Answer:
The net torque on the square plate is 2.72 N-m.
Explanation:
Given that,
Side = 0.2 m
Force ![F_{1}=18\ N](https://tex.z-dn.net/?f=F_%7B1%7D%3D18%5C%20N)
Force ![F_{2}=26\ N](https://tex.z-dn.net/?f=F_%7B2%7D%3D26%5C%20N)
Force ![F_{3}=14\ N](https://tex.z-dn.net/?f=F_%7B3%7D%3D14%5C%20N)
We need to calculate the torque due to force F₁
Using formula of torque
![\tau_{1}=-F_{1}d_{1}](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%3D-F_%7B1%7Dd_%7B1%7D)
![\tau_{1}=-F_{1}\times\dfrac{a}{2}](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%3D-F_%7B1%7D%5Ctimes%5Cdfrac%7Ba%7D%7B2%7D)
Put the value into the formula
![\tau_{1}=-18\times\dfrac{0.2}{2}](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%3D-18%5Ctimes%5Cdfrac%7B0.2%7D%7B2%7D)
![\tau_{1}=-1.8\ N-m](https://tex.z-dn.net/?f=%5Ctau_%7B1%7D%3D-1.8%5C%20N-m)
We need to calculate the torque due to force F₂
Using formula of torque
![\tau_{2}=F_{2}d_{2}](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%3DF_%7B2%7Dd_%7B2%7D)
![\tau_{2}=F_{2}\times\dfrac{a}{2}](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%3DF_%7B2%7D%5Ctimes%5Cdfrac%7Ba%7D%7B2%7D)
Put the value into the formula
![\tau_{2}=26\times\dfrac{0.2}{2}](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%3D26%5Ctimes%5Cdfrac%7B0.2%7D%7B2%7D)
![\tau_{2}=2.6\ N-m](https://tex.z-dn.net/?f=%5Ctau_%7B2%7D%3D2.6%5C%20N-m)
We need to calculate the torque due to force F₃
Using formula of torque
![\tau_{3}=F_{3}d_{3}](https://tex.z-dn.net/?f=%5Ctau_%7B3%7D%3DF_%7B3%7Dd_%7B3%7D)
![\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}](https://tex.z-dn.net/?f=%5Ctau_%7B3%7D%3D%28F_%7B3%7D%5Csin45%2BF_%7B3%7D%5Ccos45%29%5Ctimes%5Cdfrac%7Ba%7D%7B2%7D)
Put the value into the formula
![\tau_{3}=0.1(14\sin45+14\cos45)](https://tex.z-dn.net/?f=%5Ctau_%7B3%7D%3D0.1%2814%5Csin45%2B14%5Ccos45%29)
![\tau_{3}=1.92\ N-m](https://tex.z-dn.net/?f=%5Ctau_%7B3%7D%3D1.92%5C%20N-m)
We need to calculate the net torque on the square plate
![\tau=\tau_{1}+\tau_{2}+\tau_{3}](https://tex.z-dn.net/?f=%5Ctau%3D%5Ctau_%7B1%7D%2B%5Ctau_%7B2%7D%2B%5Ctau_%7B3%7D)
![\tau=-1.8+2.6+1.92](https://tex.z-dn.net/?f=%5Ctau%3D-1.8%2B2.6%2B1.92)
![\tau=2.72\ N-m](https://tex.z-dn.net/?f=%5Ctau%3D2.72%5C%20N-m)
Hence, The net torque on the square plate is 2.72 N-m.
Answer:
A
Explanation:
Please see the attached picture for the full solution.
Since we are only concerned about the decrease in gravitational potential energy of the car, we look at the decrease in height of the car as it moves from point X to point Y, instead of the distance travelled by the car.
Answer:
Mass can never be negative. Everything has mass. Just like how they ask you to find area under the graph in maths. If the area is in the 3rd and 4th quadrant, when calculated, you would get negative answer.However, area can not be negative because it is a place/ location. It's exactly the same as mass.
We getting addicted so it keeps us from getting bored
Force is the product of mass and acceleration .
The question is ask to find acceleration.
But acceleration is the ratio of the force and the mass.
where 600kg is the mass and 7kN is the force
NB: kilo is 1000
now we have to multiply 7N by 1000
by doing so you will have 7000N
which is the force.
Now to find the acceleration: force/ mass
which is 7000/600
therefore the maximum acceleration is 11.667