Explanation:
We'll need two equations.
v² = v₀² + 2a(x - x₀)
where v is the final velocity, v₀ is the initial velocity, a is the acceleration, x is the final position, and x₀ is the initial position.
x = x₀ + ½ (v + v₀)t
where t is time.
Given:
v = 47.5 m/s
v₀ = 34.3 m/s
x - x₀ = 40100 m
Find: a and t
(47.5)² = (34.3)² + 2a(40100)
a = 0.0135 m/s²
40100 = ½ (47.5 + 34.3)t
t = 980 s
Answer:
Distance, d = 0.1 m
It is given that,
Initial velocity of meson,
Finally, the meson is coming to rest v = 0
Acceleration of the meson, (opposite to initial velocity)
Using third equation of motion as :
s is the distance the meson travelled before coming to rest.
So,
s = 0.1 m
The meson will cover the distance of 0.1 m before coming to rest. Hence, this is the required solution.
By calculation, the diameter of the wire is 2.8 * 10^-3 m.
<h3>How do we obtain the length?</h3>
The following data are given in the question;
Mass of the wire = 1.0 g or 1 * 10^-3 Kg
Resistance = 0.5 ohm
Resistivity of copper = 1.7 * 10^-8 ohm meter
Density of copper = 8.92 * 10^3 Kg/m^3
V = m/d
But v = Al
Al = m/d
A = m/ld
Resistance = ρl/A
= ρl/m/ld =
l^2 = Rm/ρd
l = √ Rm/ρd
l = √0.5 * 1 * 10^-3 / 1.7 * 10^-8 * 8.92 * 10^3
l = 1.82 m
A = πr^2
Also;
A = m/ld
A = 1 * 10^-3 Kg / 1.82 m * 8.92 * 10^3 Kg/m^3
Area of the wire = 6.2 * 10^-5 m^2
r^2 = A/ π
r = √A/ π
r = √6.2 * 10^-5 m^2/3.142
r = 1.4 * 10^-3 m
Diameter = 2r = 2( 1.4 * 10^-3 m) = 2.8 * 10^-3 m
Learn more about resistivity:brainly.com/question/14547003
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Missing parts;
Suppose you wish to fabricate a uniform wire from 1.00g of copper. If the wire is to have a resistance of R=0.500Ω and all the copper is to be used, what must be (a) the length and (b) the diameter of this wire?
Answer:
Electric current is defined as the rate of flow of electric charge in a circuit from point one point to another. This is carried by electrically charged particles within the circuit. Current is represented by the symbol I and its unit measured in Amperes. It is therefore related to the voltage and resistance of the circuit. If the current in the circuit reduces, the rate at which the charge and current on the capacitor reduces also proportionally in an exponential manner.
Explanation:
Since a decrease in the flow of current in the circuit is observed, the implication for the rate at which the charge and voltage on the capacitor is also an exponential decrease in the rate of flow with time. This is because the electric current is directly proportional to the electric charge and the time.
I believe it is acceleration!
