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3 years ago

Lymphocytes are a type of __

Physics

2 answers:

soldi70 [24.7K]3 years ago

8 0

Lymphocytes are a type of _

<u> white blood cell</u>

goblinko [34]3 years ago

3 0

Lymphocytes are a type of __ white blood cells

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A person walks 60 meters in 1 minute. Then walks 30 meters in 2 minutes. What is thier average speed over the 3 minute walk?

Gennadij [26K]

Your answer would be 75 meters

8 0

3 years ago

Read 2 more answers

Sawyer launches his 180 kg raft on the Mississippi River by pushing on it with a force of 75N. How long must Sawyer push on the

Daniel [21]

Answer: 4.8 s

Explanation:

We have the following data:

$m=180 kg$ the mass of the raft

$F=75 N$ the force applied by Sawyer

$V=2 m/s$ the raft's final speed

$V_{o}=0 m/s$ the raft's initial speed (assuming it starts from rest)

We have to find the time $t$

Well, according to Newton's second law of motion we have:

$F=m.a$ (1)

Where $a$ is the acceleration, which can be expressed as:

$a=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{t-t_{o}}$ (2)

Substituting (2) in (1):

$F=m\frac{V-V_{o}}{t-t_{o}}$ (3)

Where $t_{o}=0$

Isolating $t$ from (3):

$t=\frac{m(V-V_{o})}{F}$ (4)

$t=\frac{180 kg(2 m/s-0 m/s)}{75 N}$

Finally:

$t=4.8 s$

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3 years ago

Find the mean of set values 12g 9g 13g 12g 20g 17g 15g

nalin [4]

${\huge{\boxed{\mathcal{\green{Answer}}}}} \\ \frac{12 + 9 + 13 + 12 + 20 + 17 + 15}{7} \\ = \frac{98}{7} \\ = 14 \\ {\huge{\boxed{\mathcal{\green{Hope \: it \: Helps}}}}}$

5 0

3 years ago

A common method to measure thermal conductivity of a biomaterial is to insert a long metallic probe axially into the center of a

tia_tia [17]

Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer ( $\dot Q$ ), measured in watts, in the hollow cylinder is:

$\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})$

Where:

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$L$ - Length of the cylinder, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$D_{o}$ - Outer diameter, measured in meters.

$T_{i}$ - Temperature at inner surface, measured in Celsius.

$T_{o}$ - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

$k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)$

If we know that $\dot Q = 40.8\,W$ , $L = 0.6\,m$ , $T_{i} = 50\,^{\circ}C$ , $T_{o} = 20\,^{\circ}C$ , $D_{i} = 0.01\,m$ and $D_{o} = 0.04\,m$ , the thermal conductivity of the biomaterial is: