Immediately following the arrival of the stimulus at a skeletal muscle cell, there is a short period called the latent period during which the events of excitation-contraction coupling occur.
This process is a connection between transduction in the sarcolemma and the initiation of muscle contraction. Sarcolemma is nothing but the cell membrane of skeletal muscle.
A single muscle twitch has a latent period, a contraction phase when tension increases and a relaxation phase when tension decreases.
The period of incubation, the interval preceding exposure to a pathogen toxin or radiation, and when effects occur. Muscle contracting, the time between a nerve stimulus and muscle contraction.
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Complete question is;
A copper wire has a diameter of 4.00 × 10^(-2) inches and is originally 10.0 ft long. What is the greatest load that can be supported by this wire without exceeding its elastic limit? Use the value of 2.30 × 10⁴ lb/in² for the elastic limit of copper.
Answer:
F_max = 28.9 lbf
Explanation:
Elastic limit is simply the maximum amount of stress that can be applied to the wire before it permanently deform.
Thus;
Elastic limit = Max stress
Formula for max stress is;
Max stress = F_max/A
Thus;
Elastic limit = F_max/A
F_max is maximum load
A is area = πr²
We have diameter; d = 4 × 10^(-2) inches = 0.04 in
Radius; r = d/2 = 0.04/2 = 0.02
Plugging in the relevant values into the elastic limit equation, we have;
2.30 × 10⁴ = F_max/(π × 0.02²)
F_max = 2.30 × 10⁴ × (π × 0.02²)
F_max = 28.9 lbf
Let car A's starting position be the origin, so that its position at time <em>t</em> is
A: <em>x</em> = (40 m/s) <em>t</em>
and car B has position at time <em>t</em> of
B: <em>x</em> = 100 m - (60 m/s) <em>t</em>
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They meet when their positions are equal:
(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>
(100 m/s) <em>t</em> = 100 m
<em>t</em> = (100 m) / (100 m/s) = 1 s
so the cars meet 1 second after they start moving.
They are 100 m apart when the difference in their positions is equal to 100 m:
(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m
(subtract car B's position from car A's position because we take car A's direction to be positive)
(100 m/s) <em>t</em> = 200 m
<em>t</em> = (200 m) / (100 m/s) = 2 s
so the cars are 100 m apart after 2 seconds.
Answer:
167.354 m
Explanation:
We are given;
The mass of the car with bad shock;
m = 1500 kg
The distance at which the car sinks; x =
6 cm = 6 × 10^(−2) m
The total mass of 4 people; m_t = 11 kg
The total speed in the highway; V = 65
mph = 29.058 m/s
The spring's constant can be calculated from the formula;
F = Kx
F is also equal to mg.
Thus;
m_t × g = Kx
K = (m_t × g)/x
K = (11 × 9.81)/(6 × 10^(−2))
K = 1798.5 N/m
Mass of car and four people;m_(c+t) = 1500 + 11 = 1511 kg
Thus, the period cam be calculated from the formula;
T = 2π√((m_c+t)/k)
T = 2π√(1511/1798.5)
T = 5.759 s
the distance between adjacent bumps is calculated from;
Velocity = distance/time
Distance = velocity x time
Distance = 29.058 × 5.759
Distance = 167.354 m
