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4 years ago

9

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

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

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3 years ago

Learning Goal: To practice Problem-Solving Strategy 19.1 Work in Ideal-gas Processes. A cylinder with initial volume VVV contain

Leokris [45]

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 :)

6 0

3 years ago

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