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2 years ago

Match each measurement tool with what it measures.

Physics

1 answer:

Zarrin [17]2 years ago

5 0

Thermometer-temperature
Wind vane -wind direction
Anemometer- both wind speed (and direction)
Hygrometer- humidity

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Place the following list in order of occurrence from oldest to most recent:

Andre45 [30]

Answer:

Explanation:

2.Big bang, 3.contraction of the solar nebula, 5.stellar ignition in our sun, 4.outgassing of earths secondary atmosphere, 1.appearance of the first ocean on earth ,7.evolution of photosynthesis, 6.build- up of oxygen in earths atmosphere.

4 0

2 years ago

Which statement is most likely correct?

Len [333]

it's either the first or second choice

6 0

3 years ago

Read 2 more answers

PLEASE ANSWER ASAP BEFORE MY TEACHER AND MY MOM KILLES ME PLEASE ASAP

grandymaker [24]

Answer:

the red at the bottom should not count, but the red at the top is the least dense because it floats upon the other liquids

Explanation:

hope this helps

7 0

2 years ago

Two 1.9 kg masses are 1.1 m apart (center to center) on a frictionless table. Each has + 9.6 μC of charge.

Anni [7]

Answer:

F= 0.6 N

Explanation:

Fe(electrical force)= $k q1q2/r^2$

$k=9*10^9\\q1=-9.6 *10^-6 C\\q2= -9.6*10^-6 C\\r= 1.1m$

So,

$F=\frac{9*10^9*-9.6 *10^-6 C* -9.6*10^-6 C}{1.1^2}$

$F= 0.6 N$

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:

$k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)$

$k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}$

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

8 0

3 years ago