What is the speed of a wave that has a frequency of 45hz and wavelength of 0.1 meters?

Phosphorus-32, a radioactive isotope of phosphorus-31 (atomic number 15), undergoes a form of radioactive decay whereby a neutro

dimulka [17.4K]

Answer:

The product of the decay its Sulfur-32

Explanation:

Phosphorus-32 ( lets write it $_{15}^{32}P$ , where the number above its the atomic mass and the number below the atomic number) decays turning a neutron into a proton and emitting radiation on the form of a electron. This is the beta minus decay, and, actually, an electronic antineutrino its also produced. We can write this decay for an X isotope with a Y isotope produced as:

$_{Z}^{A}X \to _{Z+1}^{A}Y + e^- + \bar{\nu_e}$

where $e^-$ its the electron, and $\bar{\nu_e}$ the electronic antineutrino . We can see that the atomic number increases by one (cause a proton it produced and retained into the nucleus), and the atomic mass is approximately the same (there is a small difference between the neutron and proton mass, but its very small).

So, Phosphorus-32 (atomic number 15) will turn to an element with atomic number 16, and atomic mass 32, as:

$_{15}^{32}P \to _{15+1}^{32}Y + e^- + \bar{\nu_e}$ .

$_{15}^{32}P \to _{16}^{32}Y + e^- + \bar{\nu_e}$ .

The Y isotope must have an atomic number of 16 and an atomic mass of 32. The element with atomic number 16 its Sulfur (S), so, our decay its

$_{15}^{32}P \to _{16}^{32}S + e^- + \bar{\nu_e}$ .

and the product of such decay its Sulfur-32

5 0

3 years ago

A 60 year old person has a threshold of hearing of 95.0 dB for a sound with frequency f=10,000 Hz. By what factor must the inten

Nadusha1986 [10]

Answer:

The intensity increased by a factor of 158489

Explanation:

Given that,

Sound level = 95.0 dB

Sound level = 43.0 dB

Frequency = 10000 Hz

We need to calculate the ratio of sound intensity

Using formula of sound level

$sound\ level =10 log\dfrac{I}{I_{0}}$

Put the value into the formula

$95.0=10\ log\dfrac{I_{1}}{I_{0}}$ ...(I)

$43.0=10\ log\dfrac{I_{2}}{I_{0}}$ .....(II)

Subtracting these equations

$52.0=10\ log\dfrac{I_{1}}{I_{2}}$

$\log\dfrac{I_{1}}{I_{2}}=5.2$

Taking inverse log

$\dfrac{I_{1}}{I_{2}}=10^{5.2}$

$\dfrac{I_{1}}{I_{2}}=158489$

Hence, The intensity increased by a factor of 158489

4 0

3 years ago

Mr. MacDougall got his vehicle stuck in the snow. Being the nice student that you are, you stop to help Mr. MacDougall out of th

zhannawk [14.2K]

Answer:

Explanation:

work done=force*displacement

=350N*15m

=5250 joule

4 0

2 years ago

Suppose the coefficient of static friction between the road and the tires on a car is 0.638 and the car has no negative lift. Wh

larisa [96]

Answer:

12.6332454263 m/s

Explanation:

m = Mass of car

v = Velocity of the car

$\mu$ = Coefficient of static friction = 0.638

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of turn = 25.5 m

When the car is on the verge of sliding we have the force equation

$\dfrac{mv^2}{r}=\mu mg\\\Rightarrow v=\sqrt{\mu gr}\\\Rightarrow v=\sqrt{0.638\times 9.81\times 25.5}\\\Rightarrow v=12.6332454263\ m/s$

The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s

4 0

3 years ago

If two cars A and B are moving with velocity 60 km/hr and 80 km/hr

Vitek1552 [10]

Answer:

VAB = 20km/hr

Explanation:

<u>Given the following data;</u>

Velocity of car A, VA = 60km/hr

Velocity of car B, VB = 80km/hr

To find the relative velocity of B w.r.t A, VAB;

Since the two cars are moving in the same direction, we have;

VAB = VB - VA

Substituting into the equation, we have;

VAB = 80 - 60

Therefore, the relative velocity of car B with respect to car A is 20 kilometers per hour.

3 0

3 years ago