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agasfer [191]
3 years ago
5

Multiple-Concept Example 13 presents useful background for this problem. The cheetah is one of the fastest-accelerating animals,

because it can go from rest to 27 m/s (about 60 mi/h) in 4.0 s. If its mass is 110 kg, determine the average power developed by the cheetah during the acceleration phase of its motion. Express your answer in (a) watts and (b) horsepower.
Physics
1 answer:
erik [133]3 years ago
5 0

Answer:

Explanation:

given,

speed of cheetah = 27 m/s

time taken  = 4 s

mass of cheetah = 110 kg

a = \dfrac{27-0}{4}

a = 6.75 m/s²

v² = u² + 2 as

27² = 0 + 2 × 6.75× s

s = 54 m

a) power = \dfrac{work\ done}{t}

power = \dfrac{F.s}{t}

power = \dfrac{m.a.s}{t}

power = \dfrac{110\times 6.75\times 54}{4}

power = 10,023\ watt

b) power = \dfrac{10,023}{746}

   power = 13.46\ horsepower

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Answer:

Ep= 3.8 10⁵ N/C

Explanation:

Conceptual analysis

The electric field at a point P due to a point charge is calculated as follows:

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E: Electric field in N/C

q: charge in Newtons (N)

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d: distance from charge q to point P in meters (m)

Equivalence

1nC= 10⁻⁹C

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Calculation of the electric fieldsat the midpoint (P) between the two charges

Look at the attached graphic:

E₁: Electric Field at point ;Due to charge q₁. As the charge q₁ is positive negative (q₁+), the field leaves the charge .

E₂: Electric Field at point : Due to charge q₂. As the charge q₂ is negative (q₂-) ,the field enters the charge

E₁ = k*q₁/d₁² = 9*10⁹ *7.5  *10⁻⁹/ ( 0.015 )² = 3*10⁵ N/C

E₂ = k*q₂/d₂²= 9*10⁹ *2*10⁻⁹/( 0.015 )² = 0.8*10⁵ N/C

The electric field at a point P due to several point charges is the vector sum of the electric field due to individual charges.

Ep= E₁ + E₂  

Ep= 3*10⁵ N/C +  0.8*10⁵ N/C

Ep= 3.8 10⁵ N/C

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