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2 years ago

5. A cheetah with a mass of 70 kg was clocked running at 72 mph (32 m's). How many joules of

Physics

1 answer:

blsea [12.9K]2 years ago

7 0

Answer:

Kinetic energy = 35840 Joules

Explanation:

Given the following data;

Mass = 70kg

Velocity = 32m/s

To find the kinetic energy;

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

$K.E = \frac{1}{2}MV^{2}$

Where, K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

$K.E = \frac{1}{2}MV^{2}$

Substituting into the equation, we have;

$K.E = \frac{1}{2}*70*32^{2}$

$K.E = \frac{1}{2}*70*1024$

$K.E = 35 * 1024$

K.E = 35840 Joules.

Therefore, the kinetic energy possessed by the cheetah is 35840 Joules.

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2 years ago

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А A pool of water of refractive index

babymother [125]

Answer:

Apparent depth = 45 cm

Explanation:

The refractive index of water in a pool, n = 4/3

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$n=\dfrac{d}{d'}\\\\d'=\dfrac{d}{n}\\\\d'=\dfrac{60}{\dfrac{4}{3}}\\\\d'=45\ cm$

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2 years ago

Which statement is true about an atom and an element?

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3 years ago

A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber

Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

$V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}$

Now calculate the bouyant force acting on the lead

$F_L = Vpg$

$F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N$

This force will act in upward direction

Gravitational force on the lead due to its mass will act in downward direction

Hence the difference of this two force

$T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N$

If V is the volume submerged in the water then bouyant force on the bobber is

$F_B=V'pg$

Equate bouyant force with the tension and gravitational force

$F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}$

Now Total volume of bobble is

$\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\$

= $\large\boxed{4.52 \%}$

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2 years ago

the jet plane travels along the vertical parabolic path. when it is at point a it has speed of 200 m/s, which is increasing at t

givi [52]

Explanation:

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→ The tangential component of acceleration is rate of increase in the speed of plane so,

$a_{t} = v = 0.8 m/s^{2}$