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ivolga24 [154]
2 years ago
14

Consider a wave along the length of a stretched slinky toy, where the distance between coils increases and decreases. What type

of wave is this
Physics
2 answers:
Ghella [55]2 years ago
7 0

Answer:

The waves are longitudinal.

Explanation:

There are two types of waves:

Longitudinal waves : The waves in which the particles of medium vibrates in the same direction a the wave travels. Ex: sound waves  

Transverse waves : The waves in which the particles of medium in the perpendicular direction of the wave propagation. Ex: light waves.

The wave is travelling along the length of the slinky toy. So, the direction of motion of particles is same as the direction of motion of wave.

So, the waves are longitudinal in nature.

GrogVix [38]2 years ago
3 0

Answer:

"Longitudinal wave" is the appropriate answer.

Explanation:

  • Generating waves whenever the form of communication being displaced in a similar direction as well as in the reverse way of the wave's designated points, could be determined as Longitudinal waves.
  • A wave running the length of something like a Slinky stuffed animal, which expands as well as reduces the spacing across spindles, produces a fine image or graphic.
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A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
Tasya [4]
<h3><u>Answer;</u></h3>

<em>Work = 125 joules </em>

<h3><u>Explanation and solution</u>;</h3>
  • Work is the product of force and the distance covered. Therefore, Work = force × distance.
  • Work is measured in joules.
  • Work is also a change in energy, such that work is done when energy changes, so when kinetic energy, or potential energy changes the there is work being done.

Thus; kinetic energy = work done

Kinetic energy = 1/2mv²

                       = 1/2 × 10× 5²

                       = 5 × 25

                       = 125 joules

Hence, work done is 125 joules.

3 0
2 years ago
g A projectile is launched with speed v0 from point A. Determine the launch angle ! which results in the maximum range R up the
Svetlanka [38]

Answer:

The range is maximum when the angle of projection is 45 degree.

Explanation:

The formula for the horizontal range of the projectile is given by

R = \frac{u^{2}Sin2\theta }{g}

The range should be maximum if the value of Sin2θ is maximum.

The maximum value of Sin2θ is 1.

It means 2θ = 90

θ = 45

Thus, the range is maximum when the angle of projection is 45 degree.

If the angle of projection is 0 degree

R = 0

It means the horizontal distance covered by the projectile is zero, it can move in vertical direction.

If the angle of projection is 30 degree.

R = \frac{u^{2}Sin60 }{9.8}

R = 0.088u^2

If the angle of projection is 45 degree.

R = \frac{u^{2}Sin90 }{g}

R = u^2 / g

5 0
2 years ago
In ideal flow, a liquid of density 850 kg/m3 moves from a horizontal tube of radius 1.00 cm into a second horizontal tube of rad
Crank

Answer:

a)   Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1] , b) Q = 3.4 10⁻² m³ / s , c)      Q = 4.8 10⁻² m³ / s

Explanation:

We can solve this fluid problem with Bernoulli's equation.

         P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

With the two tubes they are at the same height y₁ = y₂

        P₁-P₂ = ½ ρ (v₂² - v₁²)

The flow rate is given by

         A₁ v₁ = A₂ v₂

         v₂ = v₁ A₁ / A₂

We replace

         ΔP = ½ ρ [(v₁ A₁ / A₂)² - v₁²]

         ΔP = ½ ρ v₁² [(A₁ / A₂)² -1]

Let's clear the speed

         v₁ = √ 2ΔP /ρ[(A₁ / A₂)² -1]

The expression for the flow is

           Q = A v

           Q = A₁ v₁

           Q = A₁ √ 2ΔP / rho [(A₁ / A₂)² -1]

The areas are

            A₁ = π r₁

            A₂ = π r₂

We replace

        Q = π r₁ √ 2ΔP / rho [r₁² / r₂² -1]

Let's calculate for the different pressures

      r₁ = d₁ / 2 = 1.00 / 2

      r₁ = 0.500 10⁻² m

      r₂ = 0.250 10⁻² m

b) ΔP = 6.00 kPa = 6 10³ Pa

      Q = π 0.5 10⁻² √(2 6.00 10³ / (850 (0.5² / 0.25² -1))

       Q = 1.57 10⁻² √(12 10³/2550)

        Q = 3.4 10⁻² m³ / s

c) ΔP = 12 10³ Pa

        Q = 1.57 10⁻² √(2 12 10³ / (850 3)

         Q = 4.8 10⁻² m³ / s

5 0
2 years ago
Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for
Burka [1]

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

3 0
2 years ago
The number of gallons of water in a storage tank at time t, in minutes, is modeled by w(t) = 25 − t2 for 0≤t≤5. At what rate, in
Kisachek [45]

Answer:

Rate of change of water will be -6 gallon/minute

Explanation:

We have given water in the tank as the function of time as

w(t)=25-t^2

We have to find the rate of change of water in the tank at t = 3 minute

For rate of change we have to differentiate both side

So \frac{dw}{dt}=0-2t

At t = 3 minute

\frac{dw}{dt}=0-2\times 3=-6gallon/minute

8 0
2 years ago
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