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geniusboy [140]
4 years ago
6

Mary slides down a snow-covered hill on a large piece of cardboard and then slides across a frozen pond at a constant velocity o

f 2.40 m/s. After Mary has reached the bottom of the hill and is sliding across the ice, Sue runs after her at a velocity of 4.40 m/s and hops on the cardboard. How fast do the two of them slide across the ice together on the cardboard? Mary's mass is 69.0 kg and Sue's is 56.0 kg. Ignore the mass of the cardboard and any friction between the cardboard and the snow and/or ice. (Indicate the direction with the sign of your answer.) m/s
Physics
1 answer:
Tresset [83]4 years ago
3 0

Answer: 3.288 m/s

Explanation:

Given

Mass of Mary, m1 = 69 kg

Mass of Sue, m2 = 56 kg

Speed of Mary, v1 = 2.4 m/s

Speed of Sue, v2 = 4.4 m/s

Speed of the 2 of them, v = ?

We solve this using the principle of conservation of linear momentum

m1.v1 + m2.v2 = m1.v + m2.v

m1.v1 + m2.v2 = (m1 + m2) v

v = [(m1.v1) + (m2.v2)] / (m1 + m2)

v = [(69 * 2.4) + (56 * 4.4)] / (69 + 56)

v = (164.6 + 246.4) / 125

v = 411 / 125

v = 3.288 m/s

Thus, the speed at which both Mary and Sue slide together across the ice is 3.288 m/s

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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
solong [7]

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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Answer:

F = \frac{-Gm_{1}m_{2} }{r^{2} }.

Explanation:

Gravitational force between two objects of masses m_{1},  m_{2} kept at a distance r is given by the formula

F = \frac{-Gm_{1}m_{2} }{r^{2} }

Here ,m_{1} = 2m

         m_{2} = \frac{m}{2}

         

Thus , F = \frac{-G.2m.\frac{m}{2} }{r^{2} }

          F = \frac{-Gm_{1}m_{2} }{r^{2} }.

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