\displaystyle\int_0^t\left(12\,\dfrac{\mathrm m}{\mathrm s}+\left(-2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\right)\,\mathrm d\tau\right]\,\vec\imath+\left[\displaystyle\int_0^t\left(4.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\,\mathrm d\tau\right]\,\vec\jmath

\vec r=\left[\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)t+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right]\,\vec\imath+\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\,\vec\jmath

Solve for the time when the y coordinate is 18 m:

18\,\mathrm m=\left(2.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\implies t=3.0\,\mathrm s

At this point, the x coordinate is

\left(12\,\dfrac{\mathrm m}{\mathrm s}\right)(3.0\,\mathrm s)+\left(-1.0\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3.0\,\mathrm s)^2=27\,\mathrm m

so the answer is C.

7 0
3 years ago
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