Answer:
85.556metres
Explanation:
Using pythagorean theorem
C²=A²+B²
we have c as the hypotenuse vector A thus:
93.8²=A²+38.4²
93.8²-38.4²=A²
8794.44-1474.56=A²
7319.88=A²
A=85.556
Answer : The final pressure in the two containers is, 2.62 atm
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
Thus, the expression for final pressure in the two containers will be:
where,
= pressure of N₂ gas = 4.45 atm
= pressure of Ar gas = 2.75 atm
= volume of N₂ gas = 3.00 L
= volume of Ar gas = 2.00 L
P = final pressure of gas = ?
V = final volume of gas = (4.45 + 2.75) L = 7.2 L
Now put all the given values in the above equation, we get:
Thus, the final pressure in the two containers is, 2.62 atm
Here it is. *WARNING* VERY LONG ANSWER
________________________________________...
<span>11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h) </span>
<span>The change in PE =mgh=5*9.8*12=588 J </span>
<span>______________________________________... </span>
<span>12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s. </span>
<span>Commendable and jockey Pat Day had a combined mass =M= 550.0 kg, </span>
<span>Their KE as they crossed the line=(1/2)Mv^2 </span>
<span>Their KE as they crossed the line=0.5*550*(15.98)^2 </span>
<span>Their KE as they crossed the line is 70224.11 J </span>
<span>______________________________________... </span>
<span>13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m </span>
<span>She trips and drops the spare tire of mass = m = 10.0 kg, </span>
<span>The tire rolls down the hill with an intial speed = u = 2.00 m/s. </span>
<span>The height of top of the next hill = h = 5.00 m </span>
<span>Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2 </span>
<span>Initial total mechanical energy =mgH+(1/2)mu^2 </span>
<span>Suppose the final speed at the top of second hill is v </span>
<span>Final total mechanical energy =PE+KE=mgh+(1/2)mv^2 </span>
<span>As mechanical energy is conserved, </span>
<span>Final total mechanical energy =Initial total mechanical energy </span>
<span>mgh+(1/2)mv^2=mgH+(1/2)mu^2 </span>
<span>v = sq rt [u^2+2g(H-h)] </span>
<span>v = sq rt [4+2*9.8(20-5)] </span>
<span>v = sq rt 298 </span>
<span>v =17.2627 m/s </span>
<span>The speed of the tire at the top of the next hill is 17.2627 m/s </span>
<span>______________________________________... </span>
<span>14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean. </span>
<span>a.)The mass of bean = m = 2.0 g </span>
<span>Height up to which the been jumps = h = 1.0 cm from hand </span>
<span>Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg </span>
<span>b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s </span>
<span>_____________________________ </span>
<span>15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill. </span>
<span>The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s </span>
<span>__________________________________ </span>
<span>16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m, </span>
<span>The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2 </span>
<span>______________________________________... </span>
<span>EDIT </span>
<span>1.) A train is accelerating at a rate = a = 2.0 km/hr/s. </span>
<span>Acceleration </span>
<span>Initial velocity = u = 20 km/hr, </span>
<span>Velocity after 30 seconds = v = u + at </span>
<span>Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s = </span>
<span>Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr </span>
<span>Velocity after 30 seconds = v = 80 km/hr=22.22 m/s </span>
<span>_______________________________- </span>
<span>2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins. </span>
<span>His acceleration = a =11.1/9=1.233 m/s^2 </span>
<span>Distance he covered = s = (1/2)at^2=49.95 m</span>
1 year = (365 / 121) = 3.02 half-lifes. Let's call it 3 .
The amount of radioactive isotope remaining after 3 half-lifes is
(1/2) x (1/2) x (1/2) = 1/8
A year after the medical lab received the 24 kg of W-181,
there will still be 24 kg of stuff in the container.
But only 3 kg of it will still be W-181. The other 21 kg will be
whatever substances W-181 becomes when it decays.
Sadly, even the 3 kg of good stuff won't be usable anymore ...
it'll be thoroughly mixed with the 21 kg of junk. It would be harder
and more expensive to try and separate them than to buy a new
can of pure W-181, and USE it before 7/8 of it has deteriorated.
