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3 years ago

Which of the following is an example of refraction?

Physics

2 answers:

Sladkaya [172]3 years ago

7 0

I believe it will be C.

love history [14]3 years ago

5 0

You try to pick up a shell in the water, but it isn't where it appears to be would be the correct answer

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Which of the following formulas represents an olefin (aka alkaline)

pochemuha

Answer:C2H4

Explanation:

Olefins are alkenes, c2h4 is the simplest member called ethene

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4 years ago

When two waves are moving toward each other, and their crests line up with each other, it results in a wave with greater amplitu

Maurinko [17]

constructive interference

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3 years ago

An archer shot a 0.06 kg arrow at a target. The arrow accelerated at 5,000 m/s2 to reach a speed of 50.0 m/s as it left the bow.

laila [671]

Answer:

300 N

Explanation:

The net force acting on the arrow is given by Newton's Second Law:

$F=ma$

where

m = 0.06 kg is the mass of the arrow

a = 5,000 m/s^2 is the acceleration of the arrow

Substituting the numbers into the equation, we find

$F=(0.06 kg)(5,000 m/s^2)=300 N$

7 0

3 years ago

Explain why it takes much more effort to

natali 33 [55]

Answer:

Mass Kinetic Energy and Jules

Explanation:

The train in question is big and heavy and a car is decently heavy but say a train moving at 55 mph can plow through a car and a car driving at 55mph driving at a train will be stopped dead in its tracks. This is because newtons laws of motion specifically an object in motion will stay in motion unless its opposed. The train also has a payload behind it meaning it hurts with force while a car doesn't have to much mass behind it. The train takes loner to stop for as it's acceleration as well as it's deceleration are very slow because its huge and takes a lot of force to stop it while a car is very centralized and compact when it comes to weight and its brakes are usually effective at stopping at 55 mph in about 2 to 6 seconds while a train might stay moving for a good 35 seconds. The force behind the train is immense for as even if the wheels don't spin at all the train will still move since the force behind it is great and a cars tires have a lot of grip and not a lot mass which plays into the force the car has so it can stop simply.

6 0

3 years ago

Determine the moment of inertia Ixx of the mallet about the x-axis. The density of the wooden handle is 860 kg/m3 and that of th

Yuki888 [10]

Complete Question

Diagram for this shown on the first uploaded image

Answer:

The moment of inertia Ixx of the mallet about the x-axis is $I{xx}= 0.119 kg \cdot m^2$

Explanation:

From the question we are told that

The density `of wooden handle is $\rho_w = 860 kg/m^3$

The density `of soft-metal head is $\rho_s =8000kg/m^3$

Generally the mass of the wooden can be mathematically obtained with this formula

$m_w = \rho_w A_w l_w$

Where $A_w$ is mass of wooden handle which is mathematically obtain with the formula

$A_w = \frac{\pi}{4} d^2_w$

Where $d_w$ is the diameter of the wooden handle which from the diagram is

$27mm = \frac{27}{1000} = 0.027m$

So $A_w = \frac{\pi}{4} * 0.027^2$

$l_w$ is the length of the the wooden handle which is given in the diagram as $l_w = 315mm = \frac{315}{1000} = 0.315m$

Substituting these value into the formula for mass

$m_w = 860 * (\frac{\pi}{4} * 0.027^2 ) *0.315$

$= 0.155kg$

Generally the mass of the soft-metal head can be mathematically obtained with this formula

$m_s = \rho_s A_s l_s$

Where $A_s$ is mass of soft-metal head which is mathematically obtain with the formula

$A_s = \frac{\pi}{4} d^2_s$

Where $d_s$ is the diameter of the soft-metal head which from the diagram is

$36mm = \frac{36}{1000} = 0.036m$

So $A_s = \frac{\pi}{4} * 0.036^2$

$l_s$ is the length of the the soft-metal head which is given in the diagram

as $l_s = 90mm = \frac{90}{1000} = 0.090m$

Substituting these value into the formula for mass

$m_s = 8000 * (\frac{\pi}{4} * 0.036^2 ) *0.090$

$=0.733kg$

Generally the mass moment of inertia about x-axis for the wooden handle is

$(I_{xx})_w = [\frac{1}{3}m_w + l_w^2 ]$

Substituting values

$(I_{xx})_w = [\frac{1}{3}*0.155 + 0.315^2 ]$

$=5.12*10^{-3}kg \cdot m^2$

Generally the mass moment of inertia about x-axis for the soft-metal head is

$(I_{xx})_s = [\frac{1}{12}m_s l_s ^2 + b^2]$

Where b is the distance from the centroid to the axis of the head which is mathematically given as

$b=l_w +\frac{d_s}{2}$

Substituting values

$b = 0.315 + \frac{0.036}{2}$

$= 0.336m$

Now substituting values into the formula for mass moment of inertia about x-axis for soft-metal head

$(I_{xx})_s = [\frac{1}{12} *0.733* 0.090^2 + 0.336^2]$

$=0.113 kg \cdot m^2$

Generally the mass moment of inertia about x-axis is mathematically represented as

$I_{xx} = (I_{xx})_w + (I_{xx})_s$

$= [\frac{1}{3}m_w + l_w^2 ] + [\frac{1}{12}m_s l_s ^2 + b^2]$

Substituting values

$I_{xx} = 5.12*10^{-3} +0.113$

$I{xx}= 0.119 kg \cdot m^2$

8 0

3 years ago