Answer:
His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A.
Answer:
B) The lightbulb uses 4,104,000 J more than the stereo.
Answer:
1. The length is 8.35m
2. The period on the moon is 14.05 secs
Explanation:
1. Data obtained from the question. This includes the following:
Period (T) = 5.8 secs
Acceleration due to gravity (g) = 9.8 m/s2
Length (L) =...?
The length can be obtained by using the formula given below:
T = 2π√(L/g)
5.8 = 2π√(L/9.8)
Take the square of both side
(5.8)^2 = 4π^2 x L/ 9.8
Cross multiply
4π^2 x L = (5.8)^2 x 9.8
Divide both side by 4π^2
L = (5.8)^2 x 9.8 / 4π^2
L= 8.35 m
2. Data obtained from the question. This includes the following:
Acceleration due to gravity (g) = 1.67 m/s2
Length (L) = 8.35m (the length remains the same)
Period (T) =?
The period can be obtained as follow:
T = 2π√(L/g)
T = 2π√(8.35/1.67)
T = 14.05 secs
Therefore, the period on the moon is 14.05 secs
This problem uses the relationships among current
I, current density
J, and drift speed
vd. We are given the total of electrons that pass through the wire in
t = 3s and the area
A, so we use the following equation to to find
vd, from
J and the known electron density
n,
so:
<span>The current
I is any motion of charge from one region to another, so this is given by:
</span>
The magnitude of the current density is:
Being:
<span>
Finally, for the drift velocity magnitude vd, we find:
</span>
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
A. The proeutectoid phase is Fe₃c because 0.95 wt/c is greater than the eutectoid composition which is 0.76 wt/c
b. We determine how much total territe and cementite form, we apply the lever rule expressions yields.
Wx = (fe₃c-co/cfe₃ c-cx = 6.70- 0.95/6.70- 0.022 = 0.86
The total cementite
Wfe₃C = 10-Cx/ Cfe₃c -Cx = 0.95 - 0.022/6.70 - 0.022 = 0.14
The total cementite which is formed is
(0.14) × (3.5kg) = 0.49kg
c. We calculate the pearule and the procutectoid phase which cementite form the equation
Ci = 0.95 wt/c
Wp = 6.70 -ci/6.70 - 0.76 = 6.70 -0.95/6.70 - 0.76 = 0.97
0.97 corresponds to mass.
W fe₃ C¹ = Ci - 0.76/5.94 = 0.03
∴ It is equivalent to
(0.03) × (3.5) = 0.11kg of total of 3.5kg mass.
