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notka56 [123]
4 years ago
9

Which statements about tornadoes are true? Check all that apply.

Physics
2 answers:
fgiga [73]4 years ago
8 0

Answer:

The answer is B,C,D

Explanation:

I took the test

OverLord2011 [107]4 years ago
5 0

(B)

(C)

(D)

Trust me i got it right on edge 2020

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Pls help waves for physics
djverab [1.8K]

Answer:

b) 252 Hz or 260 Hz

c) 0.25 s

Explanation:

b) The frequency of the beats is 4 Hz, and one tuning fork has a frequency of 256 Hz.  Therefore, the second tuning fork is either 4 Hz lower or 4 Hz higher.

f = 252 Hz or 260 Hz

c) Period is the inverse of frequency.

T = 1/f

T = 1 / (4 Hz)

T = 0.25 s

8 0
3 years ago
Learning Goal: To practice Problem-Solving Strategy 19.1 Work in Ideal-gas Processes. A cylinder with initial volume VVV contain
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Answer:

The work done on the gas is equal to the area under the curve pv diagram w = area of triangle = 1/2 (base)(height) = 1/2 (BC)(Ac) = 1/2 (3v - v)(3p - p) = 1/2 (9 vp - 3 vp - 3vp + vp) = 4 vp/2 W = 2 vp

Check attachment for the diagrammatic representation

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4 years ago
2 b (ignore what I already completed)
posledela

Answer:

The answer for this question is it talks about the circuit containing the led, switch and all that

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3 years ago
a student compared two samples of matter. he recorded the results in the chart. Sample A: Appearance: shiny and yellow. Sample B
Vladimir [108]

Answer:

The answer are B

Explanation:the two samples haves yellow color :)

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Read 2 more answers
Titania, the largest moon of the planet Uranus, has 1/8 the radius of the earth and 1/1700 the mass of the earth.What is the acc
garri49 [273]

Answer:

(a). The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). The average density of Titania is 1656.47 kg/m³

Explanation:

Given that,

Radius of Titania R_{t}= \dfrac{1}{8}R_{e}

Mass of Titania M_{t}= \dfrac{1}{1700}M_{e}

We need to calculate the acceleration due to gravity at the surface of Titania

Using formula of the acceleration due to gravity on earth

g_{e}=\dfrac{GM_{e}}{R_{e}^2}

The acceleration due to gravity on Titania

g_{t}=\dfrac{GM_{t}}{R_{t}^2}

Put the value into the formula

g_{t}=\dfrac{G\times\dfrac{1}{1700}M_{e}}{(\dfrac{1}{8}R_{e})^2}

g_{t}=\dfrac{64}{1700}\times G\dfrac{M_{e}}{R_{e}^2}

g_{t}=0.004705 g_{e}

g_{t}=0.03764\times9.8

g_{t}=0.37\ m/s^2

The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). We assume Titania is a sphere

The average density of the earth is 5500 kg/m³.

We need to calculate the average density of Titania

Using formula of density

\rho=\dfrac{m}{V}

\rho_{t}=\dfrac{M_{t}}{\dfrac{4}{3}\pi\times R_{t}^2}

\rho_{t}=\dfrac{\dfrac{M_{e}}{1700}}{\dfrac{4}{3}\pi\times(\dfrac{R_{e}}{8})^2}

\rho_{t}=\dfrac{512}{1700}\times\rho_{e}

\rho_{t}=\dfrac{512}{1700}\times5500

\rho_{t}=1656.47\ kg/m^{3}

The average density of Titania is 1656.47 kg/m³

Hence, (a). The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). The average density of Titania is 1656.47 kg/m³

8 0
3 years ago
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