Answer:
They have three to six valence electrons, so they can lose or gain electrons.
Explanation:
I just took the quiz on Edgenuity.
Answer:
density I believe, air gets compressed when In pressure, so the air would have more density
Mass of the car = 1200 kg
Mass of the truck = 2100 kg
Total mass of car and truck = 2100 + 1200 = 3300 kg
Since, the car pushes the truck. Hence, they will move together and will have same acceleration.
Let the acceleration be a.
According to Newton's second law:
F(net) = ma
F = 4500 N
4500 = 3300 × a
a = 1.36 m/s^2
Let the force applied by the car on truck be F.
F = F(net) on the truck
F = ma
F = 2100 × 1.36
F = 2856 N
Hence, the force applied by the car on the truck is 2856 N
Answer:
Explanation:
1 ) Since it is a isochoric process , heat energy passed into gas
= n Cv dT , n is no of moles of gas , Cv is specific heat at constant volume and dT is rise in temperature .
= 7.4 x 12.47 x ( 500 - 300 )
= 18455.6 J.
2 ) Since there is no change in volume , work done by the gas is constant.
3 ) from , gas law equation
PV = nRT
P = nRT / V
= 7.4 x 8.3 x 500 / .74
= .415 x 10⁵ Pa.
4 ) Average kinetic energy of gas molecules after attainment of final temperature
= 3/2 x R/ N x T
= 1.5 x 1.38 x 10⁻²³ x 500
= 1.035 x 10⁻²⁰ J
1/2 m v² = 1.035 x 10⁻²⁰
v² = 2 x 1.035 x 10⁻²⁰ / 1.39 x 10⁻²⁶
= 1.49 x 10⁶
v = 1.22 x 10³ m /s
5 ) In this process , pressure remains constant
gas is cooled from 500 to 300 K
heat will be withdrawn .
heat withdrawn
= n Cp dT
= 7.4 x 20.79 x 200
= 30769.2 J .
6 )
gas will have reduced volume due to cooling
reduced volume = .74 x 300 / 500
= .444 m³
change in volume
= .74 - .444
= .296 m³
work done on the gas
= P x dV
pressure x change in volume
= .415 x 10⁵ x .296
= 12284 J.