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4 years ago

A mechanical wave is created when a medium ____, or moves back and forth or up and down

Physics

2 answers:

Zielflug [23.3K]4 years ago

6 0

I think it is traverse wave

spin [16.1K]4 years ago

4 0

Answer: The correct word for the blank is 'oscillates'

" a mechanical wave is created when a medium <em>oscillates</em> , or moves back and forth or up and down

Explanation:

When the matter oscillates or vibrates it transfers energy through the medium in the form of wave which termed s mechanical wave.For example: Sound wave,slinky waves, water waves are the examples of mechanical waves.

So, " a mechanical wave is created when a medium <em>oscillates</em> , or moves back and forth or up and down

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How are the raincoats waterproof? explain

Reil [10]

Answer:

Rain jackets are coated in durable water-repellant (DWR) finish, a hydrophobic glaze that allows the coats to be breathable, yet waterproof. It lets water vapor—like sweat—out, but keeps rain from getting in. ... Without a DWR, the raincoat is just a coat. So, you'll need to recoat it.

Explanation:

This is according to the web

7 0

3 years ago

What is the instantaneous acceleration of the particle at point B?

Vikentia [17]

Acceleration is a measure of how fast the velocity of a particle/object is changing at a point. Therefore, acceleration can be measured by finding the slope of the velocity of the particle at a certain point.

Since we are given the graph of velocity, we have to find the slope of the particle at point B. In the given graph, the slope of the velocity of the particle at point B is zero. Therefore, the acceleration of the particle at point B is 0.

Hope this helps!

7 0

3 years ago

The edge of a flying disc with a radius of 0.13 m spins with a tangential speed of 3.3 m/s.

Aleks [24]

By definition, centripetal acceleration is given by:

$a = \frac{v ^ 2}{r}$

Where,

v: tangential disk speed

r: disk radius

Substituting values in the given equation we have:

$a =\frac{3.3^2}{0.13}\\a = 83.76923077$

Rounding the result we have:

$a = 83.8 \frac{m}{s^2}$

Answer:

The centripetal acceleration of the disc edge in m/s^2 is:

$a = 83.8 \frac{m}{s^2}$

3 0

4 years ago

Read 2 more answers

A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the sa

aleksandr82 [10.1K]

Answer: The value of the celsius temperature of the cube is 472.2°c.

Explanation:

The expression for the power radiated is as follows;

$P=A\epsilon\sigma T^{4}$

Here, A is the area, $\sigma$ is the stefan's constant, $\epsilon$ is the emissivity and T is the temperature.

It is given in the problem that A sphere is originally at a temperature of 500°c. The sphere is melted and recast, without loss of mass, into a cube with the same emissivity as the sphere.

Then the expression for the radiated power for the cube and the sphere can be expressed as;

$A_{1}\epsilon \e\sigma T_{1}^{4}=A_{2}\epsilon \e\sigma T_{2}^{4}$

Here, $A_{1}$ is the area of the sphere, $A_{2}$ is the area of the cube, $T_{1}$ is the temperature of the sphere and $T_{2}$ is the temperature of the cube.

The radiated powers and emissivity of the cube and the sphere are same.

$A_{1}T_{1}^{4}=A_{2}T_{2}^{4}$

The area of the sphere is $A_{1}=4\pi \times r^{2}$ .

Here, r is the radius of the sphere.

The area of the cube is $A_{2}=6\times a^{2}$ .

Here, a is the edge of the cube.

Put $A_{1}=4\pi \times r^{2}$ and $A_{2}=6\times a^{2}$ .

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ ....(1)

The masses and the densities of the sphere and the cube are same. Then the volumes are also same.

$V_{1}=V_{2}$

Here, $V_{1},V_{1}$ are the volumes of the sphere and the cube.

$\frac{4}{3}\pi r^{3}=a^{3}$

$\frac{r}{a}=(\frac{3}{4\pi })^{\frac{1}{3}}$

Put this value in the equation (1).

$T_{2}=T_{1}(\frac{2\pi }{3}\times (\frac{r}{a})^{2})^{\frac{1}{4}}$ $T_{2}=T_{1}(\frac{2\pi }{3}\times ((\frac{3}{4\pi })^{\frac{1}{3}})^{2})^{\frac{1}{4}}$

Put T_{1}=500°c.

$T_{2}=(500)(\frac{2\pi }{3}\times (\frac{3}{4\pi })^{\frac{2}{3}})^{\frac{1}{4}}$

$T_{2}=472.2^{\circ}c$

Therefore, the value of the celsius temperature of the cube is 472.7°c.

5 0

3 years ago

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force exerted by the t

vodomira [7]

Complete Question:

As a fish jumps vertically out of the water, assume that only two significant forces act on it: an upward force F exerted by the tail fin and the downward force due to gravity. A record Chinook salmon has a length of 1.50 m and a mass of 48.0 kg. If this fish is moving upward at 3.00 m/s as its head first breaks the surface and has an upward speed of 6.30 m/s after two-thirds of its length has left the surface, assume constant acceleration and determine the following. (a) the salmon's acceleration m/s2 upward (b) the magnitude of the force F during this interval N

Answer:

(a) acceleration = 15.3 ms⁻²

(b) Magnitude of net force = 734.4 N

Magnitude of upward force exerted by tail fin = 1204.8 N

Explanation:

Mass of the salmon fish = 48 kg

Length of the Salmon Fish = 1.5 m

g = 9.8 ms⁻²

(a) salmon's acceleration during the time interval N:

Downward Force on the fish is equal to the Force due to gravity and is given as:

F₂ = mg

= 48 * 9.8

= 470.4 N

The direction of movement of the fish is upward and the acceleration is constant. We are given two different velocities of fish at two different instances.

- When the head breaks out of the water surface first:

Initial velocity = v₁ = 3 m/s

- When two third of its body length is out = d = 1 m

Final Velocity = v₂ = 6.3 m/s

Using the third equation of motion:

2*a*d = v₂² - v₁²

a = (6.3² - 3²)/2*1

a = 15.3 ms⁻²

(b) magnitude of force F during this interval N = ?

We are assuming that F is the net force consisting of both the upward and the downward force.

According to Newton's 2nd law of motion, Force is given as:

F = ma

F = 48 kg * 15.3

F = 734.4 N

Magnitude of upward Force = Fₓ

Force Fₓ exerted by the tail fin of the fish is given by

F = Fₓ - F₂

That is the net force is the sum of the upward and downward forces acting on the fish body. Fₓ is positive because it is in upward direction and F₂ is negative because it is in downward direction. F which is the net force here is positive as Fₓ > F₂.

=> Fₓ = F + F₂

Fₓ = 734.4 + 470.4

Fₓ = 1204.8 N

7 0

4 years ago