Ask question

Ask question

All categories

2 years ago

8

Which are parts of the middle ear? Check all that apply.

1 answer:

stellarik [79]2 years ago

5 0

Answer:

middle ear has three bones! the hammer, anvil, stirrup, and ear drum

Explanation:

You might be interested in

A long copper rod of diameter 2.0 cm is initially at a uniform temperature of 100°C. It is now exposed to an air stream at 20°C

Lisa [10]

Answer:

t = 4.0 min

Explanation:

given data:

diameter of rod = 2 cm

T_1 = 100 degree celcius

Air stream temperature = 20 degree celcius

heat transfer coefficient = 200 W/m2. K

WE KNOW THAT

copper thermal conductivity = k = 401 W/m °C

copper specific heat Cp = 385 J/kg.°C

density of copper = 8933 kg/m3

charateristic length is given as Lc

$Lc = \frac{V}{A_s}$

$Lc = \frac{\frac{\pi D^2}{4} L}{\pi DL}$

$Lc = \frac{D}{4}$

$Lc = \frac{0.02}{6} = 0.005 m$

Biot number is given as $Bi = \frac{hLc}{k}$

$Bi = \frac{200*0.005}{401}$

Bi = 0.0025

As Bi is greater than 0.1 therefore lumped system analysis is applicable

so we have

$\frac{T(t) - T_∞}{Ti - T_∞} = e^{-bt}$ ............1

where b is given as

$b = \frac{ hA}{\rho Cp V}$

$b = \frac{ h}{\rho Cp Lc}$

$b = \frac{200}{8933*385*0.005}$

b = 0.01163 s^{-1}

putting value in equation 1

$\frac{25-20}{100-20} = e^{-0.01163t}$

solving for t we get

t = 4.0 min

6 0

3 years ago

Please Help!! When do ionic bonds occur?

iren2701 [21]

A is the answer. I think

8 0

3 years ago

a ball is projected upward at time t = 0.00 s from a point on a roof 70 m above the ground. The ball rises, then falls and strik

grin007 [14]

Answer: 17.68 s

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0$ is the height of the ball when it hits the ground

$y_{o}=70 m$ is the initial height of the ball

$V_{o}=82m/s$ is the initial velocity of the ball

$t$ is the time when the ball strikes the ground

$g=9.8m/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $t$ from (1):

$0=70m+(82m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$ (2)

Rewritting (2):

$-\frac{1}{2}(9.8m/s^{2})t^{2}+(82m/s)t+70m=0$ (3)

This is a quadratic equation (also called equation of the second degree) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula:

$t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$ (4)

Where:

$a=-\frac{1}{2}(9.8m/s^{2}$

$b=82m/s$

$c=70m$

Substituting the known values:

$t=\frac{-82 \pm \sqrt{82^{2}-4(-\frac{1}{2}(9.8)(70)}}{2a}$ (5)

Solving (5) we find the positive result is:

$t=17.68 s$

7 0

2 years ago

Energy needs, in total kilocalories per day, are greater during _____________ than any other time of life.

malfutka [58]

Energy needs, in total kilocalories per day, are greater during adolescence than any other time of life. The correct option among all the options that are given in the question is the second option or option "b". Only during the time of pregnancy will a girl require more kilocalories per day than during the time of adolescence.

5 0

3 years ago

Read 2 more answers

A rocket accelerates by burning its onboard fuel, so its mass decreases with time. Suppose the initial mass of the rocket at lif

Serggg [28]

Answer:

Height of the rocket be one minute after liftoff is 40.1382 km.

Explanation:

$v(t)=-gt-v_e\times \ln \frac{m-rt}{m}$

v = velocity of rocket at time t

g = Acceleration due to gravity = $9.8 m/s^2$

$v_e$ = Constant velocity relative to the rocket = 2,900m/s.

m = Initial mass of the rocket at liftoff = 29000 kg

r = Rate at which fuel is consumed = 170 kg/s

Velocity of the rocket after 1 minute of the liftoff =v

t = 1 minute = 60 seconds'

Substituting all the given values in in the given equation:

$v(60)=-9.8 m/s^2\times 60 s-2,900m/s\times \ln (\frac{29,000 kg-170 kg/s\times 60 s}{2,9000 kg})$

$v(60) = 668.97 m/s$

Height of the rocket = h

$Velocity=\frac{Displacement}{time}$

$668.97 m/s=\frac{h}{60 s}$

$h=668.97 m/s\times 60 s=40,138.2 m = 40.1382 km$

Height of the rocket be one minute after liftoff is 40.1382 km.

4 0

3 years ago