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1 year ago

What is the period of a wave that has a frequency of 300 hz?

Physics

1 answer:

Butoxors [25]1 year ago

3 0

Answer:

T = 0.003 s

(Period is written as T)

Explanation:

Period = time it takes for one wave to pass (measured in seconds)

frequency = number of cycles that occur in 1 second

(measured in Hz / hertz / 1 second)

Period : T

frequency : f

So, if we know that the frequency of a wave is 300 Hz, we can find the period of the wave from the relation between frequency and period

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

to find the period (T) of this wave, we need to plug in the frequency (f) of 300

T = $\frac{1}{300}$

T = 0.00333333333

So, the period of a wave that has a frequency of 300 Hz is 0.003 s

[the period/T of this wave is 0.003 s]

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What happens to the equilibrium price and quantity for jelly when the price of peanut butter increases? (Assume that peanut butt

Darya [45]

Answer:

Equilibrium price will fall and quantity demanded for jelly will also fall

Explanation:

Peanut butter and jelly are consumed together. It means they are complementary goods. These are those goods which are bought together. So increase or decrease in the price of one commodity will automatically affect the demand for another commodity.

When the price of peanut butter increases than people will demand less of peanut butter. Similarly, the demand for jelly is associated directly with the demand for peanut butter. So it will also fall . Due to the fall in the price of jelly and simultaneous fall in demand, the equilibrium price will fall.

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3 years ago

Which of the following are transferred or shared when two atoms react chemically? *

xxTIMURxx [149]

Answer:

One of the basic principles of chemistry is the electrostatic attraction between atoms or compounds. Electrons are on the outside of an atoms and that's where the charges come from and the interaction between those charges is what happens during a chemical bond. Therefore the answer would be electrons.

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2 years ago

Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.110 m, calcula

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Answer:

6.1328 kg

60.16284 N

Explanation:

r = Radius of ball = 0.11 m

$\rho$ = Density of fluid = $1.1\times 10^3\ kg/m^3$ (Assumed)

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of ball

V = Volume of ball = $\frac{4}{3}\pi r^3$

The weight of the bowling ball will balance the buouyant force

$W=F_b\\\Rightarrow mg=V\rho g\\\Rightarrow m=\frac{V\rho g}{g}\\\Rightarrow m=V\rho\\\Rightarrow m=\frac{4}{3}\pi 0.11^3\times 1.1\times 10^3\\\Rightarrow m=6.1328\ kg$

The mass of the bowling ball will be 6.1328 kg

Weight will be $6.1328\times 9.81=60.16284\ N$

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3 years ago

An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

s2008m [1.1K]

Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

Latent heat of fusion is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

Specific heat capacity of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, given that:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

Asked:

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

We have the formula:

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

We have the equation: latent heat, $Q_{L}= mL_{i}$

$Q_{L}= 0.555\times334\times10^{3}= 185370 J$

Now the time required for supply of 185370 J:

$t=\frac{Q_{L}}{q}$

$t=\frac{185370}{820}$

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

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3 years ago

A bird sits on top of a 639 m tall tower. If it's gravitational potential energy up there is 2033 J, what is its mass?

iris [78.8K]

The mass of the bird is 0.32 kg.

Explanation:

Gravitational potential energy, the energy exhibited by an object at rest due to the influence of gravitational force. So the increase in distance of object from the surface of earth leads to increase in the gravitational potential energy. Thus,

$\text {Gravitational potential energy}=m \times \text { Acceleration } \times \text { Distance of bird from bottom }$

So, as the gravitational potential energy is given as 2033 J and the position of bird placed on the tall tower is 639 m away from the bottom, then the mass (m) of the bird can be found as below.

$m o f \text { bird }=\frac{\text {Gravitational potential energy}}{a \times \text {Distance}}=\frac{2033}{9.8 \times 639}=\frac{2033}{6262.2}$

So, finally we get the bird's mass as,

m of bird = 0.32 kg

7 0

3 years ago