A basketball player standing up with the hoop launches the ball straight up with an initial velocity of v_o = 3.75 m/s from 2.5
1 answer:
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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.
Distance covered by the squirrel to look for an acorn :
d = ( 3 m/s ) × 10 s = 30 m.
Time taken to eat an Acron is 5 seconds.
Time taken to cover distance of 30 m with 2 m/s speed is :
Therefore, total time take to get back to where he started is ( 10+5+15 ) = 30 s.
Hence, this is the required solution.
Answer:
The speed of ejection is
Solution:
As per the question:
Magnetic field density, B = 0.4 T
Density of the material in the sunspot,
Now,
To calculate the speed of ejection of the material, v:
The magnetic field energy density is given by:
This energy density equals the kinetic energy supplied by the field.
Thus
where
m = mass of the sunspot in =