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Karo-lina-s [1.5K]
2 years ago
5

To navigate, a porpoise emits a sound wave that has a wavelength of 2.2 cm. The speed at which the wave travels in seawater is 1

522 m/s. Find the period of the wave.
Physics
1 answer:
Alenkasestr [34]2 years ago
3 0

To solve this problem we will apply the concept related to frequency, by which it is defined as the relationship between the speed of the wave and the wavelength. Since the frequency is the inverse of the period, this relationship will also be reversed. That is to say,

f = \frac{v}{\lambda}

Here,

v = Velocity

\lambda = Wavelength

Now the period is defined as,

T = \frac{1}{f}

Therefore

T = \frac{\lambda}{v}

Replacing we have that,

T = \frac{2.2*10^{-2}}{1522}

T = 1.445*10^{-5}s

Therefore the period of the wave is 1.445*10^{-5}s

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A student is flying west on a school trip from Winnipeg to Calgary in a jet that has an air velocity of 792 km/h.The direction t
Akimi4 [234]

Answer:

The direction the plane would have to fly to compensate for a wind velocity of 62.0 km/h[N] is 4.5° S of W

Explanation:

The given parameters are;

Velocity of Jet = 792 km/h

Direction of jet velocity = West

Velocity of wind = 62.0 km/h

Direction of wind velocity = North

Therefore, the jet has to have a component of 62.0 km/h South of West to compensate for the wind velocity

The direction of the plane, θ° South of West (S of W) to compensate for the wind is given as follows;

Tan \left (\theta   \right )= \dfrac{62}{792} = \dfrac{31}{396}

Therefore;

\theta = tan^{-1}\left (\dfrac{31}{396}   \right ) = 4.476^{\circ} \approx 4.5^{\circ}

The direction the plane would have to fly to compensate for a wind velocity of 62.0 km/h[N] = 4.5° S of W.

6 0
2 years ago
Which correctly identifies the parts of a transverse wave? A: crest B: amplitude C: wavelength D: trough A: trough B: amplitude
jenyasd209 [6]

Explanation :

In transverse waves the particles are oscillating perpendicular to the direction of propagation of waves.

The uppermost part of the wave is crests and the lowermost part is troughs.

Wavelength of a transverse wave is defined as the distance between two consecutive crests or troughs.

Amplitude is the maximum distance or displacement covered by a wave.

So, crest, amplitude, trough and wavelength identifies the parts of a transverse wave.

9 0
1 year ago
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The inner and outer surfaces of a 5m x 6m brick wall of thickness 30 cm and thermal conductivity 0.69 w/m.0 c are maintained at
Volgvan
The working equation to be used here is written below:

Q = kA(T₁ - T₂)/Δx
where
Q is the rate of heat transfer
k is the heat transfer coefficient
A is the cross-sectional area of the wall
T₁ - T₂ is the temperature difference between the sides of the wall
Δx is the thickness of the wall

The solution is as follows:

Q = (0.69 W/m²·°C)(5 m × 6 m)(50°C - 20°C)/(30 cm * 1 m/100 cm)
Q = 2,070 W/m

4 0
1 year ago
Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum
Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

\frac{1}{2}mv^2 = mgh

where

m is the mass of the man

v is the speed after jumping

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s

In the acceleration phase, we know that the initial velocity is

u=0

And the force exerted on the floor is 2.3 times the gravitational force, so

F=2.3 mg

This means the net force on you is

F_{net} = F-mg=2.3mg-mg=1.3 mg

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g

Now we can use the  following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

v^2-u^2=2as

where s is the quantity we want to find. Solving for s,

s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

\Delta h = -0.77 m

This means that the lowest point reached by the center of mass above the floor during the crouch is

h'=h+\Delta h = 1.0 - 0.77 = 0.23 m

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0
1 year ago
A 480-kilogram horse runs across a field at a rate of 40 km/hr. What is the magnitude of the horse's momentum?
Andrej [43]
Momentum (p) = mass × velocity

so, 480×40 = 19,200 kg km/hr

so the answer is C !!
4 0
2 years ago
Read 2 more answers
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