Answer:OB=58.3m
Explanation:
So here cow wanders 30m in north and turns 22 degrees in right side and moves 40m more, as shown in figure given.
now take the starting point as a origin such that cow moves in x-y co-ordinate axis.
As shown in figure length OA is the length when cow moves in north or y direction. Later she takes 22 degrees turn to right and moves 40m more.
So the final displacement is the length of cow from the origin that is length OB.
now co-ordinates of B are [40cos22°,40sin22°+30] i.e [37.084,44.984]
now displacement of cow= length of OB
= ![\sqrt{[37.084]^{2}+[44.984]^{2} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5B37.084%5D%5E%7B2%7D%2B%5B44.984%5D%5E%7B2%7D%20%20%7D)
=
OB =
Answer:
v = 6i + 12j + 4k
Explanation:
Find the magnitude of the direction vector.
√(3² + 6² + 2²) = 7
Normalize the direction vector.
3/7 i + 6/7 j + 2/7 k
Multiply by the magnitude of v.
v = 14 (3/7 i + 6/7 j + 2/7 k)
v = 6i + 12j + 4k
Answer:
m1/m2 = 0.51
Explanation:
First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:
V = √F/u
This is the equation that describes the relation between speed of a pulse and a force exerted on it.
the value of "u" is:
u = m/L
Where m is the mass of the rod, and L the length.
Now, for the rod 1:
V1 = √F/u1 (1)
rod 2:
V2 = √F/u2 (2)
Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:
1.4V2 = √F/u1 (3)
Replacing (2) in (3):
1.4(√F/u2) = √F/u1 (4)
Now, let's solve the equation 4:
[1.4(√F/u2)]² = F/u1
1.96(F/u2) =F/u1
1.96F = F*u2/u1
1.96 = u2/u1 (5)
Now, replacing the expression of u into (5) we have the following:
1.96 = m2/L / m1/L
1.96 = m2/m1 (6)
But we need m1/m2 so:
1.96m1 = m2
m1/m2 = 1/1.96
m1/m2 = 0.51
You first find the mass and the volume of that object. Then you divide mass ÷ volume
