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3 years ago

Anyone willing to help?

Physics

2 answers:

grigory [225]3 years ago

5 0

Answer:

I think it is "A" because the wind systems are created by uneven heating of Earth's surface.It also may help form large global wind patterns.

Explanation:

Ierofanga [76]3 years ago

3 0

D is the answer reeeeeeeeeeetaaaaaaardddddddddddd

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A body of mass 80kg was lifted vertically through a distance of 5.0 metres. Calculate the work done on the body. ( Acceleration

sleet_krkn [62]

Answer:

80×5×10=4000J

so therefore, work done on the body is 4000J

8 0

3 years ago

A light plane must reach a speed of 33 m/s per take off. How long a runway is needed if the constant acceleration is 3.0 m/s^2.

aleksley [76]

Given the final velocity (Vf) and the acceleration (a), the distance that should be traveled by the plane is calculated through the equation,
d = (Vf² - Vi²) / 2a
V1 should be zero because the light plane started the motion from rest. Substituting the given values,
d = ((33 m/s)² - 0)) / 2(3 m/s²)
The distance is therefore equal to 181.5 meters.

3 0

3 years ago

Outside our solar system the closest star to earth is Proxima century life from the start takes 2200000 minutes to reach earth.

s2008m [1.1K]

Answer:

2200000 = 2.2E6 min for light from Proxima to reach earth

8.3 min from light sun to reach earth

2.2E6/8.3 = 2.56E5 times for light from Proxima

Proxima is about 256,000 times farther away than the sun

Since the sun is about 93,000,000 = 9.3E7 miles from earth

Proxima is then 9.3E7 * 2.56E5 = 2.4E13 miles away

Note - the speed of light is

3.00E8 m/s * 60 s/min / 1000 m/km = 1.8E7 km/min as given

5 0

2 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: