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Mandarinka [93]
2 years ago
13

Silk flexibility would be considered a trait. TRUE OR FALSE

Physics
2 answers:
wel2 years ago
6 0
The answer would be true.

Explanation:
Although some silk in is more flexible than others, there are many varieties. There are high medium and low flexibility. Even though there are different levels, it is still considered a trait. Again, the answer in True.

Hope this helped you. I did some research on this before, so this should be correct. If you aren’t satisfied then you can ask me a question! Have a nice day.
:)
dangina [55]2 years ago
6 0

Answer:

True

Explanation:

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umka2103 [35]

Answer:

Scalar quantity can never be Negative. Because scalar has only magnitude not direction. And magnitude can't be negative.

Explanation:

4 0
3 years ago
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A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
DanielleElmas [232]

Answer:

\alpha =-2.2669642\times^{-10}rad/s^2

Explanation:

Angular acceleration is defined by \alpha =\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{\Delta t}

Angular velocity is related to the period by \omega=\frac{2\pi}{T}

Putting all together:

\alpha =\frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})

Taking our initial (i) point now and our final (f) point one year later, we would have:

\Delta t=1\ year=(365)(24)(60)(60)s=31536000
s

T_i=0.0786s

T_f=0.0786s+7.03\times10^{-6}s

So for our values we have:

\alpha =\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{2\pi}{31536000s}(\frac{1}{0.0786s+7.03\times10^{-6}s}-\frac{1}{0.0786s})=-2.2669642\times^{-10}rad/s^2

Where the minus sign indicates it is decelerating.

8 0
3 years ago
You are less likely to see a total solar eclipse than a total lunar eclipse because a. the moon’s shadow covers all of Earth dur
Bad White [126]
D. Because the moons shadow during a total lunar eclipse is tinnier than the earth.
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viktelen [127]

lol what is this question

5 0
2 years ago
A water main pipe of diameter 10 cm enters a house 2 m below ground. A smaller diameter pipe carries water to a faucet 5 m above
Lemur [1.5K]

Explanation:

Given that,

Diameter = 10 cm

Distance = 2 m

Speed v_{1}= 2\ m/s

Speed v_{2}=7\ m/s

Pressure in main pipe P_{1}=2\times10^{5}\ Pa

(I). We need to calculate the diameter

Using equation of continuity

Av_{1}=Av_{2}

\pi(\dfrac{d_{1}}{2})^2\times v_{1}=\pi(\dfrac{d_{2}}{2})^2\times v_{2}

(\dfrac{10}{2})^2\times2=(\dfrac{d_{2}}{2})^2\times7

d_{2}=\sqrt{\dfrac{25\times2\times4}{7}}

d_{2}=5.345\ cm

(II). We need to calculate the pressure the gauge pressure

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}

P_{2}=P_{1}+\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)-\rho g(h_{1}-h_{2})

P_{2}=2\times10^{5}+\dfrac{1}{2}\times1000(4-49)-1000\times 9.8\times(5)

P_{2}=1.28500\times10^{5}\ Pa

(III).  If it is possible to carry water to a faucet 17 m above ground,

Using Bernoulli equation

P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh=P_{3}+\dfrac{1}{2}\rho v_{3}^2+\rho g h_{3}

P_{3}=P_{1}+\dfrac{1}{2}\rho v_{1}^2-\rho g(h_{1}-h_{3})

Here, h_{3}=0

Put the value in the equation

P_{3}=2\times10^{5}+\dfrac{1}{2}\times1000\times4-1000\times 9.8\times17

P_{3}=3.5400\times10^{5}\ Pa

Hence, This is required solution.

7 0
3 years ago
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