Answer:
Option D
Explanation:
Resistors are said to be connected in “Series”, when they are daisy chained together in a single line. Since all the current flowing through the first resistor has no other way to go it must also pass through the second resistor and the third and so on. Then, resistors in series have a Common Current flowing through them as the current that flows through one resistor must also flow through the others as it can only take one path.
Total Resistance = R₁ + R₂ +R₃ +R₄ ohm
Note then that the total or equivalent resistance, R has the same effect on the circuit as the original combination of resistors as it is the algebraic sum of the individual resistances.
Total resistance R = 3 + 3 + 3 +3
= 12 ohm
B.---A. warm water B. thermocline C. cold water
Vo = 18 m/s
angle 35 degrees
1) Components of the initial velocity
Vox = Vo*cos(35) = 18*cos(35) m/s = 14.74 m/s
Voy = Vo* sin(35) = 18*sin(35) m/s = 10.32 m/s
2) Equations of postion:
x = Vox*t
y = Voy*t - gt^2 / 2
3) Calculations
A) t = 0.5 s, t = 1.0 st = 1.5 s, t = 2.0 s
x = 14.74 * t
t = 0.5 s => x = 14.74 m/s * 0.5s = 7.37 m
t = 1.0 s => x = 14.74 m/s * 1.0s = 14.74 m
t = 1.5s => x = 22.11 m
t = 2s => x = 29.48 m
B)
y = Voy*t - gt^2 / 2
Voy = 10.32 m/s
g = 10 m/s (approximation)
y = 10.32*t - 5t^2
t = 0.5 s=> y = 3.91m
t = 1 s => y = 5.32m
t = 1.5 s => y = 4.23m
t = 2 s => y = 0.64 m
Answer:
1. greater
2. direct
3. smaller
4. inverse
Explanation:
The speed of sound in water is greater than in air; hence for the same frequency the sound wavelength in water is <u>greater </u>than in air (for the given frequency the wavelength is in the <u>direct </u>proportion with the speed of sound).
To "see" an object via the echolocation creature needs to use sound with the wavelength <u>smaller </u>than the size of an object viewed.
That means to "see" objects of the same size dolphin and bat need to use ultrasound of the same wavelength, hence dolphin needs to use higher frequency (for the given speed of sound the wavelength is in <u>inverse </u>proportion with the frequency).
Let say the height of two balls from the ground is H
now we can use kinematics
now we have
now in the same time ball on the left will cover the horizontal distance between them
![v_x = \frac{d}{ t}[/tex[tex]v_x = \frac{3}{\sqrt{\frac{2H}{g}}}](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7Bd%7D%7B%20t%7D%5B%2Ftex%3C%2Fp%3E%3Cp%3E%5Btex%5Dv_x%20%3D%20%5Cfrac%7B3%7D%7B%5Csqrt%7B%5Cfrac%7B2H%7D%7Bg%7D%7D%7D)
<em>so above is the horizontal speed of the left ball</em>
