Answer:
= 1.75 × 10⁻⁴ m/s
Explanation:
Given:
Density of copper, ρ = 8.93 g/cm³
mass, M = 63.5 g/mol
Radius of wire = 0.625 mm
Current, I = 3A
Area of the wire, 
= 
Now,
The current density, J is given as
= 2444619.925 A/mm²
now, the electron density, 
where,
=Avogadro's Number
Now,
the drift velocity,
where,
e = charge on electron = 1.6 × 10⁻¹⁹ C
thus,
= 1.75 × 10⁻⁴ m/s
Work of the force = 10 N
Time required for the work = 50 sec
Height = 7 m
We are given with the value of work and time in the question.
Substitute the values in the formula of power and then you'll get the power required.
We know that,
w = Work
p = Power
t = Time
By the formula,
Given that,
Work (w) = 7 m = 70 Joules
Time (t) = 50 sec
Substituting their values,
p = 70/50
p = 1.4 watts
Therefore, the power required is 1.4 watts.
Hope it helps!
Answer:
yes
Explanation:
this is simple
the horizontal line is adjacent
the vertical line is opposite
recall that cos x=adj/hyp
adj=hyp(cos x)
while opp=hyp(sin x)
Answer:
Y, X, Z, W
Explanation:
You know W is the most recent because it features the nucleus in the middle and the electron cloud which was shown in models after the others.
Complete Question:
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.
Answer:
m = 0.001 M
For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/
