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 = $\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]$

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

3 0

2 years ago

Two pieces of amber are hung from threads. Piece A is charged by rubbing piece A with fur. Piece B is charged by rubbing piece B

PIT_PIT [208]

Answer:

ieces A and B must also have the same type of charges

Explanation:

In electrostatics, charges of the same sign repel and charges of different signs attract.

If we apply this to our case, we have that part A and C repel each other, therefore they have the same type of charge.

Also part A and C repel each other, therefore they have the same type of charge.

If we use the transitive property of mathematics, pieces A and B must also have the same type of charges

6 0

2 years ago

0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for

Paladinen [302]

Assuming the accleration applied was constant, we have

$v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)$

$\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)$

$\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}$

Then the force applied to the ball is given by

$F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)$

$\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N$

8 0

2 years ago