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adelina 88 [10]
2 years ago
5

2) A 0.4kg ball moves in horizontal circle of radius 3 m at speed of 100m/s. What

Physics
1 answer:
sweet-ann [11.9K]2 years ago
6 0

Answer:

F = 1300 N

Explanation:

F = mv²/R = 0.4(100²)/3 = 1333.3333...

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Electrical current in a wire
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The electrical current in a wire could be from multiple things.  This is how we wire our house.  When the electrical current flows through the wire, it can power things like lights and appliances.  The things that flow through a wire are called electrons.
6 0
3 years ago
Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge
Vanyuwa [196]

Answer:

a) F_net = 30.47 N ,   θ = 10.6º

b)  Fₓ = 29.95 N

Explanation:

For this exercise we use coulomb's law

          F₁₂ = k k \frac{ q_{1}  \  q_{2} }{ r^{2} }

the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.

As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces

X axis

        Fₓ = F_{bc x}

Y axis  

       F_{y}Fy = F_{ab} - F_{bc y}

let's find the magnitude of each force

     F_{ab} = 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²

      F_{ab} = 2.82 10¹ N

      F_{ab} = 28.2 N

   

      F_{bc} = 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²

      F_{bc} = 3.75 10¹  N

       F_{bc} = 37.5 N

let's use trigonometry to decompose this force

      tan θ = y / x

      θ = tan⁻¹ and x

       θ= tan⁻¹ ¾

      θ = 37º

let's break down the force

      sin 37 = F_{bcy} / F_{bc}

      F_{bcy} = F_{bc} sin 37

      F_{bcy} = 37.5 sin 37

      F_{bcy} = 22.57 N

      cos 37 = F_{bcx} /F_{bc}

      F_{bcx} = F_{bc} cos 37

      F_{bcx} = 37.5 cos 37

      F_{bcx} = 29.95 N

let's do the sum to find the net force

X axis

        Fₓ = 29.95 N

Axis y

        Fy = 28.2 -22.57

        Fy = 5.63 N

we can give the result in two ways

a)  F_net = Fₓ i ^ + F_{y} j ^

    F_net = 29.95 i ^ + 5.63 j ^

b) in the form of module and angle

let's use the Pythagorean theorem

    F_net = \sqrt{ F_{x}^2 + F_{y}^2 }

    F_net = √(29.95² + 5.63²)

     F_net = 30.47 N

we use trigonometry for the direction

      tan θ= \frac{ F_{y}  }{  F_{x} }

       

      θ = tan⁻¹ \frac{ F_{y}  }{  F_{x} }

      θ = tan⁻¹ (5.63 / 29.95)

      θ = 10.6º

3 0
3 years ago
How many miles of MgCI2 are there in 339g of compound?
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Find the Molar Mass of MgCl2

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Find the mols in 339 grams

1 mole = 95 grams

x mol = 339

Solve

339 = 95x   Divide by 95

339/95 = x

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