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ICE Princess25 [194]
3 years ago
12

A sky diver steps from a high-flying helicopter. if there were not air resistance, how fast would she be falling at the end of a

12 second jump?
Physics
1 answer:
Rama09 [41]3 years ago
8 0

If there is no air resistance and the body, in this case the sky diver simply falls from the helicopter then, the motion can be described as a free-falling body. The equation that would allow us to determine the speed of the body at any time, t is,

<span>                                    v(t)  = vo + gt</span>

<span>where v(t) is the unknown velocity, vo is the initial velocity which is equal to zero and g is the acceleration due to gravity which is equal to 9.8 m/s2.</span>

Substituting the known values,

<span>                                    v(t) = 0 + (9.8 m/s2)(12s)</span>

<span>                                                v(t) = 117.6 m/s</span>

<span>Hence, the speed of the sky diver at the end of 12 second is approximately 117.6 m/s.  </span>

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A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i
attashe74 [19]

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
  • As per Newton's equation of motion,

V²-U² = 2aS

  • S= distance covered by the car
  • So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

4 0
1 year ago
At the local swimming hole, a favorite trick is to run horizontally off a cliff that is 8.0 m above the water. One diver runs of
Alika [10]

Answer:

Number of revolutions=1.532 revolutions

Explanation:

Given data

Distance s=8.0 m

Angular speed a=1.2 rev/s

To find

Number of revolutions

Solution

From the equation of simple motion we not that

S=ut+1/2gt^{2}\\ where\\u=0\\So\\8.0m=0+(1/2)(9.8m/s^{2} )t^{2}\\ t^{2}=\frac{8.0m}{0.5*9.8m/s^{2} } \\ t^{2}=1.63\\t=\sqrt{1.63} \\t=1.28s

So for the number of revolutions she makes is given as

n=a*t\\n=(1.2rev/s)(1.28s)\\n=1.532revolutions

8 0
3 years ago
PLEASE HELP
enyata [817]

Answer:

29 m/s.

Explanation:

Hope this helps.

8 0
3 years ago
The si unit of measurement is a way for scientists to?
Vera_Pavlovna [14]
Have a universal record base. Everyone is able to understand the data compiled since the same measurement systems are being used around the world. This is just to simplify all of the information.
7 0
3 years ago
A beverage manufacturer wants to increase the solubility of carbon dioxide (CO2) in its carbonated drinks.
Brut [27]

D. Decreasing its temperature

Explanation:

Decreasing the temperature of the carbon dioxide gas to be dissolved in the carbonated drink will most likely increase the solubility of the gas in the drink.

Temperature has considerable effects on the solubility of gases in liquids.

  • Dissolution involves the surrounding of ions by water molecules, in this case, the carbon dioxide gas is to be surrounded by the liquid beverage medium.
  • Increasing pressure increases the rate at which gases are soluble. At high pressure, the gases are brought more in contact with the liquid medium.
  • Decreasing temperature aids gas solubility.
  • If the temperature of gases are increased,  they will not want to stay in solution as they gain a high amount of kinetic energy.
  • Therefore, it will increase their randomness and the urge to leave the solution.
  • Decrease in temperature and increase in pressure makes gas solubility to be fast.  

Learn more:

Rate of chemical reactions brainly.com/question/6281756

#learnwithBrainly

6 0
3 years ago
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