Answer:
(a) 0.0178 Ω
(b) 3.4 A
(c) 6.4 x 10⁵ A/m²
(d) 9.01 x 10⁻³ V/m
Explanation:
(a)
σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹
d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m
Area of cross-section of the wire is given as
A = (0.25) π d²
A = (0.25) (3.14) (2.6 x 10⁻³)²
A = 5.3 x 10⁻⁶ m²
L = length of the wire = 6.7 m
Resistance of the wire is given as
R = 0.0178 Ω
(b)
V = potential drop across the ends of wire = 0.060 volts
i = current flowing in the wire
Using ohm's law, current flowing is given as
i = 3.4 A
(c)
Current density is given as
J = 6.4 x 10⁵ A/m²
(d)
Magnitude of electric field is given as
E = 9.01 x 10⁻³ V/m
Answer:
F=126339.5N
Explanation:
to find the necessary force to escape we must make a free-body diagram on the hatch, taking into account that we will match the forces that go down with those that go up, taking into account the above we propose the following equation,
Fw=W+Fi+F
where
Fw= force or weight produced by the water column above the submarine.
to fint Fw we can use the following ecuation
Fw=h. γ. A
h=distance
γ=
specific weight for seawater = 10074N / m ^ 3
A=Area
Fw=28x10074x0.7=197467N
w is the weight of the hatch = 200N
Fi is the internal force of the submarine produced by the pressure = 1atm = 101325Pa for this we can use the following formula
Fi=PA=101325x0.7=70927.5N
finally the force that is needed to open the hatch is given by the initial equation
Fw=W+Fi+F
F=Fw-W+Fi
F=197467N-200N-70927.5N
F=126339.5N
Answer:
Fc = 7.14N
Explanation:
First of all, let's convert everything to the same unit system:
m = 0.0031kg R = 13.1cm * 1m / 100cm = 0.131m
ω = 50000 rev/min * 1rev /( 2π rad ) * 1min / 60s = 132.63 rad/s
Now we can calculate centripetal force as:
Replacing the values we get the answer:
Fc = 7.14N
Answer:
c. 972
Explanation:
The volume of a rectangular solid is calculated as the product of its dimensions, that is, its width, its length and its height:
1 feet is equal to 12 inches, so:
Now, we calculate the volume of the object in cubic inches:
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