Answer:
Here's what I get
Explanation:
A. Distance between A and B.
h = -½gt²
The stones go faster the farther they fall.
Stone A has already reached 5 m when B is released.
When B reaches 5 m, A has dropped further and is falling even faster.
The distance between the stones increases with time.
Figure 1 shows this effect in a graph of height vs. time.
B. Speed of Stone B
v² = 2gh =2 × ( -9.81 m·s⁻²) × (-5 m) = 98.1 m·s⁻²
v = 9.9 m/s
The stone is travelling at 9.9 m/s when it reaches 5 m.
C. Velocity vs time
v = -gt
Both stones accelerate at the same rate.
When Stone B has reached 10 m at time t, Stone A is falling much faster.
Fig. 2 shows this in a graph of velocity vs time.
Answer:B. two larger, less stable nuclei
Explanation: They collied and don't combine
First, we write the SI prefixed. The SI unit for distance is meters.
Kilo = 10³
Mega = 10⁶
Giga = 10⁹
Terra = 10¹²
Because our value has ten to the power of 11, we will use the closest and lowest power prefix, which is giga.
1.5 x 10¹¹ / 10⁹
= 1.5 x 10² Gm or 150 Gm
Writing in kilometers, we simply repeat the procedure except we divide by 10³ this time.
1.5 x 10¹¹ / 10³
= 1.5 x 10⁸ km
Answer:
T = g μ_s ( M+m )
78.4 N
Explanation:
When both of them move with the same acceleration , small box will not slip over the bigger one. When we apply force on the lower box, it starts moving with respect to lower box. So a frictional force arises on the lower box which helps it too to go ahead . The maximum value that this force can attain is mg μ_s . As a reaction of this force, another force acts on the lower box in opposite direction .
Net force on the lower box
= T - mg μ_s = M a ( a is the acceleration created by net force in M )
Considering force on the upper box
mg μ_s = ma
a = g μ_s
Put this value of a in the equation above
T - m gμ_s = M g μ_s
T = mg μ_s + M g μ_s
= g μ_s ( M+m )
2 )
Largest tension required
T = 9.8 x .50 x ( 10+6 )
= 78.4 N
