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2 years ago

A doctor observe areas of pink scaly skin on patient

Physics

1 answer:

murzikaleks [220]2 years ago

6 0

Answer:

0-0 what is rhat supposed to mean. i dont think anyone can answer that lol

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The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu

BartSMP [9]

Answer:

hello your question is incomplete attached below is the missing part

answer : short period oscillations frequency = 0.063 rad / sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

Explanation:

first we have to state the general form of the equation

= $( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns} ) = 0$

where :

$w_{np} = Natural frequency of plugiod oscillation$

$\alpha _{p} = damping ratio of plugiod oscilations$

comparing the general form with the given equation

$w^{2} _{np}$ = 18.2329

$w^{2} _{ns} = 0.003969$

hence the short period oscillation frequency ( $w_{ns}$ ) = 0.063 rad/sec

phugoid oscillations natural frequency ( $w_{np}$ ) = 4.27 rad/sec

8 0

2 years ago

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

$g=-9.8 m/s^2$ is the acceleration of gravity

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

$v'^2-u^2=2gh'$

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(0.20)}=4.74 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{4.74-5.14}{-9.8}=0.04s$

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

$\Delta t = 2\cdot 0.04 =0.08 s$

8 0

3 years ago

The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:

Furkat [3]

To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is s=r(B), that has its <span>r=radius,B=angle in rad velocity v=ds/dt= w(r)
Do not forget about </span> w = angular speed in rad/s and $w1 = 1 revolution/sec = 2Pi (rad/s)$
Now we can go to proportion
$v1=v2$
$w1*r1 = w2r2$ $w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)$
$w2 = w3 (which is the angular velocity of the rear wheel) 
$
SOLVING FOR A : $v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec$
$v3 = 14.66 ft/sec(1 mile/5280 ft)( 3600 sec/h)$ $= 9.99$ or something about <span>10 mph --- SOLVING FOR B.
</span>I'm sure it helps!

7 0

2 years ago

How are the mass and weight of an object related? Include a description with words and an equation.

nikdorinn [45]

The weight is the force experienced, whereas the mass represents the actual quantity of matter inside a body..
weigh on the surface of the earth is equal to mg
mass is m
and at the centre weight is 0 due to 0 acceleration that's 0 g
but mass is always constant and remains m, no matter where you are

6 0

2 years ago

A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

Maurinko [17]

Answer:

(a). The decay constant is $1.55\times10^{-5}\ s^{-1}$