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olya-2409 [2.1K]

3 years ago

6

A plane with the speed of 210 miles per hour is flying at an angle of 40 degrees from true east and a 45 mile per hour wind is b

lowing directly from the west. What is the angle measure and the speed of the true course of the plane?

1 answer:

Yanka [14]3 years ago

3 0

Medtvtvyftfunuhyfsvghjji

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a thunderclap sends a sound wave through the air and the ocean below The thunderclap sound wave has a constant frequency of 50 H

Montano1993 [528]

Answer:

child protective services

Explanation:

amonbgus

4 0

3 years ago

Light rays bend as they pass from air into water at an angle (not 90 degrees). this is refraction. air water normal incident ray

xxTIMURxx [149]

When light ray pass from air into water, its speed and wavelength change only the frequency of the light doesn't change.

Light travels slower in a medium of higher refractive index. It bends because of this change in speed. The wavelength of light also changes in order to maintain the constant frequency.

7 0

3 years ago

Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an

bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

$\frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}}$ ......... (1)

Hence, total charge will be as follows.

$q_{1} + q_{2}$ = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

$q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

and, $q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}$

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

$q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}$

= $\frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}$

= 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

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

How far did a frog jump if he travels at a rate of 2.1 m/s for 10 seconds?

Anestetic [448]

Answer:

21 m

Explanation:

The motion of the frog is a uniform motion (constant speed), therefore we can find the distance travelled by using

$d=vt$

where

d is the distance covered

v is the speed

t is the time

The frog in this problem has a speed of

v = 2.1 m/s

and therefore, after t = 10 s, the distance it covered is

$d=(2.1)(10)=21 m$

3 0

3 years ago