Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)
• The net force in the parallel direction is
∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
• The net force in the perpendicular direction is
∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0
Solving the second equation for <em>n</em> gives
<em>n</em> = <em>mg</em> cos(21°)
<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)
<em>n</em> ≈ 1.83 N
Then the magnitude of friction is
<em>f</em> = <em>µn</em>
<em>f</em> = 0.25 (1.83 N)
<em>f</em> ≈ 0.457 N
Solve for the acceleration <em>a</em> :
-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)
<em>a</em> ≈ -5.80 m/s²
so the block is decelerating with magnitude
<em>a</em> = 5.80 m/s²
down the ramp.
First let us calculate for the moles of CH3OH formed:
moles CH3OH = 23 g / (32 g / mol) = 0.71875 mol
We see that there are 2 moles of H2 per mole of CH3OH, so:
moles H2 = 0.71875 mol * 2 = 1.4375 mol
Assuming ideal gas behaviour, we use the formula:
PV = nRT
V = nRT / P
V = 1.4375 mol * (62.36367 L mmHg / mol K) * (90 + 237.15
K) / 756 mm Hg
<span>V = 43.06 Liters</span>
<span>External respiration takes place in the </span>Bronchi to the capillaries. AND the internal respiration takes place from the capillaries to the cell tissues.
Answer:
T= 1 s
Explanation:
Given that
When x= cm ,T= 1
we know that time period of spring mas system given as

T= Time period
m= mass
k=spring constant
So from above equation we can say that time period of system does not depends on the value of x.
So when x= 10 cm ,still time period will be 1 s.
T= 1 s
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