Answer:
By Applying pressure to the brakes
Explanation:
Driving cars through deep water that is more than 10cm can make the cars to float. Most modern cars are usually water- tight so they can start to float through water that is about 30cm deep, fast moving water is very powerful so one needs to be very careful when driving.
If the brakes are wet test them by pressing or tapping on them gently.
You can as well dry brakes by driving in low gear and applying pressure to the brakes.
Because you see yourself the opposite way in a mirror. So yes your “seeing” yourself but not how everyone else sees you.
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>
Answer:
Decrease the distance between the two objects.
Explanation:
The force (F) of attraction between two masses (M₁ and M₂) separated by a distance (r) is given by:
F = GM₁M₂ / r²
NOTE: G is the gravitational force constant.
From the equation:
F = GM₁M₂ / r²
We can say that the force is directly proportional to the masses of the object and inversely proportional to the square of the distance between them. This implies that an increase in any of the masses will increase the force of attraction and likewise, a decrease in any of the masses will lead to a decrease in the force of attraction.
Also, an increase in the distance between the masses will result in a decrease in the force of attraction and a decrease in the distance between the masses, will result in an increase in the force of attraction.
Considering the question given above,
To increase the gravitational force between the two objects, we must decrease the distance between the two objects as explained above.
