D xrxtxtxt t txt yhhgedd Ed ggdfn
Answer:
x ≤ 3.6913 m
Explanation:
Given
Mrod = 44.0 kg
L = 4.90 m
Tmax = 1450 N
Mman = 69 kg
A: left end of the rod
B: right end of the rod
x = distance from the left end to the man
If we take torques around the left end as follows
∑τ = 0 ⇒ - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0
⇒ - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0
⇒ - (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0
⇒ x ≤ 3.6913 m
Answer:
1.6 x
N
Explanation:
If the force between two actual wires has this value, the current is defined to be exactly 1 A.
The force between two parallel wires carrying current can be defined as,
F= μ
L/ 2πd
where,
current
= 1Amp
current
= 2amp
Length 'L'= 1m
distance 'd'= 2.5m
permeability of free space 'μ'= 4πx
N/m
Putting the above values in the equation,
F=( 4πx
x 1 x 2 x 1 )/ 2πx2.5
F= 1.6 x
N
Let say the height of two balls from the ground is H
now we can use kinematics

now we have


now in the same time ball on the left will cover the horizontal distance between them
![v_x = \frac{d}{ t}[/tex[tex]v_x = \frac{3}{\sqrt{\frac{2H}{g}}}](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7Bd%7D%7B%20t%7D%5B%2Ftex%3C%2Fp%3E%3Cp%3E%5Btex%5Dv_x%20%3D%20%5Cfrac%7B3%7D%7B%5Csqrt%7B%5Cfrac%7B2H%7D%7Bg%7D%7D%7D)
<em>so above is the horizontal speed of the left ball</em>
Idk since it looks sooooo close ,but is it is that’s really cool