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2 years ago

What is the weight of an object with a mass of 19 kg?

Physics

2 answers:

Natalija [7]2 years ago

7 0

Answer:

Every 2.2 kg is 1 pound. So mulitply 19 * 2.2. It's gonna be equal to 41.8

Explanation:

marysya [2.9K]2 years ago

7 0

Statement:

Mass of an object is 19 Kg.

To find out:

The weight of the object.

Solution:

Mass of the object = 19 Kg

Acceleration due to gravity = $\sf9.8 {m \: {s}^{ - 2} }$

We know, weight = mass × acceleration

Therefore, weight of the object

= (19 × 9.8) N

= 186.2 N

Answer:

The weight of the object is 186.2 N.

Hope you could understand.

If you have any query, feel free to ask.

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What is the kinetic energy of a 1130 kg truck that is moving with a velocity of 40 m/s?

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3 years ago

A body oscillates with simple harmonic motion along the x axis. Its displacement varies with time according to the equation x =

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Answer:

θ = (7π / 3) rad

Explanation:

given,

displacement of simple harmonic motion along x-axis

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x = 5 sin (π t + π/3 )

general equation of simple harmonic motion

x = A sin θ

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θ = π t + π/3

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3 years ago

We are designing a crude propulsion mechanism for a science fair demonstration. One of our team members stands on a skateboardth

Scrat [10]

Answer:

greater speed will be obtained for the elastic collision,

Explanation:

To answer this exercise we must find the speed that the sail acquires after each impact.

Let's start by hitting a ball of clay.

The system is formed by the candle and the clay balls, therefore the forces during the collision are internal and the moment is conserved.

initial instant. before the crash

p₀ = m v₀

where m is the mass of the ball and vo its initial velocity, we are assuming that the candle is at rest

final instant. After the crash

the mass of the candle is M

p_f = (m + M) v

the moment is preserved

p₀ = p_f

m v₀ = (m + M) v

v = $\frac{m}{m+M} \ v_o$

for when n balls have collided

v = $\frac{m}{n \ m + M}$ v₀

Now let's analyze the case of the bouncing ball (elastic)

initial instant

p₀ = m v₀

final moment

p_f = m v_{1f} + M v_{2f}

p₀ = p_f

m v₀ = m v_{1f} + M v_{2f}

m (v₀ - v_{1f}) = M v_{2f}

this case corresponds to an elastic collision whereby the kinetic energy is conserved

K₀ = K_f

½ m v₀² = ½ m v_{1f}² + ½ M v_{2f}²

v₁ = v_{1f} v₂ = v_{2f}

m (v₀² - v₁²) = M v₂²

let's use the identity

(a² - b²) = (a + b) (a-b)

we write our equations

m (v₀ - v₁) = M v₂ (1)

m (v₀ - v₁) (v₀ + v₁) = M v₂²

let's divide these equations

v₀ + v₁ = v₂

Let's look for the final speeds

we substitute in equation 1

m (v₀ - v₁) = M (v₀ + v₁)

v₀ (m -M) = (m + M) v₁

v₁ = $\frac{m-M}{m + M}$ v₀

we substitute in equation 1 to find v₂

$\frac{M}{m}$ v₂ = v₀ - $\frac{m-M}{m+M}$ v₀

v₂ = $\frac{m}{M} ( 1 - \frac{m-M}{m+M} ) \ v_o$

v₂ = $\frac{m}{M} ( \frac{2M}{m+M} ) \ \ v_o$

v₂ = $\frac{2m}{m +M} \ v_o$

Let's analyze the results for inelastic collision with each ball that collides with the sail, the total mass becomes larger so the speed increase is smaller and smaller.

In the case of elastic collision, the increase in speed is constant with each ball since the total mass remains invariant.

Consequently, greater speed will be obtained for the elastic collision, that is, the ball will bounce.

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