The formula for work is
F*d
Therefore work=2.0N*3.0=6N*m
Answer:
0.96 m
Explanation:
First, convert km/h to m/s.
162.3 km/h × (1000 m/km) × (1 hr / 3600 s) = 45.08 m/s
Now find the time it takes to move 20 m horizontally.
Δx = v₀ t + ½ at²
20 m = (45.08 m/s) t + ½ (0 m/s²) t²
t = 0.4436 s
Finally, find how far the ball falls in that time.
Δy = v₀ t + ½ at²
Δy = (0 m/s) (0.4436 s) + ½ (-9.8 m/s²) (0.4436 s)²
Δy = -0.96 m
The ball will have fallen 0.96 meters.
I am pretty sure that the only statement which is true for particles of the medium of an earthquake P-wave is being shown in the option : b)vibrate parallel to the wave, forming compressions and rarefactions. As you know, it can be formed in two ways : from alternating compressions and rarefactions or primary wave. I bet you will agree with me.
I think it would be the scientific method.
Answer:
The distance covered by puck A before collision is 
Explanation:
From the question we are told that
The label on the two hockey pucks is A and B
The distance between the two hockey pucks is D 18.0 m
The speed of puck A is 
The speed of puck B is 
The distance covered by puck A is mathematically represented as

=> 
The distance covered by puck B is mathematically represented as

=> 
Since the time take before collision is the same

substituting values

=> 
=> 