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JulijaS [17]
3 years ago
15

Find the value of F1 + F2 + F3.​

Physics
1 answer:
Dovator [93]3 years ago
8 0

Answer:

F = 0.78[N]

Explanation:

The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.

<u>For F₁</u>

<u />F_{y}=2[N]<u />

<u>For F₂</u>

F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]

<u>For F₃</u>

<u />F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]<u />

Now we can sum each one of the forces in the given axes:

F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]

Now using the Pythagorean theorem we can find the total force.

F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]

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  • Answer:

<em>500 sec</em>

<em>8 min 20 sec</em>

  • Explanation:

<em>Hi there !</em>

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<em>8 m ................ 1 s </em>

<em>4000 m ........ x s</em>

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A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei
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Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

F = (150\ N)Sin\ 20^o

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2 years ago
a baseball is traveling (=20m/s) and us hit by a bat. it leaves the bat traveling (-30 m/s). What is the change in the velocity?
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A truck is carrying a steel beam of length 13.0 m on a freeway. An accident causes the beam to be dumped off the truck and slide
PIT_PIT [208]

Answer:

emf= 9.88 \times 10^{-3} T

Explanation:

Given:

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  • speed of the beam, v=20\,m.s^{-1}
  • magnitude of the vertical magnetic field, B=38\times 10^{-6} T

According to the Faraday's law the emf induced in a rod passing transversely through a magnetic field is given as:

emf= l.v.B

emf=13\times 20\times 38\times 10^{-6}

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3 years ago
Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea
melisa1 [442]

Answer:

m = 1.26*10²⁵ kg.

Explanation:

Assuming that the mass of the stone is much smaller than the mass of the planet, we can get the mass, applying the Universal Law of  Gravitation to both masses, as follows:

Fg = G* ms* mp / rp²

Now, if we apply Newton's 2nd Law to the mass of the stone, we can get the gravitational acceleration, as follows:

Fg = ms*a = ms*g ⇒ g = G*mp / rp²

First of all, we need to get the value of g.

Assuming that this acceleration is constant, we can appy the kinematic equations to this situation.

We know that the stone is thrown upward with an initial velocity vo = 15 m/s.

At the highest point in the trajectory, just before of changing direction, the stone comes momentarily to a stop.

At this point, applying the definition of acceleration, we can write:

vf = vo -g*t ⇒ 0 = vo -gt ⇒ g = vo/t (1)

We have the total time since the stone was thrown upwards, not the one used for the upward trajectory.

It can be showed, using the expression for the displacement (which is the same in both directions) that the time used for going up, it's the same used to go down, so the time that we need to put in (1). is just the half of the total time.

So, replacing in (1) we get the value of g, as follows:

g = 15 m/s / 4.5 s = 3.33 m/s²

Now, we can replace this value in the equation that gives us g based in the Universal Law of Gravitation, as follows:

g=G*mp / rp² (2)

Before solving for mp, however, we need to get the value of the radius of the planet.

Assuming that it's a perfect sphere, we can get this value from the value of the circumference at the planet's equator:

rp = 2*π*rp / 2*π ⇒ rp = 1.0*10⁵ km / 2*π = 15,915 km.

With this value for  rp, we can solve (2) for mp, as follows:

mp= g*rp² / G = 3.33 m/s² * (15,915 km)² / 6,67*10⁻¹¹ N.m²/kg²

mp = 1.26*10²⁵ kg.

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