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S_A_V [24]
2 years ago
12

While walking to work in Boston shortly after sunrise you notice that the water level in the bay is exceptionally low. Based on

this very low low-tide and the dark cloudless sky last night, what is the current phase of the moon

Physics
1 answer:
Andrew [12]2 years ago
5 0

While walking to work in Boston shortly after sunrise you notice that the water level in the bay is exceptionally low. Based on this very low low-tide and the dark cloudless sky last night, the current phase of the moon is a waxing crescent. Among the 8 phases of the Moon, the waxing crescent follows the new-moon phase.

The phases of Moon happen because of the presence of moon's orbit between the Earth and the Sun. The surface of the Moon reflects sunlight and this light reaches the Earth at different angles depending on the location of the Moon in its orbit. This gives us the impression of a waxing and a waning Moon.

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A woman is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follow
Ymorist [56]

Answer:

Explanation:

Given

Force P is acting upward

C is vertical contact Force

W is the weight of the crate

As P is unable to move the Block therefore Normal reaction keeps on acting on block

thus we can say that

P-W+C=0

P=W-C

                   

7 0
2 years ago
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An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th
kolbaska11 [484]

Answer:

mg=200.4 N.

Explanation:

This problem can be solved using Newton's law of universal gravitation: F=G\frac{m_{1}m_{2}}{r^{2}},

where F is the gravitational force between two masses m_{1} and m_{2}, r is the distance between the masses (their center of mass), and G=6.674*10^{-11}(m^{3}kg^{-1}s^{-2}) is the gravitational constant.

We know the weight of the astronout on the surface, with this we can find his mass. Letting w_{s} be the weight on the surface:

w_{s}=mg,

mg=8*10^{2},

m=(8*10^{2})/g,

since we now that g=9.8m/s^{2} we get that the mass is

m=81.6kg.

Now we can use Newton's law of universal gravitation

F=G\frac{Mm}{r^{2}},  

where m is the mass of the astronaut and M is the mass of the earth. From Newton's second law we know that

F=ma,

in this case the acceleration is the gravity so

F=mg, (<u>becarefull, gravity at this point is no longer</u> 9.8m/s^{2} <u>because we are not in the surface anymore</u>)

and this get us to

mg=G\frac{Mm}{r^{2}}, where mg is his new weight.

We need to remember that the mass of the earth is M=5.972*10^{24}kg and its radius is 6.37*10^{6}m.

The total distance between the astronaut and the earth is

r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6} meters.

Now we can compute his weigh:

mg=G\frac{Mm}{r^{2}},

mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}},

mg=200.4 N.

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2 years ago
An activity that is relatively short in time (&lt; 10 seconds) and has few repetitions predominately uses the _____________ ener
tensa zangetsu [6.8K]
An activity that is relatively short in time <10 seconds and has few repetitions predominantly uses the ATP/PC energy system. The cellular respiration procedure that changes food energy into ATP which is a form of energy is largely reliant on oxygen obtainability. During exercise the source and request of oxygen obtainable to muscle is unnatural by period and strength and by the individual’s cardiorespiratory suitability level. 
Steps of the ATP-PC system:


1. Primarily, ATP kept in the myosin cross-bridges which is microscopic contractile parts of muscle is broken down to issue energy for muscle shrinkage.  This action consents the by-products of ATP breakdown which are the adenosine diphosphate and one single phosphate all on its own.


2. Phosphocreatine is then broken down by the enzyme creatine kinase into creatine and phosphate.


3. The energy free in the breakdown of PC permits ADP and Pi to rejoin creating more ATP.  This newly made ATP can now be broken down to issue energy to fuel activity. 
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2 years ago
flat block is pulled along a horizontal flat surface by a horizontal rope perpendicular to one of the sides. The block measures
Lelu [443]

Answer:

Answered

Explanation:

v= 1 m/s

A= 1 m^2

m= 100 kg

y= 1 mm

μ = ?

ζ= viscosity of  SAE 20 crankcase oil of 15° C= 0.3075 N sec/m^2

forces acting on the block are  

                                 F_s   ←    ↓     →F_f

                                                mg

N= mg

F_s=  shear force = ζAv/y        F_f= friction force = μN

now in x- direction F_s= F_f

ζAv/y  =  μN

0.3075×1×1×1/1×10^{-3} = μ×100

⇒μ=0.313 (coefficient of sliding friction for the block)

Now, as the velocity is increased shear force also increases and due to this frictional force also increases.

Now, to compensate this frictional force friction coefficient must increase

as v∝μ

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2 years ago
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When Jane drives to work, she always places her purse on the passenger’s seat. By the time she gets to work, her purse has falle
LUCKY_DIMON [66]
At some time during her drive she backed up with a substantial negative. ( backwards) acceleration. Since the pocket book is not physically connected to the seat it is free to move. Upon rapid negative acceleration the pocket book remains in its position while the car accelerates backwards away from it. this demonstrates Newtons 1st law of motion. The first law is the law of inertia. Which states, an object at rest. ( pocketbook) will remain at rest and an object in motion will continue in motion at constant velocity, unless acted upon by some outside force to change its motion.
4 0
2 years ago
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