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3 years ago

Both cars start from the same point. Which describes the motion shown?

Physics

2 answers:

Alekssandra [29.7K]3 years ago

8 0

C) The longer the line, the greater the magnitude of the vector. As for the direction, just think of a compass.

timama [110]3 years ago

7 0

C: one car travels south at 4 m/s. The other car travels east at 2 m/s

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A device that converts electrical energy into kinetic energy to turn an axle?

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A motor is the devise

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A friend throws a coin into a pool. you close your eyes and dive toward the spot where you saw it from the edge of the pool. whe

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Monochromatic light of wavelength λ=136.8μ m is shone at normal incidence through a thin film of thickness t resting atop a full

inn [45]

Answer:

1.8 × 10⁻⁸ Hm

Explanation:

Given that:

The refractive index of the film = 19

The wavelength of the light = 136.8 μ m

The thickness can be calculated by using the formula shown below as:

$Thickness=\frac {\lambda}{4\times n}$

Where, n is the refractive index of the film

${\lambda}$ is the wavelength

So, thickness is:

$Thickness=\frac {136.8\ \mu\ m}{4\times 19}$

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

7 0

2 years ago

iron ball weight 400 gram inside water when it is completely impressed in water 53 gram water is displaced what will be the weig

Alika [10]

Answer:

453 gm

Explanation:

<u>Immersed </u>objects are buoyed up by force equal to mass of displaced liquid

400 + 53 = 453 gm in air

8 0

1 year ago

A 110-kg object and a 410-kg object are separated by 3.80 m.

aniked [119]

Answer:

a) Fₙ = 2,273 10⁻⁷ N and b) x₃ = 1,297 m

Explanation:

This problem can be solved using the law of universal gravitation and Newton's second law for the equilibrium case. The Universal Gravitation Equation is

F = G m₁ m₂ / r₁₂²

a) we write Newton's second law

Σ F = F₁₃ - F₃₂

Body 1 has mass of m₁ = 110 kg and we will place our reference system, body 2 has a mass of m₂ = 410 kg and is in the position x₂ = 3.80 m

Body 3 has a mass of m₃ = 41.0 kg and is in the middle of the other two bodies

x₃ = (x₂-x₁) / 2

x₃ = 3.80 / 2 = 1.9 m

Fₙ = -G m₁ m₃ / x₃² + G m₃ m₂ / x₃²

Fₙ = G m₃ / x₃² (-m₁ + m₂)

Calculate

Fₙ = 6.67 10⁻¹¹ 41.0 / 1.9² (- 110 + 410)

Fₙ = 2,273 10⁻⁷ N

Directed to the right

b) find the point where the force is zero

The distance is

x₁₃ = x₃ - 0

x₃₂ = x₂ -x₃= 3.8 -x₃

We write the park equation net force be zero

0 = - F₁₃ + F₃₂

F₁₃ = F₃₂

G m₁ m₃ / x₁₃² = G m₃ m₂ / x₃₂²

m₁ / x₁₃² = m₂ / x₃₂²

Let's look for the relationship between distances, substituting

m₁ / x₃² = m₂ / (3.8 - x₃)²

(3.8 - x₃) = x₃ √ (m₂ / m₁)

x₃ + x₃ √ (m₂ / m₁) = 3.8

x₃ (1 + √ m₂ / m₁) = 3.8

x₃ = 3.8 / (1 + √ (m₂ / m₁))

x₃ = 3.8 / (1 + √ (410/110))

x₃ = 1,297 m

When body 3 is in this position the net force on it is zero

8 0

2 years ago