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4vir4ik [10]
3 years ago
7

Somebody help please

Physics
2 answers:
Salsk061 [2.6K]3 years ago
7 0

Answer:

7

Explanation:

its 7

DerKrebs [107]3 years ago
5 0

Explanation:

I think it's c but not sure

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The __________ lobe contains the area of the cortex involved in auditory processing called the primary __________ cortex. A. par
Vlada [557]
<span>The temporal lobe contains the area of the cortex involved in auditory processing called the primary auditory cortex.</span><span> 

The temporal </span>lobe<span> is associated with </span>auditory processing<span> and olfaction.
</span>The primary auditory cortex<span> is the </span>part<span> of the temporal </span>lobe which<span> processes </span>auditory <span>information.</span>
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Which part of the larynx is the opening between the vocal folds where vibration occurs?
Lynna [10]

Glottis I believe is the answer.

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A mouse runs along a baseboard in your house. The mouse's position as a function of time is given by x(t)=pt 2+qt, with p = 0.36
Semmy [17]

Answer: the average speed of the rat from the information given above is 0.7m/s

Explanation:

position is given as

x(t) = pt² + qt

finding the diffencial of x(t) with respect to t, we have

d(x(t))/dt = 2pt + q

we substitute the p = 0.36m/s² and q= -1.10 m/s

d(x(t))/dt = 2(0.36)t + (-1.10)

so, at t= 1s

d(x(t))/dt = 2*(0.36)*1 - 1.1 = 0.72 - 1.1 = -0.38m/s

at t= 4s

d(x(t))/dt = 2*(0.36)*4 - 1.10 = 2.88 - 1.10 = 1.78 m/s

To find the average speed,

average speed = (V1 + V2)/ 2

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3 years ago
How can a dictatorship best be classified
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Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
padilas [110]

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

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3 years ago
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