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Butoxors [25]
3 years ago
9

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]3 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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Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

(c) F = 1010.62 N = 1.01 KN

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

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k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

<u>k = 1684.38 N/m = 1.684 KN/m</u>

<u></u>

<u>(</u>c<u>)</u>

Applying Hooke's Law with:

Δx  = 0.6 m

F = (1684.38 N/m)(0.6 m)

<u>F = 1010.62 N = 1.01 KN</u>

<u></u>

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

<u>Vi = 0.105 m/s</u>

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3 years ago
You measure the velocity of a drag racer that accelerates with constant acceleration. You want to plot the data and determine th
Ivenika [448]

Answer:

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2 years ago
If the work required to stretch a spring 1 ft beyond its natural length is 12 ft-lb, how much work is needed to stretch the same
Angelina_Jolie [31]

Answer:W=\frac{3}{4} ft-lb

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Given

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