Answer: The magnitude of the current in the second wire 2.67A
Explanation:
Here is the complete question:
Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?
Explanation: Please see the attachments below
Answer:
0.661 s, 5.29 m
Explanation:
In the y direction:
Δy = 2.14 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 0.661 s
In the x direction:
v₀ = 8 m/s
a = 0 m/s²
t = 0.661 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (8 m/s) (0.661 s) + ½ (0 m/s²) (0.661 s)²
Δx = 5.29 m
Round as needed.
Answer:
ω = 3.61 rad/sec
Explanation:
Firstly, we should know that the bug will not slip if friction can provide sufficient opposing force.
μmg = mv^2/r = mω^2r
Thus;
μg = ω^2r
ω^2 = μg/r
ω = √(μg/r)
ω = √(0.321 * 9.8)/0.241
ω = √(13.05)
= 3.61 rad/sec
W = force * displacement
W = 32 pounds * 10 feet
Now you need to convert it to newton and meters
W = 142 N * 3.048 m = 434 J
(I approximated the conversions- I hope it helps)
Answer:
Given values of Planck Constant are equivalent in English system and metric system.
Explanation:
Value of Planck's constant is given in English system as 4.14 x 10⁻¹⁵eV s.
Converting this in to metric system .
We have 1 eV = 1.6 x 10⁻¹⁹ J
Converting
4.14 x 10⁻¹⁵eV s = 4.14 x 10⁻¹⁵x 1.6 x 10⁻¹⁹ = 6.63 x 10⁻³⁴ Joule s
So Given values of Planck Constant are equivalent in English system and metric system.