All categories

2 years ago

The human ear is most sensitive to the sounds in what range?

Physics

1 answer:

jeyben [28]2 years ago

3 0

Answer:

The human ear is most sensitive to sounds in the range of 2000 to 3000 Hz

You might be interested in

In which environment might you expect a deposit of rock salt or sock gypsum to have formed?

wolverine [178]

Most likely in a shallow sea environment.

8 0

3 years ago

Read 2 more answers

7. A box is being pushed from the left at 15N and pushed from the right at 10N,

professor190 [17]

It is 5N to the left

8 0

2 years ago

An electron that has a velocity with x component 1.6 × 10^6 m/s and y component 2.6 × 10^6 m/s moves through a uniform magnetic

ioda

Answer:

(a)

$\overrightarrow{F}=4.58\times10^{-14}\widehat{K}N$

(b) $\overrightarrow{F}=- 4.58\times10^{-14}\widehat{K}N$

Explanation:

Vx = 1.6 x 10^6 m/s

Vy = 2.6 x 10^6 m/s

Bx = 0.024 T

By = - 0.14 T

charge of electron, q = - 1.6 x 10^-19 C

charge of proton, q = 1.6 x 10^-19 C

(a) Force on electron is given by

$\overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B})$

Substituting the values

$\overrightarrow{F}=-1.6\times10^{-19}{\left ( 1.6\times 10^{6}\widehat{i}+2.6 \times 10^{6}\widehat{j} \right )\times \left ( 0.024\widehat{i}-0.14\widehat{j} \right )}$

$\overrightarrow{F}=4.58\times10^{-14}\widehat{K}N$

(b) Force on proton is given by

$\overrightarrow{F}=q(\overrightarrow{V}\times \overrightarrow{B})$

Substituting the values

$\overrightarrow{F}=1.6\times10^{-19}{\left ( 1.6\times 10^{6}\widehat{i}+2.6 \times 10^{6}\widehat{j} \right )\times \left ( 0.024\widehat{i}-0.14\widehat{j} \right )}$

$\overrightarrow{F}=- 4.58\times10^{-14}\widehat{K}N$

7 0

2 years ago

Water is boiled at 300 kPa pressure in a pressure cooker. The cooker initially contains 3 kg of water. Once boiling started, it

Inessa [10]

Answer:

The average rate of energy transfer to the cooker is 1.80 kW.

Explanation:

Given that,

Pressure of boiled water = 300 kPa

Mass of water = 3 kg

Time = 30 min

Dryness friction of water = 0.5

Suppose, what is the average rate of energy transfer to the cooker?

We know that,

The specific enthalpy of evaporate at 300 kPa pressure

$h_{f}=561.47\ kJ/kg$

$h_{fg}=2163.8\ kJ/kg$

We need to calculate the enthalpy of water at initial state

$h_{1}=h_{f}$

$h_{1}=561.47\ kJ/kg$

We need to calculate the enthalpy of water at final state

Using formula of enthalpy

$h_{2}=h_{f}+xh_{fg}$

Put the value into the formula

$h_{2}=561.47+0.5\times2163.8$

$h_{2}=1643.37\ kJ/kg$

We need to calculate the rate of energy transfer to the cooker

Using formula of rate of energy

$Q=\dfrac{m(h_{2}-h_{1})}{t}$

Put the value into the formula

$Q=\dfrac{3\times(1643.37-561.47)}{30\times60}$

$Q=1.80\ kW$

Hence, The average rate of energy transfer to the cooker is 1.80 kW.

3 0

3 years ago

A disk with mass 1.64 kg and radius 0.61 meters is spinning counter-clockwise with an angular velocity of 17.6 rad/s. A rod of m

Masja [62]

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ = 5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁ + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6 + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

$I_{disk} = \frac{1}{2} mr^2\\\\I_{disk} = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk} = 0.305 \ kgm^2$

The moment of inertia of the rod about its center is calculated as follows;

$I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2$

The initial rotational kinetic energy of the disk and rod;

$K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i= \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \ \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J$

The final rotational kinetic energy of the disk-rod system is calculated as follows;

$K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J$

The loss in rotational kinetic energy due to the collision is calculated as follows;

$\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J$

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

8 0

2 years ago