Answer:
a 15.22 m/s
b 45.65 m
Explanation:
Using the same formula,
x = vt, where
x is now 45.65, and
t is 3 s, then
45.65 = 3v
v = 45.65/3
v = 15.22 m/s
See the attachment for the part b. We used the distance gotten in part B, to find question A
A vehicle traveling on wet or slick roads can begin to _________ as water forms a barrier between the road and the tires and tra
2 answers:
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The mechanical advantage is defined as the ratio between the force produced by a machine and the force applied in input:
For the crowbar of the problem, the force applied in input is 40 N, while the force produced in output is equal to the weight of the rock that is lifted, so 400 N. Therefore, the mechanical advantage is
For the crowbar of the problem, the force applied in input is 40 N, while the force produced in output is equal to the weight of the rock that is lifted, so 400 N. Therefore, the mechanical advantage is
Answer:
2.25 %
Explanation:
65-95-99.7 is a rule to remember the precentages that lies around the mean.
at the range of mean () plus or minus one standard deviation (
),
at the range of mean plus or minus two standard deviation,
at the range of mean plus or minus three standard deviation,
So, note that they are asking about the probability that it is greater than 0.32, that is the mean (0.3) plus two times the standard deviation (0.1) ()
So we know that the 95.5% is between and
, hence approximately the 4.5% (100%-95.5%) is greater than 0.32 or less than 0.28. But half (4.5%/2=2.25%) is greater than 0.32 and the other half is less than 0.28.
So
Answer:
(a) 1.87 x 10⁻⁴ m/s
(b) 0.013V
Explanation:
(a) Drift speed, , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;
=
----------------(i)
Where;
q = amount of charge
n = free charge density
A = cross-sectional area of the wire
But current density, J, is the electric current per unit cross-section area. This is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e
J = =
Equation (i) can then be written as follows;
=
=
=
---------------------(ii)
From the question;
E = 0.0520N/C
p = 1.72 x 10⁻⁸ Ωm
n = 8.5 x 10²⁸ electrons/m³
c = charge on electron = 1.9 x 10⁻¹⁹C
Substitute these values into equation (ii) as follows;
=
= 1.87 x 10⁻⁴ m/s
(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e
V = E x d [where d = 25.0cm = 0.25m]
V = 0.0520 x 0.25
V = 0.013V