Answer:
Distance is 50m
Displacement is 0m
Explanation:
Distance is based on the amount of length you covered, regardless of where you end.
Displacement only considered where you started and where you ended, which is at the same spot in this case. Therefore, no displacement.
The answer is GAS. In a gas, the particles are in a completely random motion in any direction. And it is not solid. Hope I helped. Good luck
Answer:
Area=1.5(1.5)=2.25m^2
Force of gravity=10N
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{Force}{Area}\end{gathered}
⟼Pressure=
Area
Force
\begin{gathered}\\ \sf\longmapsto Pressure=\dfrac{10}{2.25}\end{gathered}
⟼Pressure=
2.25
10
\begin{gathered}\\ \sf\longmapsto Pressure=4.4Pa\end{gathered}
⟼Pressure=4.4Pa
The vector perpendicular to the plane of A = 3i+ 6j - 2k and B = 4i-j +3k is 16 i - 17 j - 27 k
Let r be the vector perpendicular to A and B,
r = A * B
A = 3i + 6j - 2k
B = 4i - j + 3k
a1 = 3
a2 = 6
a3 = - 2
b1 = 4
b2 = - 1
b3 = 3
a * b = ( a2 b3 - b2 a3 ) i + ( a3 b1 - b3 a1 ) j + ( a1 b2 - b1 a2 ) k
a * b = [ ( 6 * 3 ) - ( - 1 * - 2 ) ] i + [ ( - 2 * 4 ) - ( 3 * 3 ) ] j + [ ( 3 * - 1 ) - ( 4 * 6 ) ] k
a * b = 16 i - 17 j - 27 k
The perpendicular vector, r = 16 i - 17 j - 27 k
Therefore, the vector perpendicular to the plane of A = 3i + 6j - 2k and B = 4i - j + 3k is 16 i - 17 j - 27 k
To know more about perpendicular vectors
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Answer:
Usually the coefficient of friction remains unchanged
Explanation:
The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.
Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.
