The rate constant for this second‑order reaction is 0.760 M−1⋅s−1 at 300 ∘C. A⟶products How long, in seconds, would it take for
1 answer:
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Answer:
Explanation:
When a spring is compressed, the force exerted by the spring is given by:
where
k is the spring constant
x is the compression of the spring
In this problem we have:
k = 52 N/m is the spring constant
x = 43 cm = 0.43 m is the compression
Therefore, the force exerted by the spring on the dart is
Now we can apply Newton' second law of motion to calculate the acceleration of the dart:
where
F = 22.4 N is the force exerted on the dart by the spring
m = 75 g = 0.075 kg is the mass of the dart
a is its acceleration
Solving for a,
Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find: