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4 years ago

Why can you still seethe Moon during aneclipse?

1 answer:

3 0

During a total lunar eclipse, Earth completely blocks direct sunlight from reaching the Moon. The only light reflected from the lunar surface has been refracted by Earth's atmosphere. ... Due to this reddish color, a totally eclipsed Moon is sometimes called a blood moon.

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As shown in the diagram below, a rope attached to a 500.-kilogram crate is used to exert a force of 45 newtons at an angle of 65

kaheart [24]

Answer:

19.01 N

Explanation:

F = Force being applied to the crate = 45 N

$\theta$ = Angle at which the force is being applied = $65^{\circ}$

Horizontal component of force is given by

$F_x=F\cos\theta\\\Rightarrow F_x=45\times \cos65^{\circ}\\\Rightarrow F_x=19.01\ \text{N}$

The horizontal component of the force acting on the crate is 19.01 N.

4 0

3 years ago

Scenario 2: Use the following information to answer questions 3 and 4:

NemiM [27]

Answer:

A. 52 min

.A. 47 watts

Explanation:

Given that;

jim weighs 75 kg

and he walks 3.3 mph; the objective here is to determine how long must he walk to expend 300 kcal.

Using the following relation to determine the amount of calories burned per minute while walking; we have:

$\dfrac{MET*weight (kg)*3.5}{200}$

here;

MET = energy cost of a physical activity for a period of time

Obtaining the data for walking with a speed of 3.3 mph From the standard chart for MET, At 3.3 mph; we have our desired value to be 4.3

However;

the calories burned in a minute = $\dfrac{4.3*75 (kg)*3.5}{200}$

= 5.644

Therefore, for walking for 52 mins; Jim burns approximately 293.475 kcal which is nearest to 300 kcal.

Given that:

mass m = 75 kg

intensity = 6 kcal/min

The eg ergometer work rate = ??

Applying the formula:

$V_O_2 ( intensity ) = ( \dfrac{W}{m}*1.8)+7$

where ;

$V_O_2 ( intensity ) = \dfrac{1 \ kcal min^{-1}*10^{-3}}{5}$

$V_O_2 ( intensity ) = \dfrac{6*1 \ kcal min^{-1}*10^{-3}}{5}$

$V_O_2 ( intensity ) = 0.0012$

∴ $0.0012 = (\dfrac{W}{75}*1.8)+7 \\ \\ W = \dfrac{0.0012-7}{1.8}*75 \\ \\ W = \dfrac{7*75}{1.8} \\ \\ W = 291.66 \ kg m /min$

Converting to watts;

Since; 6.118kg-m/min is = 1 watt

Then 291.66 kgm /min will be equal to 47.67 watts

≅ 47 watts

3 0

4 years ago

State Newton's second law of<br>motion and write down<br>three examples to<br>justity it.

kakasveta [241]

Answer:

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

$f = m \times a$