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vladimir2022 [97]
4 years ago
9

What is the action force of a flying bird?

Physics
1 answer:
kirill [66]4 years ago
7 0
It is called the reaction force of a bird flying.
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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
A mechanic uses a wrench that is 22 cm long to tighten a spark plug. If the mechanic exerts a force of 58 N to the end of the wr
Licemer1 [7]

The maximum torque she can apply to the spark plug is 12.76 Nm.

<h3>What is torque?</h3>

Torque is the rotational equivalent of linear force obtained by taking the product of force and the radius of the applied force.

τ = Fr

where;

  • F is the applied force
  • r is the radius

The maximum torque she can apply to the spark plug is calculated as follows;

τ = 58 x (0.22)

τ = 12.76 Nm

Thus, the maximum torque she can apply to the spark plug is 12.76 Nm.

Learn more about torque here: brainly.com/question/14839816

4 0
2 years ago
From the Bohr Model diagram, assuming the atom is neutral, what element is represented by the diagram? A) Argon B) Chlorine Elim
viktelen [127]

The actual answer is B) Chlorine

According to the Bohr Model diagram, the atom has seventeen electrons. This makes it Chlorine.


Sorry if i'm late!!

5 0
4 years ago
Read 2 more answers
A 1200-kg car accelerates it’s speed from 4 m/s to 10 m/s in 3 seconds. Find the car average speed , the car acceleration , the
dsp73

Answer:

Given: mass 1200kg

initial velocity: 4m/s

finial velocity: 10 m/s

time 3 sec

then

speed; initial velocity + final velocity/2

4+10/3

: 4.66m/s2

8 0
3 years ago
Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.
rjkz [21]

Answer:

Momentum of block B after collision =-50\ kg\ ms^{-1}

Explanation:

Given

Before collision:

Momentum of block A = p_{A1}= -100\ kg\ ms^{-1}

Momentum of block B = p_{B1}= -150\ kg\ ms^{-1}

After collision:

Momentum of block A = p_{A2}= -200\ kg\ ms^{-1}

Applying law of conservation of momentum to find momentum of block B after collision p_{B2}.

p_{A1}+p_{B1}=p_{A2}+p_{B2}

Plugging in the given values and simplifying.

-100-150=-200+p_{B2}

-250=-200+p_{B2}

Adding 200 to both sides.

200-250=-200+p_{B2}+200

-50=p_{B2}

∴ p_{B2}=-50\ kg\ ms^{-1}

Momentum of block B after collision =-50\ kg\ ms^{-1}

6 0
3 years ago
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