ANSWER:
d. remains a non-zero constant.
STEP-BY-STEP EXPLANATION:
If we consider that there is no air resistance and that the horizontal component would be at x, the velocity remains a non-zero constant
Its simple dominant and recessive inheritance patterns and organism would need two recessive alleles(rr) for the trait to be expressed but only one dominant allele for that trait to be expressed (RR or Rr) however in the case of co-dominant alleles the heterozygous state (Rr) would produce a third phenotype rather that the dominant phenotype.
Answer:
![1.57 * 10^{3} Q](https://tex.z-dn.net/?f=1.57%20%2A%2010%5E%7B3%7D%20Q)
Explanation:
The volume charge density is defined by ρ =
(Equation A), where Q is the charge and V, the volume.
The units in the S.I. are
, so we have to express the radius in meters:
inner radius = ![4 cm * \frac{1 m}{100 cm} = 0.04m](https://tex.z-dn.net/?f=4%20cm%20%2A%20%5Cfrac%7B1%20m%7D%7B100%20cm%7D%20%3D%200.04m)
outer radius = ![6 cm * \frac{1m}{100cm} = 0.06m](https://tex.z-dn.net/?f=6%20cm%20%2A%20%5Cfrac%7B1m%7D%7B100cm%7D%20%20%3D%200.06m)
Now, we know that the volume of the sphere is calculated by the formula:
, and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:
V =
, where
is the outer radius and
is the inner radius.
Replacing the volume formula in the Equation A:
ρ = ![\frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%28r%5E%7B3%7D%20_%7Bo%7D-r_%7Bi%7D%20%5E%7B3%7D%29%7D)
ρ = ![\frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }](https://tex.z-dn.net/?f=%5Cfrac%7B3Q%7D%7B4%5Cpi%20%28r_%7Bo%7D%20%5E%7B3%7D-r_%7Bi%7D%20%5E%7B3%7D%20%29%20%7D)
Replacing the values of the outer and inner radius whe have:
ρ = ![\frac{3Q}{4\pi (1.52 * 10^{-4})}](https://tex.z-dn.net/?f=%5Cfrac%7B3Q%7D%7B4%5Cpi%20%281.52%20%2A%2010%5E%7B-4%7D%29%7D)
ρ = ![1.57 * 10^{3} Q](https://tex.z-dn.net/?f=1.57%20%2A%2010%5E%7B3%7D%20Q)
Answer:
hello your question is incomplete attached below is missing part of the question
answer:
1 ) Magnetic field due to long current carrying wire : ![B = \frac{U_{0} I}{2\pi d}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7BU_%7B0%7D%20I%7D%7B2%5Cpi%20d%7D)
Therefore the net magnetic field due the both wires ; B = B
+ B
. when we adjust the current I
= I
then the Netfield (B ) = zero
2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires
3) Increase in current through the wire will lead to increase in force and this can be explained via this equation
![F = \frac{U_{0}I_{1}I_{2} }{2\pi d }](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7BU_%7B0%7DI_%7B1%7DI_%7B2%7D%20%20%20%7D%7B2%5Cpi%20d%20%7D)
Explanation:
1 ) Magnetic field due to long current carrying wire : ![B = \frac{U_{0} I}{2\pi d}](https://tex.z-dn.net/?f=B%20%3D%20%5Cfrac%7BU_%7B0%7D%20I%7D%7B2%5Cpi%20d%7D)
Therefore the net magnetic field due the both wires ; B = B
+ B
. when we adjust the current I
= I
then the Netfield (B ) = zero
2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires
3) Increase in current through the wire will lead to increase in force and this can be explained via this equation
![F = \frac{U_{0}I_{1}I_{2} }{2\pi d }](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7BU_%7B0%7DI_%7B1%7DI_%7B2%7D%20%20%20%7D%7B2%5Cpi%20d%20%7D)
Answer:
It is half the field strength at 0.5cm
Explanation: