Answer:
Part a)

Part b)
It will not able to cross the pole
Explanation:
As we know that ball is hit with speed of 110 ft/s at an angle of 35 degree
so here we will say


now at the maximum height the vertical velocity will become zero
so here we can use kinematics

here we have



now we have


Part b)
now the height of ball is related to the distance from point of projection is given as

now we know that



since its coming negative so it will not able to cross the pole
Answer:
The potential energy has a maximum when the ball is a time that is half of the time for total travel
Explanation:
Generally potential energy is a the varies directly with the height according to this formula

and the ball attains a maximum height when the time is equal to half of the total time taken to travel
Recall the definition of the cross product with respect to the unit vectors:
i × i = j × j = k × k = 0
i × j = k
j × k = i
k × i = j
and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)
Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have
a × b = (8i + j - 2k) × (5i - 3j + k)
a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)
… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)
… … … … + 8 (i × k) + (j × k) - 2 (k × k)
a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)
a × b = - 5k - 10j - 24k - 6i - 8j + i
a × b = -5i - 18j - 29k
Answer:
0.0953125 N
Explanation:
Applying,
F = kq'q/r²................. Equation 1
Where F = electrostatic force, k = coulomb's constant, q' and q = first and second charge respectively, r = distance between the charge.
From the equation,
If both charges remain constant,
Therefore,
F = C/r²
C = Constant = product of the two charge(q' and q) and k
Fr² = F'r'²................ Equation 2
From the question,
Given: F = 6.10 N
Assume: r = x m, r' = 8x
Substitute these value into equation 2
6.1(x²) = F'(8x)²
F' = 6.1/64
F' = 0.0953125 N
Hence the new force will become 0.0953125 N
Answer:
I belive it is the one at the bottom of the table visable in the picture with 27.4 radioactive age
Explanation: