Answer:
The value is 
Explanation:
From the question we are told that
The length of the guitar string considered is 
The mass is 
Generally the tension is mathematically represented as

Here f is the frequency of the
note which is 
So

=> 
Answer:
Explanation:
Given
height of wall=5.15 m
angle of launch
Launch velocity(u)=52.4 m/s
Time of flight will be sum of time of flight of projectile+time to cover 5.15 m
Time of flight of arrow

Now time require to cover 5.15 m
Here at the time of zero vertical displacement of arrow i.e. when arrow is at the same height as of building then its vertical velocity will change its sign compared to initial vertical velocity.
at zero vertical displacement
Thus time required will be 



total time =
(b)Horizontal distance=Range of arrow(R_1) + horizontal distance in 0.118 s


=263.28+3.546=266.82 m
Answer:
1. a) Draw a line towards the right side from the engine
b) This force pushes the boat forward and helps it accelerate further
2. a) Fixed volume for both solid and liquid
Compressible for only solid
Fixed shape is also for only Solid
b) The answer is 'c'
c) Solids, because they have their particles closely packed therefore they can be compressed (not so sure bout this answer)
Answer:
the string and metre rule
Explanation:
When the computer is on the circuit has positive and negative charges it will move fluenlty to get the power when it is in on position.