All categories

4 years ago

The frequency of sound wave A is 250 Hz. Find the time period.

Physics

1 answer:

olga2289 [7]4 years ago

7 0

Time period = 1 / frequency
Time period = 1 / 250 th of a second

You might be interested in

The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a

dimaraw [331]

The absolute pressure is given by the equation,

$P_{abs}=P_{atm}-P_{vac}$

Here, $P_{abs}$ is absolute pressure, $P_{atm}$ is atmospheric pressure and $P_{vac}$ is vacuum pressure.

Therefore,

$P_{abs}=98 kPa-80 kPa=18kPa$

The gage pressure is given by the equation,

$P_{gage}=P_{abs}-P_{atm}$ .

Thus,

$P_{gage}=18kPa-98 kPa=-80 kPa$ .

In kn/m^2,

The absolute pressure,

$P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2$ .

In lbf/in2

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2$

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2$

In psi,

The absolute pressure,

$P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi$ .

The gage pressure,

$P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi$

In mm Hg

The absolute pressure,

$P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg$

The gage pressure,

$P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg$

3 0

3 years ago

Two particles with charges are initially very far apart (effectively an infinite distance apart). They are then fixed at positio

ddd [48]

Answer:

potential energy increases.

Explanation:

The potential energy between the two charged particles is given by

U = k Q q / r

If they are very far apart then r tends to infinity and the potential energy is zero.

If they come closer then the potential energy between the two charged particles increases.

Thus, the potential energy increases.

3 0

4 years ago

An object of mass 6.00 kg falls with an aceleration of 8.00 m/s2. The magnitude of air resitance must be ____ N

allsm [11]

Using Newton's Second Law, we can find the air resistance. We know the net force is equal to mass times acceleration. $F_{net} = m*a = (6.00kg)(8.00 \frac{m}{s^2}) = 48N 

F_{g} - F_{d} = 48N

48N = (6.00kg)(9.81m/s^2) - F_{d} 

F_{d} = 10.86N$

7 0

3 years ago

How is volume and temperature related?

ololo11 [35]

Charles' Law: The Temperature-Volume Law. This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature. As the volume goes up, the temperature also goes up, and vice-versa.

5 0

3 years ago

How far can a person run in 13.22 minutes if he or she runs at an average speed of 12.33 km/hr?

Alenkinab [10]

Answer:

2.72 km

Explanation:

(12.33 km)/ 1 hr * (1 hr)/ 60 min

0.2055 km/ min

distance=rate * time (assuming v is constant,

a=0)

(0.2055 km/ min)*(13.22 min)

2.72 km OR 2716.71 m

4 0

3 years ago

Read 2 more answers