Answer:
8.67807 N
34.7123 N
Explanation:
m = Mass of shark = 92 kg
= Density of seawater = 1030 kg/m³
= Density of freshwater = 1000 kg/m³
= Density of shark = 1040 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
Net force on the fin is (seawater)
![F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N](https://tex.z-dn.net/?f=F_n%3Dmg-V_s%5Crho_%7Bse%7Dg%5C%5C%5CRightarrow%20F_n%3Dmg-%5Cfrac%7Bm%7D%7B%5Crho_%7Bsh%7D%7D%5Crho_%7Bse%7Dg%5C%5C%5CRightarrow%20F_n%3D92%5Ctimes%209.81-%5Cfrac%7B92%7D%7B1040%7D%5Ctimes%201030%5Ctimes%209.81%5C%5C%5CRightarrow%20F_n%3D8.67807%5C%20N)
The lift force required in seawater is 8.67807 N
Net force on the fin is (freshwater)
![F_n=mg-V_s\rho_{f}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{f}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1000\times 9.81\\\Rightarrow F_n=34.7123\ N](https://tex.z-dn.net/?f=F_n%3Dmg-V_s%5Crho_%7Bf%7Dg%5C%5C%5CRightarrow%20F_n%3Dmg-%5Cfrac%7Bm%7D%7B%5Crho_%7Bsh%7D%7D%5Crho_%7Bf%7Dg%5C%5C%5CRightarrow%20F_n%3D92%5Ctimes%209.81-%5Cfrac%7B92%7D%7B1040%7D%5Ctimes%201000%5Ctimes%209.81%5C%5C%5CRightarrow%20F_n%3D34.7123%5C%20N)
The lift force required in a river is 34.7123 N
Answer:
the acceleration of the race car is 2 m/s²
Explanation:
Given;
initial velocity of the race car, u = 44 m/s
final velocity of the race car, v = 66 m/s
time of motion of the race car, t = 11 s
The acceleration of the race car is calculated as;
![a = \frac{v-u}{t} \\\\a = \frac{66-44}{11} \\\\a = 2 \ m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv-u%7D%7Bt%7D%20%5C%5C%5C%5Ca%20%3D%20%5Cfrac%7B66-44%7D%7B11%7D%20%5C%5C%5C%5Ca%20%3D%202%20%5C%20m%2Fs%5E2)
Therefore, the acceleration of the race car is 2 m/s²
Answer:
The speed of Susan is 2.37 m/s
Explanation:
To visualize better this problem, we need to draw a free body diagram.
the work is defined as:
![W=F*d*cos(\theta)](https://tex.z-dn.net/?f=W%3DF%2Ad%2Acos%28%5Ctheta%29)
here we have the work done by Paul and the friction force, so:
![W_p=F_p*d*cos(0)\\F_p=30N*cos(30^o)=26N\\W_p=26*3*(1)=78J](https://tex.z-dn.net/?f=W_p%3DF_p%2Ad%2Acos%280%29%5C%5CF_p%3D30N%2Acos%2830%5Eo%29%3D26N%5C%5CW_p%3D26%2A3%2A%281%29%3D78J)
![W_f=F_f*d*cos(180)\\F_f=\µ*(10*9.8-30N*sin(30^o))=16.6N\\W_p=16.6*3*(-1)=50J](https://tex.z-dn.net/?f=W_f%3DF_f%2Ad%2Acos%28180%29%5C%5CF_f%3D%5C%C2%B5%2A%2810%2A9.8-30N%2Asin%2830%5Eo%29%29%3D16.6N%5C%5CW_p%3D16.6%2A3%2A%28-1%29%3D50J)
Now the change of energy is:
![W_p-W_f=\frac{1}{2}m*v^2\\v=\sqrt{\frac{2(78J-50J)}{10kg}}\\v=2.37m/s](https://tex.z-dn.net/?f=W_p-W_f%3D%5Cfrac%7B1%7D%7B2%7Dm%2Av%5E2%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B2%2878J-50J%29%7D%7B10kg%7D%7D%5C%5Cv%3D2.37m%2Fs)
![\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}](https://tex.z-dn.net/?f=%5Cmathfrak%7B%5Chuge%7B%5Cpink%7B%5Cunderline%7B%5Cunderline%7BAnSwEr%3A-%7D%7D%7D%7D%7D)
Actually Welcome to the Concept of the Projectile Motion.
Since, here given that, vertical velocity= 50m/s
we know that u*sin(theta) = vertical velocity
so the time taken to reach the maximum height or the time of Ascent is equal to
T = Usin(theta) ÷ g, here g = 9.8 m/s^2
so we get as,
T = 50/9.8
T = 5.10 seconds
thus the time taken to reach max height is 5.10 seconds.