Answer:
(a) 42 N
(b)36.7 N
Explanation:
Nomenclature
F= force test line (N)
W : fish weight (N)
Problem development
(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed
We apply Newton's first law of equlibrio because the system moves at constant speed:
∑Fy =0
F-W= 0
42N -W =0
W = 42N
(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²
We apply Newton's second law because the system moves at constant acceleration:
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
∑Fy =m*a
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
F-W= ( W/9.8 )*a
42-W= ( W/9.8 )*1.41
42= W+0.1439W
42=1.1439W
W= 42/1.1439
W= 36.7 N
Answer:
The current decreases.
Explanation:
Current and resistance are inversely proportional. The equation connecting current, resistance and voltage is , where V is voltage, I is current and R is resistance.
Rearranging this equation, you get:
and
If the value of voltage in both equations remains constant, and the value of R decreases, the value of I will increase. Conversely, if in the second equation , the value of V remains constant the value of I decreases, then the value of R, resistance will increase.
Thus, it can be seen that the current will decrease as resistance increases and vice versa.
Aristotle created and it’s credited as the creator.
Answer:
r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.
Explanation:
From a sum of forces:
where Ff = μ * N and
N - m*g = 0 So, N = m*g. Replacing everything on the original equation:
(eq2)
Solving for r:
If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.
The current intensity is defined as the amount of charge flowing through a certain point of a wire divided by the time interval:
where Q is the charge and
is the time. Re-arranging the formula, we have
for the compressor in our problem, the intensity of current is I=66.1 A, while the time is
, so the amount of charge that crosses a certain point of the circuit during this time is