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11111nata11111 [884]

3 years ago

14

An object with mass 110 kg moved in outer space. When it was at location < 13, -18, -2 > its speed was 19.5 m/s. A single

constant force < 250, 390, -220 > N acted on the object while the object moved to location < 19, -23, -5 > m. What is the speed of the object at this final location

1 answer:

Serhud [2]3 years ago

6 0

Answer:

v = 21.4m/s

Explanation:

Given r1 = < 13, -18, -2 > m,

F = < 250, 390, -220 > N, r2 = < 19, -23, -5 > m

Where r1 and r2 are position vectors in space and F is the constant force vector in space.

r2 - r1 = < 19–13, -23–(-18), -5–(-2) > m

Δr = < 6, -5, -3 > m

F = m×a

a = F/m

m = 110kg (scalar quantity)

a = < 250/110, 390/110, -220/110 >

a = < 2.27, 3.55, -2 > m/s²

Since the force is constant, the acceleration is also constant. Therefore the equations of constant acceleration motion apply here.

V² = u² +2aΔr

The magnitude of the acceleration is calculated as follows

a = √(2.27² + 3.55² +(-2)²)

a = √(21.7554)

a = 4.66m/s²

Magnitude of Δr,

Δr = √(6² + (-5)² +(-3)²) = √(70) = 8.37m

Having calculated the magnitude of the acceleration and displacement, we can now calculate the final velocity.

u = initial velocity = 19.5m/s

v² = 19.5² + 2×4.66×8.37

v² = 458.25

v = √(458.25)

v = 21.4m/s

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Answer:

La escala del termómetro ''A'' es grados Celsius.

La escala del termómetro ''B'' es grados Fahrenheit.

Explanation:

Para hallar en qué escalas están los termómetros partimos de que la mezcla a la cuál se midió su temperatura mantuvo su temperatura constante.

Esto quiere decir que los termómetros están expresando la misma temperatura pero en una escala distinta.

Sabemos que dada una temperatura en grados Celsius ''C'' si la queremos convertir a grados Fahrenheit ''F'' debemos utilizar la siguiente ecuación :

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Ahora, si reemplazamos y asumimos que la temperatura de 18° es en grados Celsius, entonces si reemplazamos $C=18$ en la ecuación (I) deberíamos obtener $F=64.4$ ⇒

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3 years ago

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Answer:

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notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

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so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

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juin [17]

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Sunny_sXe [5.5K]

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