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Sunny_sXe [5.5K]
3 years ago
13

The cheetah is one of the fastest accelerating animals, for it can go from rest to 34.5 m/s in 3.57 s. If its mass is 112 kg, de

termine 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:
marysya [2.9K]3 years ago
5 0

Answer: a) 18670 W b) 25 hp

Explanation:

a)

Average power,

P = W/t

W= 0.5 MVf^2 - 0.5 MVo^2

Therefore P =( (0.5 x 112 x 34.5^2 ) - (0.5 x 112 x 0 ) ) / 3.57

P = 1.87 x 10^4 W

b)

Power in units of horsepower is

P = 1.87 x 10^4 / 745.7 = 25hp

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A ball is spun around in circular motion such that it completes 50 rotations in 25 s.
Leokris [45]

Answer:

(A) The period of its rotation is 0.5 s (2) The frequency of its rotation is 2 Hz.

Explanation:

Given that,

a ball is spun around in circular motion such that it completes 50 rotations in 25 s.

(1). Let T be the period of its rotation. It can be calculated as follows :

T=\dfrac{25}{50}\\\\T=0.5\ s

(2). Let f be the frequency of its rotation. It can be defined as the number of rotations per unit time. So,

f=\dfrac{50}{25}\\\\f=2\ Hz

Hence, this is the required solution.

3 0
3 years ago
Two identical resistors connected in series have an equivalent resistance of 4 ohms. The same two resistors, when connected in p
irina1246 [14]

Answer:

1 ohm

Explanation:

since there are two identical resistors, one resistor will be

R = \frac{4}{2} =2ohm [ proven as in series R_{e} = 2 + 2 = 4ohm ]

to calculate the equivalent resistance when in parallel:

\frac{1}{R_{e} }  = \frac{1}{R_{1} } + \frac{1}{R_{2}}

so,

\frac{1}{R_{e} } = \frac{1}{2} + \frac{1}{2}

R_{e} = 1ohm

4 0
2 years ago
A series circuit has a capacitor of 0.25 × 10−6 F, a resistor of 5 × 103 Ω, and an inductor of 1 H. The initial charge on the ca
svetoff [14.1K]

Answer:

q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Explanation:

Given that L=1H, R=5000\Omega, \ C=0.25\times10^{-6}F, \ \ E(t)=12V, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:

E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000  \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}

#To find the particular solution:

Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

Hence the charge at any time, t is q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C

6 0
3 years ago
What is this answer
Katyanochek1 [597]
Answer: an ectomorph is a body type that struggles to gain weight and muscle
7 0
3 years ago
Read 2 more answers
Physics {deceleration}
seropon [69]
Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)
7 0
3 years ago
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