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kipiarov [429]
4 years ago
7

Which structure is found in all EUKARYOTIC cells? Large central vacuole, Golgi apparatus, flagella, cilia

Physics
1 answer:
Pavel [41]4 years ago
3 0
The answer is Golgi apparatus.
HOPE THIS HELPS!
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Blood plasma volume for adults is about 3.1 liters. It has a density of 1.020 gml. How many pounds of blood plasma are in your b
kari74 [83]

Answer:

mass is 6.97 pounds

Explanation:

given data

volume = 3.1 liters

density = 1.020 g/ml = 1.02 kg/l

to find out

How many pounds of blood plasma

solution

we know mass formula from density that is

density = mass / volume

so

mass = density × volume    ...............1

so put all value to get mass

mass = 1.02 × 3.1

mass = 3.162 kg

mass = 3.162  × 2.205 pounds

so mass is 6.97 pounds

6 0
4 years ago
0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a
klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}

(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }

where h_{fg is the change in enthalpy of saturated vapor to saturated liquid ( 250 Btu

T is the temperature ( 15 + 460 )R

m is the mass of water ( 0.5 Ibm )

V_{fg is specific volume ( 1.5 ft³ )

we substitute

(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5}) / ( (15+460)\frac{1.5}{0.5})  

(\frac{dP}{dT} )_{sat } = 272.98 Ibf-ft²/R

Now,

(\frac{dP}{dT} )_{sat } = (\frac{P_2 - P_1}{T_2 - T_1})_{sat

where P₁ is the initial pressure ( 50 psia )

P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

T₂ is the final temperature = ?

we substitute;

T_2 = ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}

T_2 = 475 + 5.2751\\

T_2 = 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

3 0
3 years ago
Use the mass and volume data to calculate the density of liquid propane to the nearest hundredth.
Black_prince [1.1K]
Density = mass/volume 
              = 4300/10000
              =    0.43 g/mL.

Therefore density of liquid propane is 0.43 g/mL.

Hope this helps!

5 0
3 years ago
Read 2 more answers
A system of inertia 0.47 kg consists of a spring gun attached to a cart and a projectile. The system is at rest on a horizontal
Vlad [161]

Answer:

v'=0.83m/a and v=10.2m/s

Explanation:

The information that we have is:

m_1=0.050kg\\m_2=0.47kg\\\theta = 40\\h=2.2m

The maximum height of the projectile is given by the equation

h=\frac{v^2sin^2\theta}{2g}

So, rearrange for the velocity,

v=\sqrt{\frac{2gh}{sin^2\theta}}\\v=\sqrt{\frac{2*9.8*2.2}{sin^2(40)}}\\v=10.2m/s

Apply the conservation of momentum,

mvcos(\theta)=m'v'

Then rearrange the recoil speed,

v'=\frac{mvcos\theta}{m_2}\\v'=\frac{0.05*10.2*cos40}{0.47}\\v'=0.83m/s\\

4 0
4 years ago
Identify two everyday phenomena that exhibit diffraction of sound and explain how diffraction of sound applies. Identify two eve
Naya [18.7K]

Answer:

Explanation:

Diffraction is the term used to describe the bending of a wave around an obstacle. It is one of the general properties of waves.

1. Diffraction of sound is the bending of sound waves around an obstacle which propagates from source to a listener. Two of the daily phenomena that exhibit diffraction of sound are:

i. The voices of people talking outside a building can be heard by those inside.

ii. The sound from the horn of a car can be heard by people at certain distances away.

When sound waves are produced, the surrounding air molecules are required for its transmission. This is because sound wave is a mechanical wave which requires material medium for its propagation. When a source produces a sound, the sound waves bend around obstacles on its path to reach listeners.

2. Light waves are electromagnetic waves which can undergo diffraction. Diffraction of light is the bending of the rays of light around an obstacle. Two of the daily phenomena that exhibit diffraction of light are:

i. The shadow of objects which has the umbra and penumbra regions.

ii. The apparent color of the sky.

A ray of light is the path taken by light, and the combination of two or more rays is called a beam. A ray or beam of light travels in a straight line, so any obstacle on its path would subject the light to bending around it during propagation. These are major applications in pin-hole cameras, shadows, rings of light around the sun etc.

Some significant differences between diffraction of light and that of the sound are:

i. Diffraction of light is not as common as that of sound.

ii. Sound propagates through a wider region than light waves.

iii. Sounds are longitudinal waves, while lights are transverse waves.

7 0
4 years ago
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