The answer true I’m guessing. It’s a 50/50 chance
A. It is a compound made of oxygen and carbon
Answer:
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
Explanation:
initial veetical speed V₀y=0
Horizontal speed Vx = Vx₀= 3.80m/s
Vertical drop height= 3.90m
Let Vy = vertical speed when it got to the water downward.
g= 9.81m/s² = acceleration due to gravity
From kinematics equation of motion for vertical drop
Vy²= V₀y² +2 gh
Vy²= 0 + ( 2× 9.8 × 3.90)
Vy= √76.518
Vy=8.747457
Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below
V= √Vy² + Vx²
V=√3.80² + 8.747457²
V=9.537m/s
The angle can also be calculated as
θ=tan⁻¹(Vy/Vx)
tan⁻¹( 8.747457/3.80)
=66.52⁰
the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal
<span>Psychological researchers must debrief human test subjects </span><span>at the end of every experiment.
The current code of ethics in p</span>sychological research states that researchers absolutely must debrief human test subjects at the end of every study regardless or whether or not harm or deception was involved.
Debriefing a subject after a study is an essential opportunity for the researcher to explain the purpose and aim of the study to the subject, make sure the subject is not harmed or mentally disturbed, clarify why deception was used (if deception was involved) and overall, to clarify any questions or doubts the subject might have.
Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
