Answer:
A) ( - 200t + 40 ) volts
B) b) anticlockwise , c) anticlockwise , d) clockwise , e) clockwise
Explanation:
Given data:
magnetic flux (Φm) = 5.0t^2 − 2.0t
number of turns = 20
<u>a) determine induced emf </u>
E = - N 
= - N ( 10t - 2 ) = - 20 ( 10t - 2 )
= - 200t + 40 volts
<u>b) Determine direction of induced current </u>
i) at t = 0
E = - 0 + 40 ( anticlockwise direction )
ii) at t = 0.10
E = -20 + 40 = 20 ( anticlockwise direction )
iii) at t = 1
E = - 200 + 40 = - 160 ( clockwise direction)
iv) at t = 2
E = -400 + 40 = - 360 ( clockwise direction )
Answer:
a. 79.1 N
b. 344 J
c. 344 J
d. 0 J
e. 0 J
Explanation:
a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force 
by the worker must be equal to the friction force 
on the crate, which is the product of friction coefficient μ and normal force N:
Let g = 9.81 m/s2
b. The work is done on the crate by this force is the product of its force and the distance traveled s = 4.35
c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35
This work is negative because the friction vector is in the opposite direction with the distance vector
d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.
e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction
C is the correct answer.
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Answer:
0.558 atm
Explanation:
We must first consider that both gases behaves like ideal gases, so we can use the following formula: PV=nRT
Then, we should consider that, whithin a mixture of gases, the total pressure is the sum of the partial pressure of each gas:
P₀ = P₁ + P₂ + ....
P₀= total pressure
P₁=P₂= is the partial pressure of each gass
If we can consider that each gas is an ideal gas, then:
P₀= (nRT/V)₁ + (nRT/V)₂ +..
Considering the molecular mass of O₂:
M O₂= 32 g/mol
And also:
R= ideal gas constant= 0.082 Lt*atm/K*mol
T= 65°C=338 K
4.98 g O₂ = 0.156 moles O₂
V= 7.75 Lt
Then:
P°O₂=partial pressure of oxygen gas= (0.156x0.082x338)/7.75
P°O₂= 0.558 atm
