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3 years ago

10

A 60-kg cheetah reaches a speed of 30 m/s as it chases its prey.What is the kinetic energy of the cheetah?

2 answers:

GuDViN [60]3 years ago

7 0

Kinetic energy = 1/2mv^2
= (1/2)(60)(30^ 2)
KE = 27,000 Joules

*just a reminder when you are solving for kinetic energy you have to make sure that the mass is given in kg and the velocity is given in m/sec.

hope this helps :)

Rama09 [41]3 years ago

4 0

The answer would be 27,000 Joules because (1/2) m v^2 =30*900 which equals 27,000 J

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If 27 J of work are needed to stretch a spring from 15 cm to 21 cm and 45 J are needed to stretch it from 21 cm to 27 cm, what i

kupik [55]

Answer:

9 cm.

Explanation:

The energy used for stretch the spring from $15 cm$ to $21 cm$ will be , $E_{1}=27J$

The energy used for stretch the spring from $21 cm$ to $27 cm$ will be , $E_{2}=45J$

using the energy of spring formula ,we find that

$27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))$

$45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))$

Dividing both the equation will get,

$\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm$

Therefore, the natural length of the spring is, 9 cm.

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3 years ago

A 2.5 kg steel gasoline tank can holds 20.0 L of gasoline when full. What is the average density (in Kg/m^3) of the full gas can

yarga [219]

Answer:

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Explanation:

<u>Average Density </u>

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$\displaystyle D=\frac{m}{V}$

If we know the density and the volume occupied by the object, the mass can be computed as

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$V_g=20*0.001=0.02\ m^3$

That's the volume of the gasoline it contains. Knowing the density of the gasoline, we get the mass of gasoline.

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To know the total mass of both, we add the 2.5 kg of the tank

$m=m_t+m_g=13.6+2.5=16.1\ kg$

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$\displaystyle V_t=\frac{2.5}{7800}=3.205\times 10^{-4}\ m^3$

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$V=V_g+V_t=0.02+3.205\times 10^{-4}=0.0203\ m^3$

The average density is

$\displaystyle D=\frac{16.1}{0.0203}$

$\boxed{D=792.3\ kg/m^3}$

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