Answer:
a. 05cm from x axis
b. 8cm from x axis
Explanation:
If the net magnetic field is zero and the currents are in the same direction then the thanks point is between the currents i1 and i2 as show in the attachment below
a. Given that i1= 5A and i2=3A
Let assume the null point is xcm from current i1, then the null point will be (4-x)cm from current i2 since the total length is 4cm.
Now the magnetic field of the current i1 from the null point= to magnetic field of current i2 from the null point
B1=B2
μi1/2πx=μi2/2π(4-x)
i1/x=i2/(4-x)
5/x=3/(4-x)
20-5x=3x
8x=20
8x=2.5cm
since from the left of x axis is 2cm, then the null point is 2.5-2 which 0.5cm from the origin x axis.
The null point is 0.5cm from the origin x axis
b. If both current are flowing in opposite direction, the null point lies outside of the current.
Then with same analysis let assume the first current i1 is xcm from the null point and since the total length is 4cm the second current i2 will be (x-4)cm from the null point.
Also the magnetic field of the current i1 from the null point = to magnetic field of current i2 from the null point
B1=B2
μi1/2πx=μi2/2π(x-4)
i1/x=i2/(x-4)
5/x=3/(x-4)
5x-20=3x
2x=20
x=10cm.
This shows that the distance of the null point from current i1 is 10cm and the current i1 is 2cm from the x axis, then the null point is 10-2=8cm from the origin x axis.
The null point is 8cm from the x axis.
Check the attachment to see the diagram of the current and the null points
Answer:
The range of the projectile is 66.7 meters.
Explanation:
The range of a projectile is given by the following expression as :
..............(1)
The range can be calculated using equation (1). Putting the values of all parameters we get :
R = 66.7 meters
So, the range of the projectile is 66.7 meters. Hence, this is the required solution.
According to Archimedes' Principle, an Object's density determines the depth it will get by displacing the liquid or gas molecules... More the density, more it's depth.
Answer:
2KOH(aq) + H2SO4(aq) ⇒ K2SO4(aq) + 2H2O(l)
Explanation:
The reaction is a neutralization reaction since an acid, aqeous H2SO4 reacts completely with an appropriate amount of alkali, aqueous KOH to produce salt, aqueous K2SO4 and liquid water, H2O only.
2KOH(aq) + H2SO4(aq) ⇒ K2SO4(aq) + 2H2O(l)
Alkali + Acid → Salt + Water.
During this reaction, 2 moles of KOH neutralize 1 mole of H2SO4 to yield 1 mole of K2SO4 and 2 moles of H2O.
