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ludmilkaskok [199]

3 years ago

10

An airplane is heading due south at a speed of 430 km/h . A wind begins blowing from the northwest at a speed of 85.0 km/h (aver

age).What should the plane's course shift be so that it will fly due south?

1 answer:

Igoryamba3 years ago

5 0

Answer:

It should fly 8° to west of south at 430km/h

Explanation:

According to the diagram. X components for both velocities must have the same magnitude in order to get the resultant velocity due south.

$V_{w}*cos(45) = V_{A}*sin(\alpha )$ Solving for α:

α = 8.03°

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Calculate the size of the image of a tree that is 8m high and 80 m a pinhole camera that is 20 cm long . what is its magnificati

Vladimir79 [104]

1) Size of the image: 2 cm

In order to calculate the size of the image, we can use the following proportion:

$p:q = h_o : h_i$

where

p = 80 m is the distance of the tree from the pinhole

q = 20 cm = 0.2 m is the distance of the image from the pinhole

$h_o$ = 8 m is the heigth of the object

$h_i$ is the height of the image

By re-arranging the proportion, we find

$h_i = \frac{h_o \cdot q}{p}=\frac{(8 m)(0.2 m)}{80 m}=0.02 m=2 cm$

2) Magnification: 0.0025

The magnification of a camera is given by the ratio between the size of the image and the size of the real object:

$M=\frac{h_i}{h_o}$

so, in this problem we have

$M=\frac{0.02 m}{8 m}=0.0025$

4 0

3 years ago

Water is being boiled in an open kettle that has a 0.52-cm-thick circular aluminum bottom with a radius of 12.0 cm. If the water

tangare [24]

Answer:

$T_b=107.3784\ ^{\circ}C$

Explanation:

Given:

thickness of the base of the kettle, $dx=0.52\ cm=5.2\times 10^{-3}\ m$

radius of the base of the kettle, $r=0.12\ m$

temperature of the top surface of the kettle base, $T_t=100^{\circ}C$

rate of heat transfer through the kettle to boil water, $\dot Q=0.409\ kg.min^{-1}$

We have the latent heat vaporization of water, $L=2260\times 10^3\ J.kg^{-1}$

and thermal conductivity of aluminium, $k=240\ W.m^{-1}.K^{-1}$

<u>So, the heat rate:</u>

$\dot Q=\frac{0.409\times 2260000}{60}$

$\dot Q=15405.67\ W$

<u>From the Fourier's law of conduction we have:</u>

$\dot Q=k.A.\frac{dT}{dx}$

$\dot Q=k\times \pi.r^2\times \frac{T_b-T_t}{5.2\times 10^{-3}}$

where:

$A=$ area of the surface through which conduction occurs

$T_b=$ temperature of the bottom surface

$15405.67=240\times \pi\times 0.12^2\times \frac{T_b-100}{5.2\times 10^{-3}}$

$T_b=107.3784\ ^{\circ}C$ is the temperature of the bottom of the base surface of the kettle.

6 0

3 years ago

What is the intensity in W/m2 of a laser beam used to burn away cancerous tissue that, when 85.0% absorbed, puts 470 J of energy

Advocard [28]

Answer:

26036485.6433 W/m²

Explanation:

E= Energy = 470 J

t = Time = 4 seconds

d = Diameter = 2.6 mm

Power is given by

$P=\dfrac{E}{t}$

Intensity is given by

$I=\dfrac{P}{\pi r^2}\\\Rightarrow 0.85I=\dfrac{E}{t\dfrac{\pi}{4} d^2}\\\Rightarrow I=\dfrac{470}{0.85\times 4\times \dfrac{\pi}{4}\times (2.6\times 10^{-3})^2}\\\Rightarrow I=26036485.6433\ W/m^2$

The intensity of the laser beam is 26036485.6433 W/m²

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mars1129 [50]

Answer:

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Harrizon [31]

The object that goes through chemical change, changes completely to where you can not change it back to its original form. Physical change you can undo

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