Hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm maybe
Answer:
The magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
Explanation:
Given;
radius of the wire, r = 0.45 m
current on the loop, I = 2.4 A
angle of inclination, θ = 36⁰
torque on the coil, τ = 1.5 N.m
The torque on the coil is given by;
τ = NIBAsinθ
where;
B is the magnetic field
Area of the loop is given by;
A = πr² = π(0.45)² = 0.636 m
τ = NIBAsinθ
1.5 = (1 x 2.4 x 0.636 x sin36)B
1.5 = 0.8972B
B = 1.5 / 0.8972
B = 1.67 T
Therefore, the magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T
<span>The jump from 1966 to 16347 is the largest one or simply we can say it is hard to lose the 3rd electron.Whereas, it is relatively easy to lose the first two electrons.
So there will be only 2 electrons in the outer most shell.
According to the information mentioned above we can conclude the </span><span>unknown element likely belongs to the second group.
</span><span>I2 = 1752 kj/mol</span>
Answer:
1 kg⋅m⋅s−2
Explanation:
I cant really explain it, but thata the answer
Answer:
down below
Explanation:
Since we aren't the given the time, lets say that an object to 25 seconds to fall 50 meters. We can use the formula [ s = d/t ] to solve.
s = 50/25
s = 2
Therefore, the object was falling at a rate of 2 meters per second.
Best of Luck!