Which statement is always false for athletes participating in team sports?
Answer: Out of all the options shown above the one that best represents the statement that is alway false for athletes participating in team sports is answer choice C) Conflict resolution is a sign of poor sportsmanship. All the other choices are true when it comes to team sports.
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Answer:
The horizontal range will be 
Explanation:
We have given initial speed of the shell u = 
Angle of projection = 51°
Acceleration due to gravity 
We have to find maximum range
Horizontal range in projectile motion is given by

So the horizontal range will be 
A)Linear motion
If there is not net force on the car, then by the Newton Second Law, the acceleration is zero, and the only valid option for zero acceleration is A).
Answer:
Length of pipe
meter
Explanation:
Speed of a transverse wave on a string

where F is the tension in string and
is the mass per unit length
Thus,

Substituting the given values we get -

Speed of a transverse wave on a string

For third harmonic wave , frequency is equal to

Substituting the given values, we get -

Length of pipe

Substituting the given values we get
for first harmonic wave

Length of pipe
meter