1) Current in the wire: 0.0875 A
The current in the wire is given by:
where
Q is the charge passing a given point in the conductor
t is the time elapsed
In this problem, we have
Q = 420 C is the total charge passing through a given point in a time of
t = 80 min = 4800 s
So, the current is
2) Drift velocity of the electrons: 
The drift velocity of the electrons in the wire is given by:
where
I = 0.0875 A is the current
is the number of free electrons per cubic meter
A is the cross-sectional area
is the charge of one electron
The radius of the wire is
So the cross-sectional area is
So, the drift velocity is
<h2>Answer:</h2>
<u>This term shows the </u><u>mass of the space shuttle</u>
<h2>Explanation:</h2>
We know that the mass of the Earth is 5.972 × 10^24 kg. Similarly the sum of mass of earth and the mass of shuttle must be a greater number as compared to the number given. It simply means that the mass of earth is itself 5.972 × 10^24 kg and the value given is 3 × 105 kg so it is obvious that if was the sum then it must be greater than the mass of earth. Therefore we can say that this not the mass of earth, neither the sum of mass of earth and shuttle, but this is only the mass of space shuttle which is the last multiple choice.
<span>First, we use the kinetic energy equation to create a formula:
Ka = 2Kb
1/2(ma*Va^2) = 2(1/2(mb*Vb^2))
The 1/2 of the right gets cancelled by the 2 left of the bracket so:
1/2(ma*Va^2) = mb*Vb^2 (1)
By the definiton of momentum we can say:
ma*Va = mb*Vb
And with some algebra:
Vb = (ma*Va)/mb (2)
Substituting (2) into (1), we have:
1/2(ma*Va^2) = mb*((ma*Va)/mb)^2
Then:
1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2
We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1:
1/2(ma) = (ma^2)/mb
Cancel the ma of the left, leaving the right one with exponent 1:
1/2 = ma/mb
And finally we have that:
mb/2 = ma
mb = 2ma</span>
Answer:
3.5 cm
Explanation:
mass, m = 50 kg
diameter = 1 mm
radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m
L = 11.2 m
Y = 2 x 10^11 Pa
Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3
= 7.85 x 10^-7 m^2
Let the wire is stretch by ΔL.
The formula for Young's modulus is given by
ΔL = 0.035 m = 3.5 cm
Thus, the length of the wire stretch by 3.5 cm.
Answer:
44 N/m
Explanation:
The extension, e, of the spring = 2.9 m - 1.4 m = 1.5 m
The work needed to stretch a spring by <em>e</em> is given by
where <em>k</em> is spring constant.
Using the appropriate values,
