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3 years ago

A trunk rotation is a common dyamic flexibility assessment ture or false

1 answer:

6 0

True. It would be false if the statement was "trunk rotation is the most common <em>static</em> flexibility assessment."

So, you're answer should be "true". Hope that helped!

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Help it’s actually physical science but plz help

Butoxors [25]

The correct answer is B

3 0

2 years ago

Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun

Dominik [7]

Answer:

a. $K_{Axis}=2.574x10^{29}J$

b. $K_{Orbit}=2.6577x10^{33}J$

Explanation:

$K_{Axis}=\frac{1}{2}I*w^2$

$I_{Sphere}=\frac{2}{5}*m*r^2$

$w=\frac{2\pi }{T}$ , $T=24hrs*\frac{3600s}{1hr} =86400s$

radius earth = 6371 km

mass earth = 5,972*10^24 kg

$K_{Axis}=\frac{1}{2}*\frac{2}{5}*m*r^2*(\frac{2\pi}{T})^2$

$K_{axis}=\frac{4\pi^2}{5}*5.98x10^{24}kg*(6.38x10^6m)^2*(\frac{1}{86400s})^2$

$K_{Axis}=2.574x10^{29}J$

$T=1year*\frac{365day}{1year}*\frac{24hr}{1day}*\frac{3600s}{1hr}=31536000s$

$K_{Orbit}=\frac{1}{2}*I*w$

$I=m*r^2$

$K_{Orbit}=\frac{1}{2}*m*r^2*(\frac{2\pi}{T})^2$

$K_{Orbit}=\frac{4\pi^2}{5}*5.98x10^{24}*6.38x10^6m*(\frac{1}{31.536x10^6s})^2$

$K_{Orbit}=2.6577x10^{33}J$

4 0

3 years ago

A house is lifted from its foundations onto a truck for relocation. The house is pulled upward by a net force of 2850 N. This fo

Gwar [14]

Answer:

m = 95000 kg

Explanation:

Given that,

Net force acting on the house, F = 2850 N

Initial speed, u = 0

Final speed, v = 15 cm/s = 0.15 m/s

We need to find the mass of the house. Let the mass be m. We know that the net force is given by :

F = ma

Where

a is the acceleration of the house.

So,

$F=m\dfrac{v-u}{t}\\\\m=\dfrac{Ft}{(v-u)}\\\\m=\dfrac{2850\times 5}{(0.15-0)}\\\\m=95000\ kg$

So, the mass of the house is equal to 95000 kg.

3 0

2 years ago

Which statements accurately describes the waste generated by nuclear reactions ? check all that apply.

Neko [114]

A) its 's hazardous for hundreds to thousand of year B) Some waste can be recycled and used again is the answer

4 0

2 years ago

A normal human body temperature can range from 33.2∘C to 38.2∘C. This range includes variations with gender, time of day, and wh

Tpy6a [65]

To develop this problem it is necessary to use the expression that allows us to convert the degrees Celsius to Fahrenheit. The expression that allows to realize it is given mathematically by:

$\°F = (\°C* \frac{9}{5}) + 32$

In this way for 33.2 °C:

$\°F = (\°C * \frac{9}{5}) + 32$

$\°F = (32.2* \frac{9}{5}) + 32$

$\°F = 91.76$

In this way for 38.2°C

$\°F = (\°C * \frac{9}{5}) + 32$

$\°F = (38.2* \frac{9}{5}) + 32$

$\°F = 91.76$

Expressed in a range term, we can say that the measure in degrees Fahrenheit is:

$[91.76\° F , 100.76\° F]$

8 0

3 years ago