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jolli1 [7]
3 years ago
9

A substance did not change it's chemical nature in reaction which mostly likely describe the reaction

Physics
1 answer:
Ad libitum [116K]3 years ago
8 0
If the substance doesn't change chemically, it is a physical reaction.
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Sphere A with mass 80 kg is located at the origin of an xy coordinate system; sphere B with mass 60 kg is located at coordinates
IRINA_888 [86]

Answer:

Fc = [ - 4.45 * 10^-8 j ] N  

Explanation:

Given:-

- The masses and the position coordinates from ( 0 , 0 ) are:

       Sphere A : ma = 80 kg , ( 0 , 0 )

       Sphere B : ma = 60 kg , ( 0.25 , 0 )

       Sphere C : ma = 0.2 kg , ra = 0.2 m , rb = 0.15

- The gravitational constant G = 6.674×10−11 m3⋅kg−1⋅s−2

Find:-

what is the gravitational force on C due to A and B?

Solution:-

- The gravitational force between spheres is given by:

                       F = G*m1*m2 / r^2

Where, r : The distance between two bodies (sphere).

- The vector (rac and rbc) denote the position of sphere C from spheres A and B:-

 Determine the angle (α) between vectors rac and rab using cosine rule:

                   cos ( \alpha ) = \frac{rab^2 + rac^2 - rbc^2}{2*rab*rac} \\\\cos ( \alpha ) = \frac{0.25^2 + 0.2^2 - 0.15^2}{2*0.25*0.2}\\\\cos ( \alpha ) = 0.8\\\\\alpha = 36.87^{\circ \:}

 Determine the angle (β) between vectors rbc and rab using cosine rule:

                   cos ( \beta  ) = \frac{rab^2 + rbc^2 - rac^2}{2*rab*rbc} \\\\cos ( \beta  ) = \frac{0.25^2 + 0.15^2 - 0.2^2}{2*0.25*0.15}\\\\cos ( \beta  ) = 0.6\\\\\beta  = 53.13^{\circ \:}

- Now determine the scalar gravitational forces due to sphere A and B on C:

       Between sphere A and C:

                  Fac = G*ma*mc / rac^2

                  Fac = (6.674×10−11)*80*0.2 / 0.2^2  

                  Fac = 2.67*10^-8 N

                  vector Fac = Fac* [ - cos (α) i + - sin (α) j ]

                  vector Fac = 2.67*10^-8* [ - cos (36.87°) i + -sin (36.87°) j ]

                  vector Fac = [ - 2.136 i - 1.602 j ]*10^-8 N

       Between sphere B and C:

                  Fbc = G*mb*mc / rbc^2

                  Fbc = (6.674×10−11)*60*0.2 / 0.15^2  

                  Fbc = 3.56*10^-8 N

                  vector Fbc = Fbc* [ cos (β) i - sin (β) j ]

                  vector Fbc = 3.56*10^-8* [ cos (53.13°) i - sin (53.13°) j ]

                  vector Fbc = [ 2.136 i - 2.848 j ]*10^-8 N

- The Net gravitational force can now be determined from vector additon of Fac and Fbc:

                  Fc = vector Fac + vector Fbc

                  Fc = [ - 2.136 i - 1.602 j ]*10^-8  + [ 2.136 i - 2.848 j ]*10^-8

                  Fc = [ - 4.45 * 10^-8 j ] N  

3 0
4 years ago
What color of light can plants "see"?
mars1129 [50]
As you know, plants are usually green<span>, which means that most other colors are absorbed. One of the most common pigments is called chlorophyll, and one of the varieties is responsible for the </span>green<span> color of plants; it strongly absorbs </span>blue<span> and </span>red<span>light, which leaves only the </span>green<span> light to make it to our eyes.</span>
6 0
3 years ago
How to represent milligram in kilogram by standard formula?
Anettt [7]

Answer:

0.000001 kg

Explanation:

because 1 kg equal 1,000,000 milligrams

we take   \frac{1}{1,000,000} which equals 0.000001 kg

4 0
3 years ago
As a train enters the station it slows down from 40 m/s to a 10 m/s in 5 seconds
lora16 [44]

Answer:

See the answers below.

Explanation:

To solve this problem we must use the following equation of kinematics.

v_{f}=v_{o}-a*t

where:

Vf = final velocity = 10 [m/s]

Vo = initial velocity = 40 [m/s]

t = time = 5 [s]

a = acceleration [m/s²]

Now replacing:

10=40-a*5\\40-10=a*5\\30=5*a\\a=6[m/s^{2}]

Note: The negative sign in the above equation means that the velecity is decreasing.

2)

To solve this second part we must use the following equation of kinematics.

v_{f}^{2} =v_{o}^{2} -2*a*x\\

where:

x = distance [m]

(10)^{2} =(40)^{2} -2*6*x\\100=1600-12*x\\12*x=1600-100\\12*x=1500\\x=125[m]

7 0
3 years ago
Distance between Ranjan and Gomes house is 9km. Ranjan has to attend Gomes birthday party at 7 o'clock. He started from his home
Nataliya [291]

Answer:

Kindly check explanation

Explanation:

Given the following :

Distance between Ranjan and Gomez house = 9km

Start time = 6'o clock

Time he arrived at Gomes house = 7'o clock

Time used for chatting during this period = 5 minutes

Distance covered in 40minutes = 6km

Time spend for second part of the journey = (60 - (40 + 5))minutes = 15 minutes = 15/60 = 0.25 hoir

Distance covered during second part of journey = 9 - 6 = 3 km

Speed = distance / time

Speed = 3km / 0.25 hr = 12km/hr

Average speed for entire journey :

First part :

Speed = distance / time

Time = 40 minutes = (40/60) = 0.667 hour

Speed = 6km / 0.667 hour = 9km / hr

Average speed :

(First part + second part) / 2

(9km/hr + 12km/hr) / 2 = 21km/hr / 2 = 10.5 km/hr

= 10.5km/hr

5 0
3 years ago
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