Models show how the atoms in a compound are connected.
Conservation of linear momentum:
m*v inital = m*v final
0.06*0.7 + 0.03*0 = 0.06*(-0.2) + 0.03*v
(my algebra, or use ur calculator: 0.06*.07=0.042, etc ... or ur teacher may think you got some help)
0.06*(0.7+0.2)=0.03*v, v = 0.06*0.9/0.03=1.8 m/s
Answer 1.8 m/s (positive, to the right).
Answer:
from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.
Explanation:
This can be explained on the basis of conservation of angular momentum.
This means the initial and the final angular velocity is conserved. Consider initial position (1)in the pike and final position in the be truck position. So there inertia's will also be different.
⇒

also,


since, 

therefore,

So, from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.
Answer:
No, both the thermometers will give the different reading.
Explanation:
Given,
- Both thermometer has same ice point =

- Both thermometer has same steam point =

- Distance between the ice point and steam point in both the thermometer is same of 100 division,
All the data given in both the thermometers are same, but the material in the thermometer is different due to this the reading at 60^o C will differ in both the thermometer. Because the reading on both the thermometer is depended upon the thermal expansion of the material inside it, but both the materials are different. Due to this the rise of fluid in the thermometer, i,e,. the volume of the fluid material in the thermometer will depend upon the thermal expansion. Hence both the material alcohol and mercury have the different thermal expansion, therefore the rise of the fluid in the thermometer also differ in both the thermometer.
Answer: 80m
Explanation:
Distance of balloon to the ground is 3150m
Let the distance of Menin's pocket to the ground be x
Let the distance between Menin's pocket to the balloon be y
Hence, x=3150-y------1
Using the equation of motion,
V^2= U^s + 2gs--------2
U= initial speed is 0m/s
g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s
40m/s is contant since U (the coin is at rest is 0) hence V =40m/s
Slotting our values into equation 2
40^2= 0^2 + 2 * 10* (3150-y)
1600 = 0 + 63000 - 20y
1600 - 63000 = - 20y
-61400 = - 20y minus cancel out minus on both sides of the equation
61400 = 20y
Hence y = 61400/20
3070m
Hence, recall equation 1
x = 3150 - 3070
80m
I hope this solve the problem.