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Crazy boy [7]
4 years ago
9

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

for it can go from rest to 30.4 m/s in 5.40 s. If its mass is 108 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:
Andre45 [30]4 years ago
7 0

Answer:

9241.6 W or 12.39318 hp

Explanation:

u = Initial velocity = 0

v = Final velocity

m = Mass

t = Time taken

Energy

KE=\frac{1}{2}m(v^2-u^2)\\\Rightarrow KE=\frac{1}{2}108(30.4^2-0^2)\\\Rightarrow KE=49904.64\ Joules

Power

P=\frac{KE}{t}\\\Rightarrow P=\frac{49904.64}{5.4}\\\Rightarrow P=9241.6\ W

Converting to hp

1\ W=\frac{1}{745.7}\ hp

\\\Rightarrow 9241.6\ W=\frac{9241.6}{745.7}\ hp=12.39318\ hp

The power developed by the cheetah is 9241.6 W or 12.39318 hp

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An L−R−C series circuit has 91.5 Ω , and the amplitude of the voltage across the resistor is 36.0 V . What is the average power
sesenic [268]

Answer:

14.16 W

Explanation:

We have given impedance Z=91.5 ohm

Voltage across the resistor = 36 V

We have to find the power delivered by the source

We know that according to telegens theorem , the absorbed = power delivered

So the voltage across the resistor = 36 V

Current in the RLC circuit i=\frac{V}{Z}=\frac{36}{91.5}=0.3934A this current will be same in the whole circuit as it is a series RLC circuit

Now the power absorbed by the resistor P=i^2R=0.3934^2\times 91.5=14.16W

As the power absorbed = power delivered

So power delivered by the source =14.16 W

3 0
3 years ago
An ant travels 2.78 cm [W] and then turns and travels 6.25 cm [S 40 degrees E]. What is the ant's total displacement?
Sunny_sXe [5.5K]
Answer is 6.84 approx
reason:-
            (2.78^2+6.25^2)^1/2=6.84 approx
4 0
3 years ago
A batter hits a fly ball which leaves the bat 0.89 m above the ground at an angle of 62 ∘ with an initial speed of 29 m/s headin
KatRina [158]

consider the motion in Y-direction

v₀ = initial velocity = 29 Sin62 = 25.6 m/s

a = acceleration = - 9.8 m/s²

t = time of travel

Y = vertical displacement = - 0.89 m

using the equation

Y = v₀ t + (0.5) a t²

- 0.89 = (25.6) t + (0.5) (- 9.8) t²

t = 5.3 sec


consider the motion along the horizontal direction :

v₀ = initial velocity = 29 Cos62 = 13.6 m/s

a = acceleration = 0 m/s²

t = time of travel = 5.3 sec

X = horizontal displacement =?

using the equation

X = v₀ t + (0.5) a t²

X = (13.6) (5.3) + (0.5) (0) t²

X = 72.1 m


d = distance traveled by the center fielder to catch the ball = 107 - x = 107 - 72.1 = 34.9 m

t = time taken = 5.3 sec

v = speed of center fielder

using the equation

v = d/t

v = 34.9/5.3

v = 6.6 m/s

3 0
3 years ago
A runner exerts a net force of 335 n to accelerate at a rate of 2.5 m/s what is the runners mass
Olegator [25]
F=ma
M=F/a
M=335/2.5
M=134 kg
8 0
3 years ago
A package of aluminum foil contains 50. ft2 of foil, which weighs approximately 6.0 oz. Aluminum has a density of 2.70 g/cm3. Wh
SIZIF [17.4K]

Answer:

1.36 x 10^-3 cm

Explanation:

Area = 50 ft^2 = 46451.5 cm^2

mass = 6 oz = 170.097 g

density = 2.70 g/cm^3

Let t be the thickness of foil in cm.

mass = volume x density

mass = area x thickness x density

170.097 = 46451.5 x t x 2.70

t = 1.36 x 10^-3 cm

Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.

3 0
3 years ago
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