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3 years ago

Plz help

Physics

1 answer:

katrin2010 [14]3 years ago

4 0

The particles of the medium (slinky in this case) move up and down (choice #2) in a transverse wave scenario.

This is the defining characteristic of transverse waves, like particles on the surface of water while a wave travels on it, or like particles in a slack rope when someone sends a wave through by giving it a jolt.

The other kind of waves is longitudinal, where the particles of the medium move "left-and-right" along the direction of the wave propagation. In the case of the slinky, this would be achieved by giving a tensioned slinky an "inward" jolt. You would see that such a jolt would give rise to a longitudinal wave traveling along the length of the tensioned slinky. Another example of longitudinal waves are sound waves.

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A machinist turns on the power on to a grinding wheel at time t= 0 s. The wheel accelerates uniformly from rest for 10 s and rea

Ugo [173]

Answer:

θt = 514.3 revolutions

Explanation:

(1)The wheel accelerates uniformly from rest for 10 s and reaches the operating angular speed of 58rad/s.

The uniformly accelerated circular movement a circular path movement in which the angular acceleration is constant.

We apply the equations of circular motion uniformly accelerated

ωf = ω₀ + α*t Formula (1)

θ = ω₀*t + (1/2)*α*t² Formula (2)

ωf² = ω₀² +2*α*θ Formula (3)

Where:

θ : angle that the body has rotated in a given time interval (rad)

α : angular acceleration (rad/s²)

t : time interval (s)

ω₀ : initial angular speed ( rad/s)

ωf : final angular speed ( rad/s)

Number of revolutions made by the wheel from t = 0 to t = 10 s

Data

ω₀ = 0

t = 10 s

ωf = 58 rad/s

We replace data in the formula (1) to calculate α

ωf = ω₀ + α*t

58 = 0 + α*(10)

α = 58 /10

α = 5.8 rad/s²

We replace data in the formula (2) to calculate θ

θ = ω₀*t + (1/2)*α*t²

θ = 0 + (1/2)*( 5.8)*(10)²

θ₁ = 290 rad

(2)The wheel is run at that angular velocity for 30 s, and then power is shut off.

The movement of the wheel is circular with constant angular speed and the formula to calculate θ is:

θ = ω*t

ω = 58 rad/s , t= 30s

θ = (58 rad/s)*(30)

θ ₂= 1740 rad

(3)The wheel slows down uniformly at 1.4 rad/s² until the wheel stops.

ω₀ = 58 rad/s

α = -1.4 rad/s²

ωf = 0

We replace data in the formula (3) to calculate θ

(ωf)² = (ω₀)² + (2)*(α )*θ

0 = (58)² + (2)*(-1.4)*θ

(2)*(1.4)*θ = (58)²

θ = (58)² / (2.8)

θ₃ = 1201.42 rad

Total number of revolutions made by the wheel (θt)

θt =θ₁+θ₂+θ₃

θt = 290 rad+ 1740 rad + 1201.42 rad

θt = 3231.42 rad

1 revolution = 2π rad

θt = 3231.42 rad* ( 1revolution/2π rad)

θt = 514.3 revolutions

7 0

3 years ago

Why is it so difficult for astronomers to see new stars in the process of birth?

vekshin1

Birth happens very quickly, so it is hard to "catch" stars in the act. - most stars are born inside dusty clouds, which block any light that may be coming from the stars.

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1 year ago

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3 years ago

B. Projectile on cliff (range)

dimulka [17.4K]

Answer:

x = 41.28 m

Explanation:

This is a projectile launching exercise, let's find the time it takes to get to the base of the cliff.

Let's start by using trigonometry to find the initial velocity

cos 25 = v₀ₓ / v₀

sin 25 = Iv_{oy} / v₀

v₀ₓ = v₀ cos 25

v_{oy} = v₀ sin 25

v₀ₓ = 22 cos 25 = 19.94 m / s

v_{oy} = 22 sin 25 = 0.0192 m / s

let's use movement on the vertical axis

y = y₀ + v_{oy} t - ½ g t²

when reaching the base of the cliff y = 0 and the initial height is y₀ = 21 m

0 = 21 + 0.0192 t - ½ 9.81 t²

4.905 t² - 0.0192 t - 21 = 0

t² - 0.003914 t - 4.2813 =0

we solve the quadratic equation

t = $\frac{ 0.003914\ \pm \sqrt{0.003914^2 + 4 \ 4.2813 } }{2}$

t = $\frac{0.003914 \ \pm 4.13828}{2}$

t₁ = 2.07 s

t₂ = -2.067 s

since time must be a positive scalar quantity, the correct result is

t = 2.07 s

now we can look up the distance traveled

x = v₀ₓ t

x = 19.94 2.07

x = 41.28 m

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3 years ago

is it possible to breed a black haired long tailed retriever? If so, what percentage is possible? PLEASE answer the corrects ans

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Answer: it can be considered a genetic mutation with a history of a Golden Retriever in their blood but it is very rare. and there our some black retrievers you can buy too. i hope i helped

Explanation:

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2 years ago

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