both it velocity and acceleration is zero.
Answer:
θ = Cos⁻¹[A.B/|A||B|]
A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result
Explanation:
We can use the formula of the dot product, in order to find the angle between two non-zero vectors. The formula of dot product between two non-zero vectors is written a follows:
A.B = |A||B| Cosθ
where,
A = 1st Non-Zero Vector
B = 2nd Non-Zero Vector
|A| = Magnitude of Vector A
|B| = Magnitude of Vector B
θ = Angle between vector A and B
Therefore,
Cos θ = A.B/|A||B|
<u>θ = Cos⁻¹[A.B/|A||B|]</u>
Hence, the correct answer will be:
<u>A. The angle between two nonzero vectors can be found by first dividing the dot product of the two vectors by the product of the two vectors' magnitudes. Then taking the inverse cosine of the result</u>
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>
Answer:
288.0 units; that is the electrostatic force of attraction become quadruple of its initial value.
Explanation:
If all other parameters are constant,
Electrostatic Force of attraction ∝ (1/r²)
F = (k/r²) = 72.0
If r₁ = r/2, what happens to F₁
F₁ = (k/r₁²) = k/(r/2)² = (4k/r²) = 4F = 4 × 72 = 288.0 units
