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Finger [1]
3 years ago
6

Find the linear speed of the bottom of a test tube in a centrifuge if the centripetal acceleration there is 5.4×104 times the ac

celeration of gravity. The distance from the axis of rotation to the bottom of the test tube is 7.2 cm .
Engineering
1 answer:
Luda [366]3 years ago
4 0

Answer: v= 195.2 m/s

Explanation:

The centripetal acceleration acts on the bottom of the test tube trying to take it closer to the axis of rotation, and it is defined by the following expression:

ac = v² / r

We know that ac = 5.4. 10⁴ . g = 5.4.10⁴.9.8 m/s²

Expressing r in m ⇒ r = 0.072 m.

So, the only unknown that remains is v, which is the linear speed of the bottom of the test tube, which we want to find out:

v = √ac. r = √5.4.10⁴.9.8 m/s². 0.072 m = 195.2 m/s

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Sandwich materials typically use a high density core with non-structural cover plates. a)True b)- False
Lorico [155]

Answer: False

Explanation: Sandwich materials are usually in composite material form which has a fabrication of two thin layers which are stiff in nature and have  light weighing and thick core .The construction is based on the ratio that is of stiffness to the weight .Therefore, the density of the material in the core is not high and are only connected with the skin layer through adhesive .So the given statement is false that sandwich materials typically use a high density core with non- structural cover plates.

4 0
3 years ago
Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
viktelen [127]

Answer:

\theta_1=15^o\\\theta_2=75^o

Explanation:

<u>Projectile Motion</u>

In projectile motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration (assuming no friction), and the acceleration in the vertical direction is always the acceleration of gravity. The basic formulas are shown below:

V_x=V_{ox}=V_ocos\theta

Where \theta is the angle of launch respect to the positive horizontal direction and Vo is the initial speed.

V_y=V_{oy}-gt=V_osin\theta-gt

The  horizontal and vertical distances are, respectively:

x=V_{o}cos\theta t

\displaystyle y=y_o+V_{o}sin\theta t-\frac{gt^2}{2}

The total flight time can be found as that when y = 0, i.e. when the object comes back to ground (or launch) level. From the above equation we find

\displaystyle t_f=\frac{2V_osin\theta}{g}

Using this time in the horizontal distance, we find the Range or maximum horizontal distance:

\displaystyle R=\frac{V_o^2sin2\theta}{g}

Let's solve for \theta

\displaystyle sin2\theta=\frac{R.g}{V_o^2}

This is the general expression to determine the angles at which the projectile can be launched to hit the target. Recall the angle can have to values for fixed positive values of its sine:

\displaystyle \theta_1=\frac{asin\left(\frac{R.g}{V_o^2}\right)}{2}

\displaystyle \theta_2=\frac{180^o-asin\left(\frac{R.g}{V_o^2}\right)}{2}

Or equivalently:

\theta_2=90^o-\theta_1

Given Vo=37 m/s and R=70 m

\displaystyle \theta_1=\frac{asin\left(\frac{70\times 9.8}{37^2}\right)}{2}

\theta_1=15^o

And

\theta_2=90^o-15^o=75^o

5 0
3 years ago
Rope BCA passes through a pulley at point C and supports a crate at point A. Rope segment CD supports the pulley and is attached
djverab [1.8K]

Answer:

363 pounds 32 degrees

Explanation:

Express your answers numerically in pounds and degrees to three significant figures separated by a comma. slader

6 0
3 years ago
A long rod of 60-mm diameter and thermophysical properties rho= 8000 kg/m3, c= 500 J/kg·K, and k= 50 W/m·K is initially at a uni
Dvinal [7]

Answer:

Tc =    = 424.85 K

Explanation:

Data given:

D = 60 mm = 0.06 m

\rho = 8000 kg/m^3

k = 50 w/m . k

c = 500 j/kg.k

h_{\infty} = 1000 w/m^2

t_{\infity} = 750 k

t_w = 500 K

surface area = As = \pi dL

\frac{As}{L} = \pi D = \pi \timeS 0.06

HEAT FLOW Q  is

Q = h_{\infty} As (T_[\infty} - Tw)

 = 1000 \pi\times 0.06 (750-500)

  = 47123.88 w per unit length of rod

volumetric heat rate

q = \frac{Q}{LAs}

  = \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}

q = 1.66\times 10^{7} w/m^3

Tc = \frac{- qR^2}{4K} + Tw

= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 56} +  500

   = 424.85 K

7 0
3 years ago
_____ braking is when you squeeze the brake pedal until just before the wheels lock, then ease off the pedal, then squeeze again
grin007 [14]
The answer should be d) threshold
6 0
3 years ago
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