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2 years ago

The height of a transverse waves is known as its?

Physics

1 answer:

allochka39001 [22]2 years ago

4 0

Answer:

amplitude

Explanation:

wave is a called the crest, and the low point is called the trough. For longitudinal waves, the compressions and rarefactions are analogous to the crests and troughs of transverse waves. The distance between successive crests or troughs is called the wavelength. The height of a wave is the amplitude.…

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A golf ball has a mass of 46 grams, and when struck by a driver reaches a speed of 42 m/s in a time of 0.00050 seconds. what was

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Impulse in physics is the integral of force, F, with respect with time, t. This value is a vector quantity since force is a vector quantity as well. It can be calculated from the product of force and time. We do as follows:

Impulse = Ft
= m(a)(t)
= m(v/t)t
= 0.046 (42/0.0005) (0.0005)
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2 years ago

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3 years ago

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Water flows through a 4.50-cm inside diameter pipe with a speed of 12.5 m/s. At a later position, the pipe has a 6.25-cm inside

jek_recluse [69]

Given,

The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m

The initial speed of the water, v₁=12.5 m/s

The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m

From the continuity equation,

$\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}$

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

$\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}$

Thus, the flow rate of the water at the later position is 5.99 m/s

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8 months ago

Active listening includes all of the following EXCEPT: A. paraphrasing B. clarifying C. ignoring D. empathizing Please select th

scoray [572]

Answer:

C: Ignoring

Explanation:

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2 years ago

How fast would the International Space Station (ISS) have to travel to maintain a circular orbit a distance of 1400 km above the

professor190 [17]

Answer:

The International Space Station move at 7.22 km/s.

Explanation:

Orbital speed of satellite is given by $v=\sqrt{\frac{GM}{r}}$ , where G is gravitational constant, M is mass of Earth and r is the distance to satellite from centre of Earth.

r = R + h = 6350 + 1400 = 7750 km = 7.75 x 10⁶ m

G = 6.673 x 10⁻¹¹ Nm²/kg²

M = 5.98 x 10²⁴ kg

Substituting

$v=\sqrt{\frac{6.673\times 10^{-11}\times 5.98\times 10^{24}}{7.75\times 10^6}}=7223.86m/s=7.22km/s$

The International Space Station move at 7.22 km/s.

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2 years ago