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m=24 kg, Ff=53 N, g=9.8 m/s^2,
use the formula Fg<span> = mg, F</span>g=24*9.8,
solve for Fg, Fg=235.2 N,
then use the formula Ff<span> = ต</span>kFn, 53=235.2ตk,
solve for ตk, ตk<span>=.23</span>
Answer:
Explanation:
Length of copper is L = 10m
Cross sectional area
A = 1.25×10^-4m²
Potential difference V = 0.25V
Magnetic field B = 0.19T
Resistivity ρ = 1.70 × 10^-8 Ω·m.
The resistance of the wire can be calculated using
R = ρL/A
R = 1.70 × 10^-8 × 10/1.25×10^-4
R = 1.36×10^-3 ohms
Using ohms law
v= IR
Then, I = V/R
I = 0.25/1.36×10^-3
I = 183.82 Amps
Maximum torque is give as
τ = NiAB
If the wire form a square
Then each side is L/4 = 10/4 = 2.5m
Then, Area of a square is L²
A = 2.5² = 6.25m²
Number of turn is assume to be 1
N=1
Therefore
τ = 1×183.82×6.25×0.19
τ = 218.29 Nm
Answer:
D) intermolecular attraction deceased
Radiation as it can go through a vacuum
Answer:
a)n= 3.125 x
electrons.
b)J= 1.515 x
A/m²
c)
=1.114 x
m/s
d) see explanation
Explanation:
Current 'I' = 5A =>5C/s
diameter 'd'= 2.05 x
m
radius 'r' = d/2 => 1.025 x
m
no. of electrons 'n'= 8.5 x
a) the amount of electrons pass through the light bulb each second can be determined by:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
As we know that: Q= ne
where e is the charge of electron i.e 1.6 x
C
n= Q/e => 5/ 1.6 x 
n= 3.125 x
electrons.
b) the current density 'J' in the wire is given by
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x
)²)
J= 1.515 x
A/m²
c) The typical speed'
' of an electron is given by:
=
=1.515 x
/ 8.5 x
x |-1.6 x
|
=1.114 x
m/s
d) According to these equations,
J= I/A
=
=
If you were to use wire of twice the diameter, the current density and drift speed will change
Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.
Also drift velocity will decrease as it is inversely proportional to the area