Answer:
1.97×10⁻²¹ J
Explanation:
Use ideal gas law to find temperature.
PV = nRT
(9 atm) (9 L) = (83.3 mol) (0.0821 L·atm/mol/K) T
T = 11.9 K
The average kinetic energy per atom is:
KE = 3/2 kT
KE = 3/2 (1.38×10⁻²³ J/K) (11.9 K)
KE = 2.46×10⁻²² J
For a mass of 5.34×10⁻²⁶ kg, the kinetic energy is:
KE = (5.34×10⁻²⁶ kg) (1 mol / 0.004 kg) (6.02×10²³ atom/mol) (2.46×10⁻²² J)
KE = 1.97×10⁻²¹ J
Answer:
a. 11 m/s at 76° with respect to the original direction of the lighter car.
Explanation:
In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

The magnitude of the final velocity of the wreck can be found as:
![v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bfx%7D%5E%7B2%7D%2B%20v_%7Bfy%7D%5E%7B2%7D%5C%5Cv_%7Bf%7D%3D%5Csqrt%5B%5D%7B2.6%5E%7B2%7D%20%2B%2010.4%5E%7B2%7D%7D%20%5C%5Cv_%7Bf%7D%3D%2010.72)
The final velocity has an intensity of roughly 11 m/s
As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.
The object that reaches the ground first is thebowling ball i think because it weighs heavier but when the tennis ball hits the ground it will bounce again and again till it stops and the bowling ball will just stop it wont bounce
Answer:
0.34
Explanation:
2.5 Mg = 2500 kg
The change in speed from 100 km/h to 40 km/h is

The deceleration caused by friction force is the change in speed per unit of time

Using Newton 2nd law we can calculate the friction force that caused this deceleration:
F = ma = 2500 * 3.33 = 8333 N
Let g = 9.8m/s2. Friction force is the product of normal (gravity) force and friction coefficient

