Answer:
a) Therefore 2.6km is greater than 2.57km.
Statement A is greater than statement B.
b) Therefore 5.7km is equal to 5.7km
Statement A is equal to statement B
Explanation:
a) Statement A : 2.567km to two significant figures.
2.567km 2. S.F = 2.6km
Statement B : 2.567km to three significant figures.
2.567km 3 S.F = 2.57km
Therefore 2.6km is greater than 2.57km.
Statement A is greater than statement B.
b) statement A: (2.567 km + 3.146km) to 2 S.F
(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km
Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).
2.567km to 2 S.F = 2.6km
3.146km to 2 S.F = 3.1km
2.6km + 3.1km = 5.7km
Therefore 5.7km is equal to 5.7km
Statement A is equal to statement B
One light-year is the distance that light travels in vacuum
in one year. It's a unit of distance.
Answer:
D. Molecules of a gas slow down and change to a liquid state.
Explanation:
- Condensation refers to a process by which a gas changes from gaseous state to liquid state. For example, water vapor changes to from the state of being a gas to liquid state water.
- Condensation is the opposite of evaporation and occurs when gaseous particles slow down and change into liquid state.
- Heat energy is lost during condensation and gaseous molecules lose kinetic energy making them to slow down and thus changing to liquid state,
To solve this problem, we must remember about the law of
conservation of momentum. The initial momentum mist be equal to the final
momentum, that is:
m1 v1 + m2 v2 = (m1 + m2) v’
where v’ is the speed of impact
Since we are not given the masses of each car m1 and m2,
so let us assume that they are equal, such that:
m1 = m2 = m
Which makes the equation:
m v1 + m v2 = (2 m) v’
Cancelling m and substituting the v values:
50 + 48 = 2 v’
2 v’ = 98
v ‘ = 49 km/h
<span>The speed of impact is 49 km/h.</span>
The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s