If a wave has a length of 1.2 m and a speed of 2000 m/s, what is the frequency?

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity o

MariettaO [177]

Complete question:

while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?

Answer:

The frequency reflected by the stationary wall to the bat is 41 kHz

Explanation:

Given;

frequency emitted by the bat, = 39 kHz

velocity of the bat, $v_b$ = 8.32 m/s

speed of sound in air, v = 340 m/s

The apparent frequency of sound striking the wall is calculated as;

$f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz$

The frequency reflected by the stationary wall to the bat is calculated as;

$f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz$

$f_s\approx 41 \ kHz$

3 0

3 years ago

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed

gregori [183]

Answer:

the rate of the change of the length of the shadow is - 0.8625 m/s.

The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.

Explanation:

Given the data in the question;

Let x represent the man's distance from building,

initially x = 1m2

dx/d t= -2.3 m/s

Also Let y represent shadow height

so we determine dy/dt when x is 4m from the building

form the image description of the problem, we see two-like triangles with the same base and height ratios

so

2 / (12-x) = y / 12

24 = y(12 - x )

y = 24 / (12-x)

dy/dt = 24/(12-x)² × dx/dt

Now at x = 4,

we substitute

dy/dt will be;

⇒ 24/(12 - 4)² × -2.3

= 24/64 - 2.3

= 0.375 × -2.3

dy/dt = - 0.8625 m/s

Therefore, the rate of the change of the length of the shadow is - 0.8625 m/s.

The negative(-) sign means the length of the shadow decreases at a rate of 0.8625 m/s.

5 0

3 years ago

what is the density of a substance that has a mass of 2.0 g , and when placed in a graduated cylinder the volume changed from 70

lubasha [3.4K]

A material with a mass of 2.0 g when placed in a graduated cylinder the volume changed from 70 ml to 75 ml has a density of 0.4 g/mL.

How do I calculate the substance's density?

We'll start by getting the substance's volume. This is attainable as follows:

Water volume: 70 mL

75 mL = volume of material + water.

Substance volume =?

Substance volume equals (substance volume plus water) - (Volume of water)

Substance volume = 75 - 70

5 mL is the substance's volume.

Finally, we will calculate the substance's density. Below is an example to help:

2.0 g is the substance's mass.

5 mL is the substance's volume.

Substance density =?

Mass / volume equals density.

Substance density = 2/5

0.4 g/mL is the substance's density.

The density is therefore 0.4 g/mL.

To know more about density, visit:

brainly.com/question/952755

#SPJ4

6 0

1 year ago

Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used

natali 33 [55]

Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is $T =8391.6 N$

Explanation:

From the question we are told that

The strain on the strain on the head is $\Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m$

The contact area is $A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2$

Looking at the first diagram

At 600 MPa of stress

The strain is $0.3mm/mm$

At 450 MPa of stress

The strain is $0.0015 mm/mm$

To find the stress at $\Delta l$ we use the interpolation method

$\frac{\sigma_{\Delta l} - \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l } - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}$

Substituting values

$\frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}$

$\sigma _{\Delta l} -450 = 49.50$

$\sigma _{\Delta l} = 499.50 MPa$

Generally the force on each head is mathematically represented as

$F = \sigma_{\Delta l} * A$

Substituting values

$F = 499.50*10^{6} * 2.8*10^{-6}$

$=1398.6N$

Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as

$T = 6 * F$

$= 6* 1398.6$

$T =8391.6 N$

6 0

3 years ago

Consider the system consisting of the box and the spring, but not Earth. How does the energy of the system when the spring is fu

BabaBlast [244]

Answer:

the energy when it reaches the ground is equal to the energy when the spring is compressed.

Explanation:

For this comparison let's use the conservation of energy theorem.

Starting point. Compressed spring

Em₀ = K_e = ½ k x²

Final point. When the box hits the ground

Em_f = K = ½ m v²

since friction is zero, energy is conserved

Em₀ = Em_f

1 / 2k x² = ½ m v²

v = $\sqrt{ \frac{k}{m} }$ x

Therefore, the energy when it reaches the ground is equal to the energy when the spring is compressed.

5 0

3 years ago