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mote1985 [20]
4 years ago
11

Two Forces are exerted on an Object In a Vertical Direction: A 20N force downwards, and 10 N force upwards. The mass of the obje

ct is 25KG. WHAT IS THE NET FORCE EXPRESSION?
WORTH 30 POINTS!!! HELP ME PLZ.
Physics
1 answer:
SVEN [57.7K]4 years ago
6 0
When you add up (20N down) and (10N up), 
you get a sum of (10N down).

The mass of the object has no effect on the forces.


Now ... I see 5 points for the answer.
Where are the other 25 coming from ?
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HELP !! Maura is deciding which hose to use to water her outdoor plants. Maura noticed that the water coming out of her garden h
MA_775_DIABLO [31]
THE GREEN HOSE:
Define the (x,y) coordinate at a height of 4 feet up from the ground to match where Majra is holding the green hose.
This means that the equation for the green hose is of the form
y = a(x - h)² + 4          (1)

Because water from the green hose lands on the ground 10 feet from where Majra is standing, therefore
y(10) = -4                    (2)

Because the curve passes through (0,0), therefore
ah² + 4 = 0
ah² = - 4                     (3)

To satisfy (2), obtain
a(10 - h)² + 4 = -4
a(10 - h)² = - 8            (4)

Divide (3) by (4).
h²/(10-h)² = 1/2
2h² = (10 - h)² = 100 - 20h + h²
h² + 20h - 100 = 0             

Solve with the quadratic formula.
x = 0.5[-20 +/- √(8400)] = 4.142, - 24.142
Reject the negative solution.
The vertex is at (4.142, 4).

From (3), obtain
a = -4/4.142² = -0.2332

The equation for the green hose is
y = 0.2332(x - 4.142)² + 4

THE RED HOSE
The red hose has a vertex at (3,7), according to the equation y = -(x-3)² + 7.

A graph of y(x) for both hoses is shown in the attached figure.

Answers:
a. The red hose will throw the water higher. 

b. The equation for the green hose is
     y = -0.2332(x - 4.124)² + 4,
     with the origin at a height of 4 feet above ground level.

c. The domain for the green hose that makes sense is 0 ≤ x ≤ 10 feet.
     The corresponding range is -4 ≤ y ≤ 4 feet.


3 0
3 years ago
Distance
34kurt

I honestly dont know

8 0
3 years ago
The resultant force acting on an object of mass 5.0kg varies with time as shown. the object is initially at rest.
il63 [147K]
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8 0
3 years ago
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
Balanced Forces acting on an object will not change the object's motion. Unbalanced Forces acting on an object will change the c
ale4655 [162]

Answer:

TRUE

Explanation:

The answer is true.

Balance forces acting on a body will not change the motion of the body because the body experiences no net resultant force in one direction. When any body experiences equal forces with opposite directions, the net force or the resultant force experience by the body is zero.

In case of an unbalanced forces, there is a net force acting in one direction and so it causes the body to change in its state of motion in the direction of the net force.

4 0
3 years ago
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