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Leona [35]
4 years ago
7

Why can you still seethe Moon during aneclipse?​

Physics
1 answer:
love history [14]4 years ago
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.

4.

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

f= force

m=mass

a=acceleration

Explanation:

examples:

riding your bicycle

•your bicycle is the mass, your leg pushing in pedals of your bicycle is the force

pushing a box

•the box is the mass, you are pushing the box

setting a pencil down in a table

•the pencil is the mass, you are puting the pencil down

3 0
3 years ago
The pressure of a 1.0 liter balloon is at 1 atm at 293 K if you were to take the balloon into a freezer that is still at One ATM
Fofino [41]

Here we can say that number of moles as well as pressure will remain same for the balloon

so here we can say by ideal gas equation

PV = nRT

so here for constant pressure and constant number of moles we have

\frac{V_1}{T_1} = \frac{V_2}{T_2}

\frac{1}{293} = \frac{V}{263}

now by rearranging the equation we have

V = \frac{263}{293}

V = 0.89 Liter

so final volume will be 0.89 liter

5 0
3 years ago
The amount of energy in the universe is...
g100num [7]
The amount of energy in the universe is remaining constant .

So , option (D) is right ..

3 0
3 years ago
Read 2 more answers
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