Answer:
<em>The net force that acts on the sailboat has a magnitude of 217 N and is applied at an angle of 64° north of east.</em>
Explanation:
<u>Mechanical Force</u>
The second Newton's law states the net force exerted by an external agent on an object of mass m is:
![\vec F = m.\vec a](https://tex.z-dn.net/?f=%5Cvec%20F%20%3D%20m.%5Cvec%20a)
Where
is the acceleration of the object. Note both the force and the acceleration are vectors. The relationship between them is the mass, a scalar.
This means the net force and the acceleration have the same direction.
The sailboat has a mass m= 350 Kg and moves at a=0.62 m/s^2. The magnitude of the net force acting on the boat is:
![F=0.62 * 350=217\ N](https://tex.z-dn.net/?f=F%3D0.62%20%2A%20350%3D217%5C%20N)
As stated above its direction is the same as the acceleration, thus:
The net force that acts on the sailboat has a magnitude of 217 N and is applied at an angle of 64° north of east.
Answer:
Explanation:
Given that:
distance (z) = 7.86 m
mass of the person = 81.7 kg
Acceleration (a) = 0.729 m/s²
By using Newton's second law along the vertical axis:
![T - mg = ma\\ \\ T = ma + mg \\ \\ T = m(a+g) \\ \\](https://tex.z-dn.net/?f=T%20-%20mg%20%3D%20ma%5C%5C%20%5C%5C%20T%20%3D%20ma%20%2B%20mg%20%5C%5C%20%5C%5C%20%20T%20%3D%20m%28a%2Bg%29%20%5C%5C%20%5C%5C)
T = 81.7 (0.729 +9.8)
T = 860.22 N
The work done now is:
![W_T = T\times z \\ \\](https://tex.z-dn.net/?f=W_T%20%3D%20T%5Ctimes%20z%20%5C%5C%20%5C%5C)
![W_T = 860.22 \times 7.86](https://tex.z-dn.net/?f=W_T%20%3D%20860.22%20%5Ctimes%207.86)
![\mathbf{W_T =6761.3292\ N}](https://tex.z-dn.net/?f=%5Cmathbf%7BW_T%20%3D6761.3292%5C%20N%7D)
Answer:
Explanation:
Given that
Height of the tree is 3.7m
Therefore yo=3.7m
yox= 0m
The tiger lands 4.8m from the tree
Then, Range x=4.8m
Since she flies horizontally and she lands away from the bottom of the tree, her path will be trajectory as shown in the attachment
Let know the time of flight, using the equation of motion
voy is the initial velocity of the vertical motion of the is the tiger, which is zero at the beginning.
y = y0 + voy*t + ½*g*t²
Given that,
y=3.7m
yo=0m
voy=0m/s
g=9.81m/s²
y = y0 + voy*t + ½*g*t²
3.7=0+0•t+½×9.81×t²
3.7=0+0+4.905t²
3.7=4.905t²
t²=3.7/4.905
t²=0.7543
t=√0.7543
t=0.87sec
Time to reach the ground is
Now to know the initial velocity of the horizontal motion, using equation of motion
x=xo+Voxt
Vox is the horizontal initial velocity of the tiger.
x=xo+Voxt
4.8=0+Vox×0.87
4.8=0+0.87Vox.
4.8=0.87Vox.
Then, Vox=4.8/0.87
Vox=5.52m/s
Then, her initial velocity Vo
Vo=√voy²+vox²
Vo=√0²+5.52²
Vo=√5.52²
Vo=5.52m/s
The initial velocity is 5.52m/s