Answer:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong):</em>
a) a×b = 34.27k
b) a·b = 128.43
c) (a + b)·b = 305.17
d) The component of a along the direction of b = 9.66
Explanation:
<u>Assuming b = 9.3i + 9.5j</u> <em>(b = 931 + 9.5 is wrong)</em> we can proceed as follows:
a) The vectorial product, a×b is:
b) The escalar product a·b is:
c) <u>Asumming (a</u><u> </u><u>+ b)·b</u> <em>instead a+b·b</em> we have:
![(a + b)\cdot b = [(8.6 + 9.3)i + (5.1 + 9.5)j]\cdot (9.3i + 9.5j) = (17.9i + 14.6j)\cdot (9.3i + 9.5j) = 305.17](https://tex.z-dn.net/?f=%28a%20%2B%20b%29%5Ccdot%20b%20%3D%20%5B%288.6%20%2B%209.3%29i%20%2B%20%285.1%20%2B%209.5%29j%5D%5Ccdot%20%289.3i%20%2B%209.5j%29%20%3D%20%2817.9i%20%2B%2014.6j%29%5Ccdot%20%289.3i%20%2B%209.5j%29%20%3D%20305.17)
d) The component of a along the direction of b is:
I hope it helps you!
In this question force is measured in g cm/s2 so we know that to get the answer we times g by cm/s2
50 × 20 = 1000
Answer:
The force is 
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The weight of the gate is 
The vertical component of F is 
From the diagram , taking moment about the pivot we have
Where 
is the weight of the gate evaluated as
=> 
=> 
=>
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
ANSWER:
The easiest way to get a fairly accurate measure of your water flow rate is to time yourself filling up a bucket. So for example if you fill up a 10 litre bucket in 1.5 minutes, then your flow rate will be: 10/1.5 = 6.66 Litres per minute.
