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MArishka [77]
2 years ago
15

Need help ASAP, 1 MC

Physics
1 answer:
stepladder [879]2 years ago
3 0

Answer:

The first one is the only one that is true all the time

Explanation:

The second one may be true if friction is high enough.

The other three are false all the time

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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
3 years ago
Some expressways are curved, banked and designed to maximize ___________ at higher speeds.
Liono4ka [1.6K]

Answer:

Safety

Explanation:

Expressways are banked to resist centifugal action

6 0
3 years ago
Maggie's truck decelerates from 18.00 m/s to rest in 3.30 s. If the Maggie's mass is 100.0 kg and the truck is 1810 kg, calculat
Free_Kalibri [48]

Answer:

look at my Explanation

Explanation:

If the Maggie's mass is 100.0 kg and the truck is 1810 kg, calculate the magnitude of the net (unbalanced) force that can cause the acceleration.

8 0
3 years ago
Two vectors are shown. Which statement best compares the vectors?
DiKsa [7]

Answer:

B

Explanation:

Because vector Y is longer than vector X

so when you take a magnitude( without minus ) each other

you see that Y>X

8 0
3 years ago
Read 2 more answers
Aluminum has a density of
agasfer [191]

<h3><u>Volume is 0.1848 m³</u></h3><h3 />

Explanation:

<h2>Given:</h2>

m = 49.9 kg

ρ = 270 kg/m³

<h2>Required:</h2>

volume

<h2>Equation:</h2>

ρ \:= \:\frac{m}{v}

where: ρ - density

m - mass

v - volume

<h2>Solution:</h2>

Substitute the value of ρ and m

ρ \:= \:\frac{m}{v}

270\: kg/m³\:= \:\frac{49.9\:kg}{v}

(v)\:270\: kg/m³\:= \:49.9\:kg

v\:= \:\frac{49.9\:kg}{270\: kg/m³}

v\:= \:0.1848\:m³

<h2>Final Answer:</h2><h3><u>Volume is 0.1848 m³</u></h3>
3 0
3 years ago
Read 2 more answers
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