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kiruha [24]
3 years ago
13

Calculate the magnitude of the following vector components in OX and OY directions​

Physics
1 answer:
12345 [234]3 years ago
3 0

Answer:

Very easy question

For x-component, we use Fₓ = Fcosα

For y-component, we use Fy = Fsinα

The major problem is with angle α.

Here α = 90+30

Because the angle is measured from +ve X-axis.

By putting values if you know co-ratios that how Sin and cos Convert to each other then you have got it.

So for further calculation,

Fₓ = Fcos(90+30) = 25(-sin30)

Fₓ = -25(1/2) = 25/2 = 12.5  - will be eliminate because force in magnitude is not "-ve"

As it is

Fy = Fsin(90+30) = Fcos30

Fy = 25 (√3 /2) = 25√3/2

......................................................................................

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Answer:

3

Explanation:

-The formula for kinetic energy is the following:

K.E.=\frac{1}{2}mv^{2}

where:

K.E. is kinetic energy

m is mass

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Here we can see that kinetic energy is proportional to the velocity squared, so if the speed or velocity of the scooter doubles then this would quadruple its kinetic energy.

-The acceleration is the rate of change of velocity per unit of time. A scooter could double its speed with a constant acceleration, so an increase in speed doesn't necessarily mean an increase in acceleration.

-The formula for momentum is the following:

p=mv

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p is momentum

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Here we can see that the momentum is directly proportional to the velocity, so if the scooter where to double its velocity then it would also double its momentum.

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4 years ago
Car A of mass 1200kg traveling at 10m/s , collided in too the back of the car B, which is stationary. Following the collision,.
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Answer:

Car B has a mass of 800 kg.

General Formulas and Concepts:

<u>Momentum</u>

Law of Conservation of Momentum: \displaystyle m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_{f}

Explanation:

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] m₁ = 1200 kg

[Given] v₁i = 10 m/s

[Solve] m₂

[Given] v₂i = 0 m/s

[Given] vf = 6 m/s

<u>Step 2: Solve for m₂</u>

  1. Substitute in variables [Law of Conservation of Momentum]:                       (1200 kg)(10 m/s) + m₂(0 m/s) = (1200 kg + m₂)(6 m/s)
  2. Multiply:                                                                                                             12000 kg · m/s = (1200 kg + m₂)(6 m/s)
  3. Isolate m₂ term:                                                                                                2000 kg = 1200 kg + m₂
  4. Isolate m₂:                                                                                                         800 kg = m₂
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<h3>Electric field on the master charge</h3>

E = kq/r²

where;

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E = (9 x 10⁹ x 0.63)/(0.75²)

E = 1.008 x 10¹⁰ N/C

<h3>Force on the test charge</h3>

F = Eq

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F = (1.008 x 10¹⁰) x (0.5)

F = 5.04 x 10⁹ N

Thus, the magnitude of the electric field on the master charge is 1.008 x 10¹⁰ N/C, and the force on the test charge is 5.04 x 10⁹ N.

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Answer:

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