What do you call a molecule composed of two atoms of the same element?

What is the period that corresponds to a frequency of 39.5 Hz? Answer in units of s.

tatiyna

The answer for the following problem is mentioned below.

Therefore the time period is 0.02 seconds.

Explanation:

Frequency:

The number of waves that pass a fixed place in a given amount of time. (or)

The number of waves that pas by per second.

The SI unit of the frequency is Hertz(Hz).

Time period:

The time taken for one complete cycle of vibration to pass a given point.

The SI unit of time period is seconds. (s)

Given:

Frequency (f) = 39.5 Hz

To calculate:

Time period (T)

We know;

According to the problem;

From the problem;

f = $\frac{1}{T}$

Where;

f represents the frequency

T represents the time period

f = $\frac{1}{39.5}$

f = 0.02 seconds

Therefore the time period is 0.02 seconds.

5 0

3 years ago

One cubic meter (1.0 m3

hammer [34]

Because the masses that you give are for blocks that are 1 cubic meter in volume, they also serve as the densities for the two metals that you are comparing.

mass = density*volume 

volume = (4/3)*pi*r^3 

volume of iron sphere = (4/3)*3.14*0.0201^3 = 3.40*10^-5 m^3 

mass of iron sphere = 7860* 3.40*10^-5 m^3 = 0.27 kg = mass of Aluminum Sphere 

Volume of Al Sphere = 0.27/2700 = 9.90*10^-5 m^3 

Radius = cube root (volume / (4/3) / pi) = 2.87 cm. 

I did this using the MS calculator, and I'm not 100% sure on the numerical answer, but the process is what you need to do to solve the problem. You should double check my answer.

hope this helped :)

8 0

2 years ago

Find the work w1 done on the block by the force of magnitude f1 = 65.0 n as the block moves from xi = -3.00 cm to xf = 3.00 cm .

inn [45]

W = F•dx
F = 65 N,
dx = xf - xi = 0.03 m - (-0.03 m) = 0.06 m
W = 65 N × 0.06 m = 3.9 J

5 0

2 years ago

What is transmitted by all waves

Nesterboy [21]

<h3>Answer;</h3>

Energy

<h3>Explanation;</h3>

A wave is a transmission of disturbance from one point to another. All waves involve transmission of energy from one point called the source to another point.
Waves describes various ways in which energy can be transferred from a point source.
In electromagnetic waves, for instance, energy transmission occurs as a result of vibrations of electric and magnetic fields.
In mechanical waves energy transmission is as a result of vibration of particles in the medium used. For example in sound waves, energy is transferred through vibration of particles of air or particles of a solid or medium through which sound travels through.

8 0

3 years ago

Read 2 more answers

A large rock of mass me materializes stationary at the orbit of Mercury and falls into the sun. Itf the Sun has a mass ms and ra

son4ous [18]

Answer:

The answer is $v = \sqrt{2G\frac{M_s}{R^2}(R-r_s)}$ .

Explanation:

From the law of gravity,

$F = G \frac{Mm}{r^2}$

considering F as a conservative force, $F = - \nabla U$ ,

the general expression for gravitational potential energy is

$U = -G \frac{Mm}{r}$ ,

where G is the gravitational constant, M and m are the mass of the attracting bodies, and r is the distance between their centers. The negative sign is because the force approaches zero for large distances, and we choose the zero of gravitational potential energy at an infinite distance away.

However, as the mass of the Sun is much greater than the mass of the rock, the gravitational acceleration is defined as

$g = -G \frac{M}{r^2}$ ,

(the negative sign indicates that the force is an attractive force), and the potential energy between the rock and the Sun is

$U = g M_e R$ ,

which is actually the total energy of the system, because the rock materializes stationary at this point (there is no radial kinetic energy).

When the rock hits the surface of the Sun, almost all potential energy is converted to kinetic energy, but not all because the Sun is not a puntual mass. So the potential energy converted to kinetic energy is

$U_p = g M_e(R- r_s)$ ,

then, the kinetik energy when the rock hits the surface is

$U_k =\frac{1}{2}M_e v^2 = g M_e(R- r_s)$ ,

so

$v = \sqrt{2g(R-r_s)}$

where g is the gravitational acceleration generated by the Sun at R,

$g = G \frac{M_s}{R^2}$ .

8 0

2 years ago