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2 years ago

A 104 horse-power suv speeds up from 21 m/s to 29 m/s in 3 seconds, what is the mass of the suv?

Physics

1 answer:

AleksAgata [21]2 years ago

4 0

Answer:

Correct answer: m = 1,160 kg

Explanation:

Given:

P = 104 hp = 77.48 kW = 77.48 · 10³ W

V₁ = 21 m/s Initial speed (velocity)

V₂ = 29 m/s Final speed (velocity)

t = 3 seconds The time interval during which the work was performed

m = ?

The formula for calculating power is:

P = A / Δt

Since work is a measure of energy change in this case kinetic:

P = ΔEk / Δt = (Ek₂ - Ek₁) / Δt = (m V₂²/ 2 - m V₁²/ 2) / Δt

P = m · (V₂² - V₁²) / 2 · Δt ⇒ m = (2 · P · Δt) / (V₂² - V₁²)

m = (2 · 77.48 · 10³ · 3) / (29² - 21²) = 464.88 · 10³ / (841 - 441)

m = 464.88 · 10³ / 400 = 1.16 · 10³ = 1,160 kg

m = 1,160 kg

God is with you!!!

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An object of mass, m1 with a velocity, v1 collides with another object at rest (v2 = 0) with a mass, m2. After the collision, m1

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$v"_{1} = v_{1} tan$ Θ

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Explanation:

Applying the law of conservation of momentum, we have:

Δ $p_{x = 0}$

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$m_{1}v_{1} = m_{2}v"_{2} cos$ Θ (Equation 1)

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$p_{y} = p"_{y}$

$0 = m_{1} v"_{1} - m_{2} v"_{2} sin$ Θ (Equation 2)

From Equation 1:

$v"_{2} = \frac{m_{1}v_{1}}{m_{2}cos}$ Θ

From Equation 2:

$m_{2} v"_{2}$ sinΘ = $m_{1} v_{1}$

$v"_{1} = \frac{m_{2} v"_{2}sinΘ}{m_{1} }$

Replacing Equation 3 in Equation 4:

$v"_{1}=\frac{m_{2}\frac{m_{1}v_{1}}{m_{2}cosΘ}sinΘ}{m_{1}}$

$v"_{1}=v_{1}\frac{sinΘ}{cosΘ}$

$v"_{1}=v_{1}tan$ Θ (Equation 5)

And we found Θ from the Equation 5:

tanΘ= $\frac{v"_{1}}{v_{1}}$

Θ= $tan^{-1}(\frac{v"_{1}}{v_{1}})$

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3 years ago

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Answer:

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