Answer:
the correct answer is 273.2 k
Here is the answer of the given problem above.
Use this formula: <span>P = FV = ma*at = ma^2 t
</span><span>Substitute the values, and therefore, we got m(a0)^2t = m(x)^2 (2t)
then, solve for x which is the acceleration at 2t.
</span>The <span>answer would be a0/sqrt(2).
Hope this answers your question. Thanks for posting.
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Answer:
Explanation:
initial momentum = .36 kg.m.s⁻¹
negative impulse = force x time = .02 x 12 = .24 kg.m.s⁻¹
final momentum - initial momentum = impulse
final momentum = initial momentum + impulse
= .36 - .24
= .12 kg.m.s⁻¹
Here it is. *WARNING* VERY LONG ANSWER
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<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>
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<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>
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<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>
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<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>
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<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>
Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.
<h3>What is electrical force?</h3>
Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.
The given data in the problem is
q₁ is the negative charge = 6 µC=6×10⁻⁶ C
q₂ is the positive charge = 3 µC=3×10⁻⁶ C
r is the distance between the charges=0.002 m
is the electric force =?
The value of electric force will be;
Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.
To learn more about the electrical force refer to the link;
brainly.com/question/1076352
