Answer:
The force parallel to the horizontal is 26.24 N
Explanation:
She pulls on the leash with a force F = 30 N, this force, since its at an angle of 29° (i will cal this angle 
), it has a force component on x (the horizontal, i will call this force 
) and a force component on y (the vertical, i will call this 
).
This can be seen in the attached picture.
Since we are asked about the force parallel to the horizontal, we need to find the component of the force , since is the adjacent angle, we need to use cosine:
since 
and 
The force parallel to the horizontal is 26.24 N
Answer:
v₁ = 3.5 m/s
v₂ = 6.4 m/s
Explanation:
We have the following data:
m₁ = mass of trailing car = 400 kg
m₂ = mass of leading car = 400 kg
u₁ = initial speed of trailing car = 6.4 m/s
u₂ = initial speed of leading car = 3.5 m/s
v₁ = final speed of trailing car = ?
v₂ = final speed of leading car = ?
The final speed of the leading car is given by the following formula:
<u>v₂ = 6.4 m/s</u>
The final speed of the leading car is given by the following formula:
<u>v₁ = 3.5 m/s</u>
Answer:
Applying enough force in one direction to move the object, making kinetic energy.
Explanation:
Simpleness
Answer:
F = 0.78[N]
Explanation:
The given values correspond to forces, we must remember or take into account that the forces are vector quantities, that is, they have magnitude and direction. Since we have two X-Y coordinate axes (two-dimensional), we are going to decompose each of the forces into the X & y components.
<u>For F₁</u>
<u />![F_{y}=2[N]](https://tex.z-dn.net/?f=F_%7By%7D%3D2%5BN%5D)
<u />
<u>For F₂</u>
![F_{x}=2*cos(60)\\F_{x}=1[N]\\F_{y}=-2*sin(60)\\F_{y}=-1.73[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D2%2Acos%2860%29%5C%5CF_%7Bx%7D%3D1%5BN%5D%5C%5CF_%7By%7D%3D-2%2Asin%2860%29%5C%5CF_%7By%7D%3D-1.73%5BN%5D)
<u>For F₃</u>
<u />![F_{x}=-1*sin(60)\\F_{x}=-0.866[N]\\F_{y}=1*cos(60)\\F_{y}=0.5 [N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D-1%2Asin%2860%29%5C%5CF_%7Bx%7D%3D-0.866%5BN%5D%5C%5CF_%7By%7D%3D1%2Acos%2860%29%5C%5CF_%7By%7D%3D0.5%20%5BN%5D)
<u />
Now we can sum each one of the forces in the given axes:
![F_{x}=1-0.866=0.134[N]\\F_{y}=2-1.73+0.5\\F_{y}=0.77[N]](https://tex.z-dn.net/?f=F_%7Bx%7D%3D1-0.866%3D0.134%5BN%5D%5C%5CF_%7By%7D%3D2-1.73%2B0.5%5C%5CF_%7By%7D%3D0.77%5BN%5D)
Now using the Pythagorean theorem we can find the total force.
![F=\sqrt{(0.134)^{2} +(0.77)^{2}}\\F= 0.78[N]](https://tex.z-dn.net/?f=F%3D%5Csqrt%7B%280.134%29%5E%7B2%7D%20%2B%280.77%29%5E%7B2%7D%7D%5C%5CF%3D%200.78%5BN%5D)
