The 12-lead ECG test that shows st-segment elevation in leads v2 and v3 of ≥ 0.2 mv, could be a diagnostic for myocardial infarction.
What is 12 - lead ECG test?
A 12-lead electrocardiogram (ECG) is a medical test that is recorded using leads, or nodes, attached to the body.
Electrocardiograms, sometimes referred to as ECGs, capture the electrical activity of the heart and transfer it to graphed paper.
When a 12-lead ECG shows st-segment elevation in leads v2 and v3 of ≥ 0.2 mv, it could be a diagnostic for myocardial infarction.
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Particles q1, 92, and q3 are in a straight line.
Particles q1 = -5.00 x 10-6 C,q2 = -5.00 x 10-6 C,
and q3 = -5.00 x 10-6 C. Particles q₁ and q2 are
separated by 0.500 m. Particles q2 and q3 are
separated by 0.500 m. What is net force on 93?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-5.00 x 10-6 C
91
0.500 m-
-5.00 x 10-6 C
92
-5.00 x 10-6 C
93
0.500 m-
The answer to the question is 1.125 Newton
Formula for electrostatic force is F = ( K q1 q2 )/ r²
where q1 and q2 represent the charges and r represents the distance between them and the value of K is 9 × 10⁹.
Total net force on q3 will be the summation of electrostatic force between q1 and q3 and electrostatic force between q2 and q3, as all the three charges are of same sign and lie in the same line.
Electrostatic force between q1 and q3
r will be 0.500 + 0.500 = 1 m
F = ( K q1 q3 )/ r²
F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 1²
F₁₃ = 2.25 × 10⁻¹ N
Electrostatic force between q2 and q3
r will be 0.500 m
F = ( K q1 q3 )/ r²
F = ( (9 × 10⁹ ) x (5.00 x 10⁻⁶) x (5.00 x 10⁻⁶) )/ 0.5²
F = (225 × 10⁻³) / (25 × 10⁻²)
F₂₃ = 9 × 10⁻¹ N
Total force on q3 will be F₁₃ + F₂₃
Total force on q3 = ( 2.25 × 10⁻¹ ) + (9 × 10⁻¹ ) N
Total force on q3 = ( 2.25 × 10⁻¹ ) + (9 × 10⁻¹ ) N
Total force on q3 = ( 11.25 × 10⁻¹ ) N
Total force on q3 = 1.125 N
Thus after solving we got the net force on q3 as 1.125 Newton
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The vertical component of 9N is found to be four-thirds of the horizontal component.
a) Find the horizontal component.
b)Find the magnitude of the vector
Answer:
a) Let horizontal component be h
4/3×h= 9
4h = 9x3
h = 27/4
b) Magnitude = √ [9²+ (27/4)²
= 11.25
A ball is shot from the ground straight up into the air with initial velocity of 40 ft/sec. Assuming that the air resistance can be ignored, how high does it go
The height to which the ball attained, given the data is 7.58 m
Data obtained from the questionInitial velocity (u) = 40 ft/s = 40 × 0.3048 = 12.192 m/sFinal velocity (v) = 0 m/s (at maximum height) Acceleration due to gravity (g) = 9.8 m/s²Maximum height (h) =? How to determine the maximum heightv² = u² – 2gh (since the ball is going against gravity)
0² = 12.192² – (2 × 9.8 × h)
0 = 12.192² – 19.6h
Collect like terms
0 – 12.192² = –19.6h
–12.192² = –19.6h
Divide both side by –19.6
h = –12.192² / –19.6
h = 7.58 m
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