Answer: 2/-2 x is always on the top y is always on the bottom on the x axis it moves 2 spaces in the positive direction in the t axis it moves 2 spaces down in the negative direction
Answer: 5, because if x=5 then the denominator would be equal to 0 which is not possible.
Answer:
Step-by-step explanation:
bearing is the angle which a line makes with the north.
Answer:
7/10
Step-by-step explanation:
Answer:
Therefore the magnitude of the applied force is
N.
Step-by-step explanation:
Torque: Torque is the cross product of force and the distance of applied force from the rotational axis.
![\tau=\vec F\times \vec r= |F||r| sin\theta](https://tex.z-dn.net/?f=%5Ctau%3D%5Cvec%20%20F%5Ctimes%20%5Cvec%20r%3D%20%7CF%7C%7Cr%7C%20sin%5Ctheta)
Where
is angle between F and r.
Newton-meter is the S.I unit of torque.
Along the positive y axis, the wrench lies .
The length of wrench is 0.1 m.
[ since the vector
lies on y axis]
The direction of applied force is along the vector <0,2,-4>.
Then the angle between
and <0,2,-4> = the angle between F and
.
We know that,
![=0+(0.1).2+0.(-4)=0.2](https://tex.z-dn.net/?f=%3D0%2B%280.1%29.2%2B0.%28-4%29%3D0.2)
![\therefore\sqrt{(0.1)^2}.\sqrt{0^2+2^2+(-4)^2} \ cos \theta= 0.2](https://tex.z-dn.net/?f=%5Ctherefore%5Csqrt%7B%280.1%29%5E2%7D.%5Csqrt%7B0%5E2%2B2%5E2%2B%28-4%29%5E2%7D%20%5C%20%20cos%20%5Ctheta%3D%200.2)
![\Rightarrow (0.1).2\sqrt5 cos \theta =0.2](https://tex.z-dn.net/?f=%5CRightarrow%20%280.1%29.2%5Csqrt5%20cos%20%5Ctheta%20%3D0.2)
![\Rightarrow cos \theta =\frac{0.2}{0.2\sqrt 5}](https://tex.z-dn.net/?f=%5CRightarrow%20%20cos%20%5Ctheta%20%3D%5Cfrac%7B0.2%7D%7B0.2%5Csqrt%205%7D)
![\Rightarrow cos \theta =\frac{1}{\sqrt 5}](https://tex.z-dn.net/?f=%5CRightarrow%20%20cos%20%5Ctheta%20%3D%5Cfrac%7B1%7D%7B%5Csqrt%205%7D)
We know that,
![sin \theta =\sqrt{1-cos^2\theta}=\sqrt{1-(\frac{1}{\sqrt5})^2}](https://tex.z-dn.net/?f=sin%20%5Ctheta%20%3D%5Csqrt%7B1-cos%5E2%5Ctheta%7D%3D%5Csqrt%7B1-%28%5Cfrac%7B1%7D%7B%5Csqrt5%7D%29%5E2%7D)
![=\sqrt {1-\frac15}](https://tex.z-dn.net/?f=%3D%5Csqrt%20%7B1-%5Cfrac15%7D)
![=\sqrt {\frac{5-1}{5}](https://tex.z-dn.net/?f=%3D%5Csqrt%20%7B%5Cfrac%7B5-1%7D%7B5%7D)
![=\frac{2}{\sqrt 5}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B2%7D%7B%5Csqrt%205%7D)
Here
= 100 N-m,
and ![sin \theta = \frac{2}{\sqrt 5}](https://tex.z-dn.net/?f=sin%20%5Ctheta%20%3D%20%5Cfrac%7B2%7D%7B%5Csqrt%205%7D)
Now
![\tau=|F||r| sin\theta](https://tex.z-dn.net/?f=%5Ctau%3D%7CF%7C%7Cr%7C%20sin%5Ctheta)
![\Rightarrow 100= |F|(0.1) \frac{2}{\sqrt 5}](https://tex.z-dn.net/?f=%5CRightarrow%20100%3D%20%7CF%7C%280.1%29%20%5Cfrac%7B2%7D%7B%5Csqrt%205%7D)
![\Rightarrow |F|=\frac{100\times \sqrt5}{0.2}](https://tex.z-dn.net/?f=%5CRightarrow%20%7CF%7C%3D%5Cfrac%7B100%5Ctimes%20%5Csqrt5%7D%7B0.2%7D)
![\Rightarrow |F|=500\sqrt 5](https://tex.z-dn.net/?f=%5CRightarrow%20%7CF%7C%3D500%5Csqrt%205)
Therefore the magnitude of the force is
N.