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babymother [125]

3 years ago

11

When kicking a football, the kicker rotates his leg about the hip joint. If the velocity of the tip of the kicker’s shoe is 35.0

m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity?

1 answer:

kkurt [141]3 years ago

6 0

Answer:

33.33 rad / s

Explanation:

Linear velocity = 35 m/s

Radius = 1.05 m

The relation between the linear velocity and the angular velocity is given by

Linear velocity = radius × angular velocity

Angular velocity = linear velocity / radius

Angular velocity = 35 / 1.05

= 33.33 rad/ s

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A laundry detergent
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A sample contains 10 g of a radioactive isotope. How much radioactive isotope will remain in the sample after 2 half-lives?

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A 50-kg copper block initially at 140Â°c is dropped into an insulated tank that contains 90 l of water at 10Â°c. Determine the f

xxMikexx [17]

Answer:

$T_f=24.71$

Explanation:

From the question we are told that:

Mass of block $m=50$

Temperature of block $T_b =140 \textdegree C$

Volume of water $V= 90L$

Temperature of water $T_w=10 \textdegree C$

Density of water $\rho=1000kg/m^3$

Specific heat of water $C_w=4.18KJ/kg-k$

Specific heat of copper $C_p=0.96KJ/kg-k$

Generally the equation for equilibrium stage is mathematically given by

$mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)$

$50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)$

$48(140-T_f)=376.2(T_f-10)$

$140-T_f=7.8375(T_f-10)$

$140-T_f=7.8375T_f-78.375$

$-8.8375T_f=-218.375$

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3 years ago

(50 POINTS) please answer each question:

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3 years ago

A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons

Vlada [557]

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

$K.E\geq P.E$

or

$\dfrac{1}{2}mv^2 \geq G\dfrac{Mm}{R}$

where $m$ is the mass of the asteroid, $R= 24,000,000\:m$ is its distance form earth's center, $M = 5.9*10^{24}kg$ is the mass of the earth, and $G = 6.7*10^{-11}m^3/kg\: s^2$ is the gravitational constant.

Solving for $v$ we get:

$v \geq \sqrt{\dfrac{2GM}{R} }$

putting in numerical values gives

$v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }$

$\boxed{v\geq 5739.5m/s}$

in kilometers this is

$v\geq5.7395m/s.$

Hence, the minimum speed required is 5.7395km/s.

5 0

3 years ago