The initial kinetic energy of the boat and its rider is
![K_i = \frac{1}{2} mv_i^2 = \frac{1}{2}(900 kg)(1.4 m/s)^2=882 J](https://tex.z-dn.net/?f=K_i%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20mv_i%5E2%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%28900%20kg%29%281.4%20m%2Fs%29%5E2%3D882%20J%20%20%20)
After Sam stops it, the final kinetic energy of the boat+rider is
![K_f = 0 J](https://tex.z-dn.net/?f=K_f%20%3D%200%20J)
because its final velocity is zero.
For the law of conservation of energy, the work done by Sam is the variation of kinetic energy of the system:
![W=K_f-K_i =0-882 J=-882 J](https://tex.z-dn.net/?f=W%3DK_f-K_i%20%3D0-882%20J%3D-882%20J)
where the negative sign is due to the fact that the force Sam is applying goes against the direction of motion of the boat.
Answer:
56 m
Explanation:
if you do the math it will add up
Since you did not enumerate the features to choose from, I'll just give you a brief information about acidic water.
>Acidic water<span> has a pH below 7 or less than 7.0.
</span>> Acidic water can dissolve a limestone rock.
> The <span>land features which are formed in limestone by reaction with acid water are the stalactites and the stalagmites.</span>
Answer:
The horizontal distance traveled by the ball before it hits the ground is 48.85 meters.
Explanation:
It is given that,
Speed of golf ball, v = 25 m/s
Angle above horizontal or angle of projection, θ = 65°
We need to find the distance travelled by the ball before it hots the ground or in other words we need to find the range. It is given by R.
![R=\dfrac{v^2\ sin2\theta}{g}](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7Bv%5E2%5C%20sin2%5Ctheta%7D%7Bg%7D)
![R=\dfrac{(25\ m/s)^2\ sin2(65)}{9.8\ m/s^2}](https://tex.z-dn.net/?f=R%3D%5Cdfrac%7B%2825%5C%20m%2Fs%29%5E2%5C%20sin2%2865%29%7D%7B9.8%5C%20m%2Fs%5E2%7D)
R = 48.85 m
So, the distance travelled by the ball before it hots the ground is 48.85 meters. Hence, this is the required solution.
Answer:
Explanation:
Given
mass of duck=2.40 kg
=0.150 N
=0.160 N in a direction south of east
Net force in x- direction
![F_x=0.150+0.160\cos 45=0.263 N](https://tex.z-dn.net/?f=F_x%3D0.150%2B0.160%5Ccos%2045%3D0.263%20N)
![a_x=\frac{0.263}{2.4}=0.109 N](https://tex.z-dn.net/?f=a_x%3D%5Cfrac%7B0.263%7D%7B2.4%7D%3D0.109%20N)
net force in Y direction
![F_y=0.160\times \sin 45=0.113 N](https://tex.z-dn.net/?f=F_y%3D0.160%5Ctimes%20%5Csin%2045%3D0.113%20N)
![a_y=\frac{0.113}{2.4}=0.047 m/s^2](https://tex.z-dn.net/?f=a_y%3D%5Cfrac%7B0.113%7D%7B2.4%7D%3D0.047%20m%2Fs%5E2)
Thus displacement in x direction
![x=ut+\frac{at^2}{2}](https://tex.z-dn.net/?f=x%3Dut%2B%5Cfrac%7Bat%5E2%7D%7B2%7D)
![x=0.15\times 2.8+\frac{0.109\times 2.8^2}{2}=0.847 m](https://tex.z-dn.net/?f=x%3D0.15%5Ctimes%202.8%2B%5Cfrac%7B0.109%5Ctimes%202.8%5E2%7D%7B2%7D%3D0.847%20m)
Displacement in Y direction
![Y=ut+\frac{at^2}{2}](https://tex.z-dn.net/?f=Y%3Dut%2B%5Cfrac%7Bat%5E2%7D%7B2%7D)
here u=0
![Y=0+\frac{0.047\times 2.8^2}{2}=0.184 m](https://tex.z-dn.net/?f=Y%3D0%2B%5Cfrac%7B0.047%5Ctimes%202.8%5E2%7D%7B2%7D%3D0.184%20m)
net displacement ![\sqrt{0.847^2+0.184^2}=\sqrt{0.7513}](https://tex.z-dn.net/?f=%5Csqrt%7B0.847%5E2%2B0.184%5E2%7D%3D%5Csqrt%7B0.7513%7D)
=0.866 m
Direction of displacement
![tan\theta =\frac{0.184}{0.847}](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%5Cfrac%7B0.184%7D%7B0.847%7D)
south of east
![\theta =-12.25^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%20%3D-12.25%5E%7B%5Ccirc%7D)