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2 years ago

11. The primary job of the cell wall is to

Physics

2 answers:

Lina20 [59]2 years ago

6 0

Answer:

11. A. Protect the cell and keep its shape

13. C. Chloroplast

Hope this helps :)

hodyreva [135]2 years ago

3 0

11. protect the cell and keep its shape.

12. chloroplast

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The temperature of water in a beaker is 45°C. What does this measurement represent

andrezito [222]

Based on my information, this would actually be representing "the average kinetic energy of water particles". So, as you take notice that where this temperature is being located, and also, how this would be $45$ °C, this would make more sense for this to be representing as <span>the average kinetic energy of water particles.</span>

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3 years ago

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A typical running track is an oval with 74-mm-diameter half circles at each end. A runner going once around the track covers a d

lisabon 2012 [21]

The centripetal acceleration a is 4.32 $\times$ 10^-4 m/s^2.

<u>Explanation:</u>

The speed is constant and computing the speed from the distance and time for one full lap.

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Computing the v = 0.4 m / 100 s

v = 4 $\times$ 10^-3 m/s.

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= (4 $\times$ 10^-3)^2 / 0.037

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3 years ago

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Nat2105 [25]

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2 years ago

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A centrifuge in a medical laboratory rotates at an angular speed of 3,500 rev/min, clockwise (when viewed from above). When swit

Ratling [72]

Answer:

The magnitude of angular acceleration is $232.38\ rad/s^2$ .

Explanation:

Given that,

Initial angular velocity, $\omega_i=3500\ rev/min=366.5\ rad/s$

When it switched off, it comes o rest, $\omega_f=0$

Number of revolution, $\theta=46=289.02\ rad$

We need to find the magnitude of angular acceleration. It can be calculated using third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta\\\\\alpha =\dfrac{-\omega_i^2}{2\theta}\\\\\alpha =\dfrac{-(366.51)^2}{2\times 289.02}\\\\\alpha =-232.38\ rad/s^2$

So, the magnitude of angular acceleration is $232.38\ rad/s^2$ . Hence, this is the required solution.

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3 years ago

In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21

Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

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$z = 20km$