"An object weighing 10.01 grams is placed in a graduated cylinder containing 3.90 mL of water. If the total volume of the object
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This question is incomplete, the complete question is;
A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?
Answer:
the current flows through the outer loop is 1.3 × 10⁻⁵ A
Explanation:
Given the data in the question;
Length = 11.34 cm = 0.1134 m
radius a = 1.85 cm = 0.0185 m
turns N = 1627
Net resistance = 144.9 Ω
radius b = 3.77 cm = 0.0377 m
= 1651.6 Ω
ε = 34.95 V
t = 2.58 μs = 2.58 × 10⁻⁶ s
Now, Inductance; L = μ₀N²πa² /
so
L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134
L = 0.003576665 / 0.1134
L = 0.03154
Now,
ε = d∅/dt = ( BA ) =
[ (μ₀In)πa² ]
so
ε = μ₀n ( πa² )
ε = [ μ₀Nπa² / ]
ε = [ μ₀Nπa² / ] [ (ε/L)e^( -t/
) ]
I = ε/ = [ μ₀Nπa² /
] [ (ε/L)e^( -t/
) ]
so we substitute in our values;
I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]
I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]
I = 1.3 × 10⁻⁵ A
Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A
<u>Answer</u>:
(a) magnitude of F = 797 N
(b)the total work done W = 0
(c)work done by the gravitational force = -1.55 kJ
(d)the work done by the pull = 0
(e) work your force F does on the crate = 1.55 kJ
<u>Explanation</u>:
<u>Given</u>:
Mass of the crate, m = 220 kg
Length of the rope, L = 14.0m
Distance, d = 4.00m
<u>(a) What is the magnitude of F when the crate is in this final position</u>
Let us first determine vertical angle as follows
=>
=> =
Now substituting thje values
=> =
=>
=>
=>
Now the tension in the string resolve into components
The vertical component supports the weight
=>
=>
=>
=>
=>
=>T =2391N
Therefore the horizontal force
F = 797 N
b) The total work done on it
As there is no change in Kinetic energy
The total work done W = 0
<u>c) The work done by the gravitational force on the crate</u>
The work done by gravity
Wg = Fs.d = - mgh
Wg = - mgL ( 1 - Cosθ )
Substituting the values
=
=
=
=
=
= -1552.55 J
The work done by gravity = -1.55 kJ
<u>d) the work done by the pull on the crate from the rope</u>
Since the pull is perpendicular to the direction of motion,
The work done = 0
e)Find the work your force F does on the crate.
Work done by the Force on the crate
WF = - Wg
WF = -(-1.55)
WF = 1.55 kJ
<u>(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)</u>
Here the work done by force is not equal to F*d
and it is equal to product of the cos angle and F*d
So, it is not equal to the product of the horizontal displacement and the answer to (a)