Answer:128 N
Explanation:
Sample of 3 cm and 4 mm diameter found to break under a minimum force of 128 N .
If sample is 1.5 cm long with same cross-sectional area then minimum force required to break is also 128 N because the applied force is same for any length and diameter of tendon.
With gravitational acceleration at 9.8, initial height at 3.5m and distance at 22m the initial horizontal velocity is 26.03 ms and the flight time is .845 seconds
To solve this problem it is necessary to apply the concepts related to Newton's second law and the kinematic equations of movement description.
Newton's second law is defined as
Where,
m = mass
a = acceleration
From this equation we can figure the acceleration out, then
From the cinematic equations of motion we know that
Where,
Final velocity
Initial velocity
a = acceleration
x = displacement
There is not Final velocity and the acceleration is equal to the gravity, then
From the equation of motion where acceleration is equal to the velocity in function of time we have
Therefore the time required is 0.0705s
Answer:
(a): The frequency of the waves is f= 0.16 Hz
Explanation:
T/4= 1.5 s
T= 6 sec
f= 1/T
f= 0.16 Hz (a)
Answer: 13.1° from the horizontal
Explanation: For projectile
Horizontal distance R of a projectile is:
R= U×Usin2x/g
Where U is the speed of projectile, x is angle of projectile and g= acceleration due to gravity=9.8m/s
R= 15×15sin(2×31)/9.8= 225sin(62)/9.8=20.27m
Therefore if R is halved.
R/2 = 20.27/2=10.14m
Hence the angle for this case is.
Making sin(2x) the subject of formula
Sin2x= Rg/U×U
R is now 10.14 in this case.
Sin2x= 10.14×9.8/15×15=0.4415
2x=arcsin0.4415=26.2
x= 26.2/2
x= 13.1°
Good luck...
