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Bogdan [553]
3 years ago
5

Is this diagram important in exams?

Physics
1 answer:
pickupchik [31]3 years ago
6 0
Yes it might come as a blank diagram and ask you to fill in the blanks or it might ask you to say how much energy is wasted
                                       good luck on your exam
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How is work defined in physics
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Explanation:

Work is the dot product of the force and displacement vectors.

W = F · d

In other words, it is the force times the parallel component of the distance.

W = F d cos θ, where θ is the angle between the force and distance.

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The force a magnet exerts on another magnet, on iron or a similar metal, or on moving charges is
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A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a
Stella [2.4K]

Answer:

y = 67.6 feet,   y = 114.4/ (22 - 3t)

Explanation:

For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram

Large triangle Projector up to the screen

         tan θ = y / L

For the small triangle. Projector up to the person

         tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

         y₀ / (L -d) = y / L

         y = y₀  L / (L-d)

The distance from the screen (d), we look for it with kinematics

         v = d / t

        d = v t

we replace

         y = y₀ L / (L - v t)

         y = 5.2 22 / (22 - 3 t)

         y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

For the suggested distance the shadow has a height of

           y = 114.4 / (22-13)

           y = 67.6 feet

7 0
3 years ago
(b) The distance of mass from mass A if there is no gravitational force acted on C
shepuryov [24]

Answer:

(a) The force, acting on object 'C' is approximately 2.66972 × 10⁻¹⁰ Newtons

(b) The distance of 'C' from 'A', in the direction particle 'B' if there is no  meters gravitational force acting on 'C' is appromimately 0.829 meters or 1.877 meters

Explanation:

The given parameters are;

The mass of particle, A, m₁ = 2 kg

The mass of particle, B, m₂ = 0.3 kg

The mass of particle, C, m₃ = 0.05 kg

The distance between particle 'A' and particle 'B', r₁ = 0.15 m

The distance between particle 'B' and particle 'C', r₂ = 0.05 m

(a) The gravitational force, 'F', is given as follows;

F =G \times \dfrac{m_{1} \times m_{2}}{r^{2}}

Where;

F = The force between the two masses

G = The gravitation constant = 6.67430 × 10⁻¹¹ N·m²/kg²

m₁ = The mass of object 1

m₂ = The mass of object 2

If 'C' is placed at 0.05 m from 'B', we have;

F₂₃ =  6.67430 × 10⁻¹¹ × 0.05 × 0.3/(0.05²) ≈ 4.00458 × 10⁻¹⁰

The gravitational force between force between particle 'B' and particle 'C', F₂₃ = 4.00458 × 10⁻¹⁰ N (towards the right)

F₁₃ =  6.67430 × 10⁻¹¹ × 0.05 × 2/(0.1²) ≈ × 10⁻¹⁰

The gravitational force between force between particle 'A' and particle 'B', F₁₃ = 6.6743 × 10⁻¹⁰ N (towards the left)

The force, 'F', acting on object 'C' = F₁₃ - F₂₃

F = (6.6743 - 4.00458) × 10⁻¹⁰ = 2.66972 × 10⁻¹⁰ N

The force, acting on object 'C' ≈ 2.66972 × 10⁻¹⁰ N

(b), When there is no gravitational force acting on 'C', let the distance of 'C' from 'A' = x

We have;

F₂₃ = F₁₂

F_{23} =G \times \dfrac{m_{1} \times m_{2}}{r_1^{2}} = F_{13} =G \times \dfrac{m_{1} \times m_{3}}{r_2^{2}}

By plugging in the values and removing like terms, we get;

\dfrac{0.3 \times 0.05}{(1.15 - x)^{2}}  = \dfrac{2 \times 0.05}{x^2}

(1.15 - x)² × 2 × 0.05 = 0.3 × 0.05 × x²

0.1·x² - 0.23·x + 1.3225 = 0.015·x²

0.1·x² - 0.23·x + 1.3225 - 0.015·x² = 0

0.085·x² - 0.23·x + 0.13225= 0

x = (0.23± √((-0.23)² - 4 × 0.085 × ( 0.13225)))/(2 × 0.085))

x ≈ 0.829, or x ≈ 1.877

Therefore, the distance of 'C' from 'A', if there is no gravitational force acting on 'C', x ≈ 0.829 m, or x = 1.877 m, in the direction of 'B'

7 0
2 years ago
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