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3 years ago

Match the organs to the functions they perform during digestion. passes food to esophagus

Physics

1 answer:

mixas84 [53]3 years ago

8 0

Answer:

the mouth I think

Because: the oesophagus is the throat and the mouth is how the food gets to the oesophagus. might be wrong because I'm not sure about the organ part but I'm pretty sure that's all it could be

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The main reason that most professional research telescopes are reflectors is that

GuDViN [60]

<h3><u>Answer;</u></h3>

Large mirrors are easier to build than large lenses.

<h3><u>Explanation;</u></h3>

<em><u>Reflector telescopes have a number of advantages as compared to refracting telescopes and other types of telescopes. </u></em>
<em><u>Reflector telescopes do not suffer from chromatic aberration because all wavelengths will reflect off the mirror in the same way. The support for the objective mirror is all along the back side so they can be made very large.</u></em>
Additionally, reflector telescopes are cheaper to make than refractors of the same size. Also since in reflector telescopes light is reflecting off the objective, rather than passing through it, only one side of the reflector telescope's objective needs to be perfect.

7 0

3 years ago

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

The table shows the decay in a 59 g sample of bismuth 212 overtime.

Vikki [24]

Answer: C

14.75g

Explanation:

Given that the half life time = 60.5s

Let No = initial mass = 59g

N = decayed mass

At time t = 0, No = 59g

At time t = 60.5s,

N = No/2 = 59/2

= 29.5g

At time t = 121

N = 29.5/2 = 14.75g

Therefore N = 14.75g

7 0

2 years ago

What two factors could the students change to investigate how to increase and decrease the magnetic force between the paperclip

scoundrel [369]

..............................................A

4 0

2 years ago

White light is shining on a green grape. The grape __.

I am Lyosha [343]

White light is shining on a green grape. The grape would absorb all colors except green and will reflect the green color around and into our eyes. That's why the grape is green. Hope this helped!

5 0

3 years ago

Read 2 more answers