Answer:
M₀ = 5i - 4j - k
Explanation:
Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e
M₀ = r x F
From the question,
r = i + j + k
F = 1i + 0j + 5k
Therefore,
M₀ = (i + j + k) x (1i + 0j + 5k)
M₀ = ![\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C1%261%261%5C%5C1%260%265%5Cend%7Barray%7D%5Cright%5D)
M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)
M₀ = i(5) - j(4) + k(-1)
M₀ = 5i - 4j - k
Therefore, the moment about the origin O of the force F is
M₀ = 5i - 4j - k
Answer:
"where crests and troughs have their maxima at the same time"
Crests and troughs are 180 deg out of phase and when they have their maxima at the same time and place, their net contribution will be zero"
<h3>No:1</h3>
The object is moving with constant or uniform acceleration and in average speed
<h3>No:-2</h3>
The object is de accelerating
<h3>No:-3</h3>
The object deaccelerated and came to rest so fast.
<h3>No:-4</h3>
The object moves slowly first then accelerated.
<h3>No:-5</h3>
The object accelerated at first so fast then move with constant acceleration then again accelerated .
A steering wheel, a wrench, a screwdriver, and the back wheel of a bike are all examples of tools with a wheel and axle.
