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sattari [20]
3 years ago
12

Which two bones constitute for forearm

Physics
1 answer:
JulsSmile [24]3 years ago
7 0
<span>The bones of your forearm are the radius and therefore the arm bone. If you hold your arms at your aspect along with your palms facing up, the arm bone is that the bone nighest to your body and therefore the radius is highest to your thumb The arm bone is larger at the elbow - it forms the "point" of your elbow and therefore the radius is larger</span>
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39 m/s

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Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra
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Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

s^2 = (x_A - x_B)^2+(y_A - y_B)^2  (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h

with this time, we know can now calculate the distance at which B is:

\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

and applying Pythagoras:

\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

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3 years ago
The place theory of hearing is better than frequency theory at explaining pitch perception for __________ sounds
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An Astronaut lands on an Earthlike planet and drops a small lead ball with a mass from the top of her spaceship. The point of re
bearhunter [10]

First we have to find out the gravity on that planet. We use Newton second equation of motion. It is given as,

s = ut +(gt^2)/2

Distance s = 25m

Time t = 5 s

Velocity u = 0

By putting these values,

25 = 1/2.g.(5)²

g = 2

So the gravity on that planet is 2. Lets find out the weight of the astronaut.

Mass of the astronaut on earth m = 80 kg

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A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
Keith_Richards [23]

Answer

given,

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mass of small sphere = 0.0200 kg

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I =0.032\ kg.m^2

using conservation of energy

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\omega=\sqrt{\dfrac{(m_1-m_2)gL}{I}}

\omega=\sqrt{\dfrac{(0.05-0.02)\times 9.8 \times 0.8}{0.032}}

\omega=2.71 \rad/s

we know,

v = r ω

v = \dfrac{L}{2} \times 2.71

v = \dfrac{0.8}{2} \times 2.71

v = 1.084 m/s

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4 years ago
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