Answer:
e. all of these
Explanation:
The fatigue strength is improved by then high alloy steels , high yield steels , high hardened steel , high ultimate steel .
Due to the formation of the improved materials in alloy steels will increase the fatigue strength . Similarly for a high yield steels and hardened steels these cycles to failure will improve .
<span>they are travelling at right angles to each other.
At any given instant they form a right triangle with their starting point
</span>South bound <span>= x [mi/h]
</span> East bound <span> = x+1 [mi/h]
after five hours they will be
d=5x
and
d=5(x+1)
miles away from the starting point
(5x)^2+(5(x+1))^2=625
25x^2+(5x+5)^2=625
25x^2+25x^2+50x+25=625
50</span>x^2+50x-600=0
<span> x^2+ x - 12=0
(x+4)(x-3)=0
take the postive value
x= 3 mph the speed of south bound
4mph east bound </span>
The ears are superior and posterior to the shoulders and lateral to the nose.
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.