Answer:
Explanation:
It is given that,
Speed of one quoll around a curve, v = 3.2 m/s (maximum speed)
Radius of the curve, r = 1.4 m
On the curve, the centripetal force is balanced by the frictional force such that the coefficient of frictional is given by :



So, the coefficient of static friction between the quoll's feet and the ground in this trial is 0.74. Hence, this is the required solution.
Answer:
293.7 degrees
Explanation:
A = - 8 sin (37) i + 8 cos (37) j
A + B = -12 j
B = a i+ b j , where and a and b are constants to be found
A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j
- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j
Comparing coefficients of i and j:
a = 8 sin (37) = 4.81452 m
b = -12 - 8cos(37) = -18.38908
Hence,
B = 4.81452 i - 18.38908 j ..... 4 th quadrant
Hence,
cos ( Q ) = 4.81452 / 12
Q = 66.346 degrees
360 - Q = 293.65 degrees from + x-axis in CCW direction
Answer:
72.98 km
Explanation:
Her displacement is simply the distance from her final position to her initial position.
Now, I've drawn and attached a triangle diagram to depict this her movement.
Point O is her initial starting point.
Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.
From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.
Thus;
x² = 62² + 26² - 2(62 × 26)cos 120
x² = 4520 + 806
x² = 5326
x = √5326
x = 72.98 km
Answer:
Diameter of the cylinder will be 
Explanation:
We have given young's modulus of steel
Change in length 
Length of rod 
Load F = 11100 KN
Strain is given by 
We know that young's modulus 
So 

We know that stress 
So 

So 