Answer:
(a) The current should be in opposite direction
(b) The current needed is 39.8 A
Explanation:
Part (a)
Based, on right hand rule, the current should be in opposite direction
Part (b)
given;
strength of magnetic field, B = 370 µT
distance between the two parallel wires, d = 8.6 cm

At the center, the magnetic field strength is twice

R = d/2 = 8.6/2 = 4.3 cm = 0.043 m

Therefore, current needed is 39.8 A
Answer:
Sorry cant find the answer but i hope you got it right and if you didn't you'll still do great. :)
Explanation:
Answer:
Selenium dichloride
Explanation:
Selenium (Se) and Chlorine (Cl) are both elements capable of combining together to form a compound with the chemical formula; SeCl2. Since the chlorine atom is more electronegative than selenium atom, the chlorine pulls more electrons towards itself to form an IONIC bond.
The SeCl2 compound formed is called Selenium dichloride as two atoms of Chlorine are needed to combine with one atom of Selenium to form the compound.
Answer: KE = 62.5J
Explanation:
Given that
Mass of object = 5kg
kinetic energy KE = ?
velocity of object = 5m/s
Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.
i.e K.E = 1/2mv^2
KE = 1/2 x 5kg x (5m/s)^2
KE = 0.5 x 5 x 25
KE = 62.5J
Thus, the object has 62.5 joules of kinetic energy.
Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d=
* tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d=
* tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip