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Rasek [7]
2 years ago
5

After clearing the bar on a high jump, you land softly on a giant mattress. landing on a mattress is more comfortable than landi

ng on a sand heap of equal size because:_______
a. you transfer less momentum to the mattress in coming to a stop than you would have transferred to the sand heap.
b. the force the mattress exerts on you to stop your descend is much less than the force the sand heap would exert on you.
c. you transfer less energy to the mattress in coming to a stop than you would have transferred to the sand heap.
d. you transfer more momentum to the mattress in coming to a stop than you would have transferred to the sand heap.
e. your velocity is less as you land on the mattress than it would be if you landed on the sand heap.
Physics
1 answer:
madam [21]2 years ago
3 0

Answer:

The right approach is Option b (the force..................exert on you).

Explanation:

  • Even before you fall on something like a soft object, users eventually slow to a halt. You are still giving away all the downward momentum, but progressively although with small powers, you are doing so.  
  • Although you can get injured by massive powers, this gradual displacement is a positive thing. And that is why you have a mattress you would like to settle on.  

The other options given are not connected to the situation described. So, the solution here was the right one.

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An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com
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Answer:

(a) 3.81\times 10^5\ Pa

(b) 4.19\times 1065\ Pa

Explanation:

<u>Given:</u>

  • T_1 = The first temperature of air inside the tire = 10^\circ C =(273+10)\ K =283\ K
  • T_2 = The second temperature of air inside the tire = 46^\circ C =(273+46)\ K= 319\ K
  • T_3 = The third temperature of air inside the tire = 85^\circ C =(273+85)\ K=358 \ K
  • V_1 = The first volume of air inside the tire
  • V_2 = The second volume of air inside the tire = 30\% V_1 = 0.3V_1
  • V_3 = The third volume of air inside the tire = 2\%V_2+V_2= 102\%V_2=1.02V_2
  • P_1 = The first pressure of air inside the tire = 1.01325\times 10^5\ Pa

<u>Assume:</u>

  • P_2 = The second pressure of air inside the tire
  • P_3 = The third pressure of air inside the tire
  • n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)

Part (a):

Using the above equation for this part of compression in the air, we have

\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa

Hence, the pressure in the tire after the compression is 3.81\times 10^5\ Pa.

Part (b):

Again using the equation for this part for the air, we have

\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa

Hence, the pressure in the tire after the car i driven at high speed is 4.19\times 10^5\ Pa.

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