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Mice21 [21]
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?

Physics
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.

4 0
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:

D=792.3\ kg/m^3

Explanation:

<u>Average Density </u>

The density of an object of mass m and volume V is

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

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

m=D.V

The tank can hold 20 L of gasoline when full. Converting to cubic meters

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.

m_g=680*0.02=13.6\ kg

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

The volume of the tank is computed solving for V

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

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

The total volume is

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}

7 0
3 years ago
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AVprozaik [17]
<span>The answer would be A)  Bohr model

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Likurg_2 [28]

Answer:

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Explanation:

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3 years ago
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