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3 years ago

12

A plane is traveling North at 80km/hr into a 20 km/hr headwind (South). What is the plane's resultant velocity?

2 answers:

frosja888 [35]3 years ago

8 0

Answer:

60km/hr

Explanation:

R=80 - 20

=60km/hr, north

Alexxandr [17]3 years ago

5 0

Answer:

Consider our plane flying at a speed of 80 km/hr into a 40 km/hr crosswind. Its speed, relative to the ground, can be calculated by adding the two vectors as shown on the right. What happens to the speed of the plane as the crosswind increases in magnitude? Look at the animation on the right. It is obvious that the speed of the plane, relative to the ground, increases in magnitude.

Explanation:

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The inner and outer surfaces of a 0.5-cm thick 2-m by 2-m window glass in winter are 10°C and 3°C, respectively. If the thermal conductivity of the glass

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You start pushing a suitcase full of clothes in the horizontal direction with a force of 25.0 newtons. The weight of the suitcas

sasho [114]

Answer:

Distance traveled will be 5.6307 m

Explanation:

Time t = 3 sec

We have given force F = 25 N

We know that force is given by F = ma

So ma = 25 -----------eqn 1

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We know that weight is given by W = mg

So mg = 196 -----------------eqn 2

From equation 1 and equation 2 $\frac{a}{g}=\frac{25}{196}$

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3 years ago

The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov

son4ous [18]

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The eyepiece focal length is $f_e = 0.027 \ m$

Explanation:

From the question we are told that

The focal length is $f_o = 5.5 \ mm = -0.0055 \ m$

This negative sign shows the the microscope is diverging light

The angular magnification is $m = 300$

The distance between the objective and the eyepieces lenses is $Z = 19 \ cm = 0.19 \ m$

Generally the magnification is mathematically represented as

$m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]$

Where $f_e$ is the eyepiece focal length of the microscope

Now making $f_e$ the subject of the formula

$f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }$

substituting values

$f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }$

$f_e = 0.027 \ m$

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