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Crazy boy [7]
3 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]3 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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-- The potential energy of a jar full of gas does depend on the
temperature of the gas.  The warmer it is, the greater its pressure
is, and the more work it can do if you let it out through a little hole
in the jar.  If it gets hot enough, it'll have enough potential energy
to blow the jar to smithereens.
6 0
3 years ago
Read 2 more answers
Pls answer ASAP pls bc I’m tryna get my grade up please
Sidana [21]

Answer:

The right answer for this question is 85%.

(I had the same question.)

6 0
2 years ago
Two objects (42.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h
Kazeer [188]

Answer:

a = 3.27 m/s²

T = 275 N

Explanation:

Given that:

Mass m₁ = 42.p0 kg

Mass m₂ = 21.0 kg

Consider both masses to be in a whole system, then:

The acceleration can be determined as:

(m_1+m_2)a = g(m_1-m_2)

Making acceleration the subject in the above formula;

a =\dfrac{g(m_1-m_2)}{(m_1+m_2)}

a =\dfrac{9.8(42.0-21.0)}{(42.0+21.0)}

a =\dfrac{9.8(21.0)}{(63.0)}

a =\dfrac{205.8}{(63.0)}

a = 3.27 m/s²

in the string, the tension is calculated using the formula:

T = \dfrac{2m_1m_2g}{(m_1+m_2)}

T = \dfrac{2(42)(21)(9.81)}{(42+21)}

T = \dfrac{17304.84}{63}

T = 274.68 N

T ≅ 275 N

8 0
3 years ago
A 1022kg Caprice car stopped at an intersection is rear-ended by a 1620kg ranger truck moving with a speed of 14.5m/s. If the ca
Alika [10]

Answer:

Explanation:

mass of car, m = 1022 kg

mass of truck, M = 1620 kg

initial velocity of truck, U = 14.5 m/s

initial velocity of car, u = 0 m/s

Let the final velocity of car is v and the final velocity of truck is V.

Collision is elastic, so the coefficient of restitution, e = 1

Use conservation of momentum

initial momentum of car + initial momentum of truck = final momentum of car + final momentum of truck

m x u + M x U = m x v + M x V

0 + 1620 x 14.5 = 1022 v + 1620 V

23490 = 1022 v + 1620 V ..... (1)

Use the formula of coefficient of restitution

e = \frac{V_{1}-V_{2}}{u_{2}-u_{1}}

1 (14.5 - 0) = v - V

14.5 = v - V

V = v - 14.5 .... (2)

Put in equation (1)

23490 = 1022 v + 1620 (v - 14.5)

23490 = 1022 v + 1620 v - 23490

46980 = 2642 v

v = 17.8 m/s

Put in equation (2)

V = 17.8 - 14.5

V = 3.3 m/s

Thus, the speed of car is 17.8 m/s and the velocity of truck is 3.3 m/s after collision.

8 0
3 years ago
When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play
saw5 [17]

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

6 0
3 years ago
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