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3 years ago

Which of the following is an important ethical behavior for scientists?

Physics

2 answers:

frozen [14]3 years ago

6 0

In my opinion is <span>B. Logic</span>

DedPeter [7]3 years ago

4 0

B. Logic.

They want an explanation for everything.

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Two go carts, A and B race eachother around a 1.0 km track. Go cart A travels at a constant speed of 20.0 m/s. Go cart B acceler

jarptica [38.1K]

We have that Since Cart A spends t=77.5secs and Cart B spends t=50sec

Therefore

The Cart A wins the Race and by 25.5( 77.7-50)secs

From the question we are told

Two go carts, A and B race each other around a 1.0 km track.
Go cart A travels at a constant speed of 20.0 m/s.
Go cart B accelerates uniformly from rest at a rate of 0.333 m/s

For Cart A

Generally the equation for the Velocity is mathematically given as

$v=\frac{d}{t}\\\\t=\frac{d}{v}\\\\t=\frac{1000}{20}\\\\t=50sec$

For Cart B

Generally the Newtons equation for the Motion is mathematically given as

$S=ut+1/2at^2\\\\Therefore\\\\S=ut+1/2at^2\\\\1000=0+1/2*(0.33)t^2\\\\t=\frac{1000}{1/2*(0.33}}\\\\t=77.5secs$

Since

Cart A spends t=77.5secs and Cart B spends t=50sec

Therefore

The Cart A wins the Race and by 25.5( 77.7-50)secs

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brainly.com/question/12319416?referrer=searchResults

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3 years ago

A potter's wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/srev/s . The wheel can be consi

vagabundo [1.1K]

Answer:1.7 rev/s

Explanation:

Given

Frequency of wheel $N_1=2\ rev/s$

angular speed $\omega_1=2\pi N_1=4\pi\ rad/s$

mass of wheel $m_1=4.5\ kg$

diameter of wheel $d_1=0.30\ m=30\ cm$

radius of wheel $r_1=\frac{d_1}{2}=\frac{30}{2}=15\ cm$

mass of clay $m_2=2.8\ kg$

the radius of the chunk of clay $r_2=8\ cm$

Moment of inertia of Wheel

$I_1=\dfrac{m_1r_1^2}{2}=\dfrac{4.5\times 15^2}{2}\ kg-cm^2$

Combined moment of inertia of wheel and clay chunk

$I_2=\dfrac{m_1r_1^2}{2}+\dfrac{m_2r_2^2}{2}=\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2}\ kg-cm^2$

Conserving angular momentum

$\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \dfrac{4.5\times 15^2}{2}\cdot 4\pi=(\dfrac{4.5\times 15^2}{2}+\dfrac{2.8\times 8^2}{2})\omega_2\\\\\Rightarrow \omega _2=\dfrac{4\pi }{1+\dfrac{2.8}{4.5}\times (\dfrac{8}{15})^2}=\dfrac{4\pi}{1+0.1769}=0.849\times 4\pi$