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GrogVix [38]
3 years ago
11

Aaron leaves one city at noon. He has to be at another city 186km away at 3:00p.m. The speed limit the entire way is 65 km/h. Ca

n he arrive at the second city on time? Explain. NEED ANSWER IMMEDIATELY! 99 points available for answer!
Physics
2 answers:
Zina [86]3 years ago
7 0

186 km/ 65km/h = 2.86 hours for him to reach his destination
I hope this helps;)
If you have any questions or concerns please PM me!
MA_775_DIABLO [31]3 years ago
6 0
Since Aaron left the city at 12pm (noon) and needs to be arrive at the next city at 3pm he has exactly 3 hours. Since he is going 65 km/h we need to divide 185 by 65 to find out how long it will take for Aaron to travel 185km.

186 / 65 =
2.86

Aaron will arive at the next city in 2.86 hours meaning he will not arrive late and will arive on time. I hope this helps you :D
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Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
ElenaW [278]
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
6 0
3 years ago
this stationary wave is what we call the first harmonic of the first normal mode of the system. in units of l, the length of the
pentagon [3]

First harmonic of a closed pipe is determined as velocity, v, to four times length (4L), F₀ v/4L.

<h3>First harmonic of a closed pipe</h3>

The first harmonic of a closed pipe is the fundamental frequency of the closed of the closed pipe.

L = λ/4

where;

  • L is the length of the pipe
  • λ is the wavelength of sound

λ = 4L

But, v = F₀λ

v = F₀(4L)

F₀ = v/4L

where;

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  • v is speed of sound

Thus, first harmonic of a closed pipe is determined as velocity, v, to four times length (4L), F₀ v/4L.

Learn more about fundamental frequency here:  brainly.com/question/1967686

#SPJ11

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Which subgenre would include stories about talking animals or magical powers? Image result for fable animal O fable O realistic
Furkat [3]

Answer:

fable

Explanation:

8 0
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A 328-kg car moving at 19.1 m/s in the + x direction hits from behind a second car moving at 13.0 m/s in the same direction. If
Kisachek [45]

Answer:

+14.0 m/s

Explanation:

we have mV + Mv = mU + Mu we solve for U = (mV + Mv - Mu)/m = (328*19.1 + 790*13 - 790*15.1)/328 = 14. Answer

8 0
3 years ago
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