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frez [133]
3 years ago
6

The weight (W) of an object is the product of its mass (m) in kilograms and g, the acceleration due to gravity in meters/second2

. If the acceleration due to gravity is 9.8 meters/second2, find the corresponding weights for these masses.
mass in kilograms = {22.1, 33.5, 41.3, 59.2, 78}


The weights (in newtons) for the given masses are
Physics
1 answer:
Nookie1986 [14]3 years ago
7 0
The weights in newtowns for the given masses are

<span> masses        22.1,                 33.5,             41.3,           59.2,            78
 weights        216.58N           328.3N       404.74N      580.16N     764.4N

e.g,   for m=22.1kg, W=22.1kgx9.8N/kg =216.58N</span>
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The average velocity of blood flowing in a certain 4-mm-diameter artery in the human body is 0.28 m/s. The viscosity and density
OLga [1]

Answer:

V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s

Explanation:

The volume flow rate of the blood in the artery can be given by the following formula:

V = Av

where,

V = Volume flow rate = ?

A = cross-sectional area of artery = πd²/4 = π(0.004 m)²/4 = 1.26 x 10⁻⁵ m²

v = velcoity = 0.28 m/s

Therefore,

V = (1.26\ x\ 10^{-5}\ m^2)(0.28\ m/s)

<u>V = 3.5 x 10⁻⁶ m³/s = 3.5 cm³/s</u>

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Skid Marx set the world record for acceleration as he went from zero to 10 m/s in 3 seconds. How far did he travel in that time?
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12000 miles per hour

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If skid Marx went from zero to 10 miles in 3 seconds the his speed would have been 12000 miles per hour.

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Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.
Ksivusya [100]

|v| =\sqrt{ G \cdot M / r}, where

  • M the mass of the planet, and
  • G the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

<em>Equation 1</em>  (see below) relates net force the object experiences, \Sigma F to its orbit velocity v and its mass m required for it to stay in orbit :

\Sigma F = m \cdot v^{2} / r <em>(equation 1)</em>

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force \Sigma F acting on the soccer ball shall equal to its weight, W = m \cdot g where g the gravitational acceleration constant. Thus

\Sigma F = W = m \cdot g <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> and solve for v; note how the mass of the soccer ball, m, cancels out:

m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0) <em>(equation 3)</em>

<em>Equation 4 </em> gives the value of gravitational acceleration, g, a point of negligible mass experiences at a distance r from a planet of mass M (assuming no other stellar object were present)

g = G \cdot M/ r^{2} <em>(equation 4)</em>

where the universal gravitation <em>constant</em> G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}

Thus

\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}

3 0
3 years ago
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