Answer:
See description
Explanation:
This is an example where we need Tornicelli's law, which states that the horizontal speed of a fluid that starts falling from an orifice is the same speed that an object acquires from free-falling.

we are given:
![h_{cilinder} = 0.2 [m]\\h = 0.05 [m]\\d=0.15[m]](https://tex.z-dn.net/?f=h_%7Bcilinder%7D%20%3D%200.2%20%5Bm%5D%5C%5Ch%20%3D%200.05%20%5Bm%5D%5C%5Cd%3D0.15%5Bm%5D)
the horizontal velocity of the water at the start is:
![v = \sqrt{2(9.8)(0.05)}=0.989949 [m/s]=1[m/s]](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2%289.8%29%280.05%29%7D%3D0.989949%20%5Bm%2Fs%5D%3D1%5Bm%2Fs%5D)
now we need to find the time for the water drops to fall d:
as the gravity is the only force interacting with the water we have:

replace for y = d
![0.15 = \frac{1}{2} g*t^2=>t=\sqrt{\frac{2*0.15}{9.8}}=0.1749[s]](https://tex.z-dn.net/?f=0.15%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20g%2At%5E2%3D%3Et%3D%5Csqrt%7B%5Cfrac%7B2%2A0.15%7D%7B9.8%7D%7D%3D0.1749%5Bs%5D%20)
now that we have t we notice that there are no horizontal forces interacting with the water, so the horizontal position is given by:

Finally, we replace v and t:
![x(2.45) = 1*0.1749 = 0.1749 [m]=17.49[cm]](https://tex.z-dn.net/?f=x%282.45%29%20%3D%201%2A0.1749%20%3D%200.1749%20%5Bm%5D%3D17.49%5Bcm%5D)
Answer:
Waxing Gibbous
Third quarter
Waning Gibbous
Explanation:
If moon rises at 3:00 pm then the phase of the moon will be "Waxing Gibbous".
This is because, the moon is actually not fully illuminated but has achieved more than half of its full illumination.
If the moon is highest in the sky at sunrise then the phase of the Moon will be the "Third quarter"
This is because of the fact that at this position moon will rise at midnight, thus it will be at the highest point at the time of the sunrise.
If the moon sets at 10:00 am then the phase of the Moon is "Waning Gibbous"
This is because of the fact that at this position the Moon is moving towards becoming new Moon but at the same time, the moon is illuminated more than its half illumination.
Answer:
the cat is 0.4238 m in front of the dog as it leaps through the window
Explanation:
Given that;
acceleration a = 0.85 m/s²
speed v = 1.40 m/s
the cat is at rest, so initial velocity u = 0
we know that, since the cat is sleeping on the floor in the middle of a 2.8-m-wide room, it needs to cover (2.8 m / 2 ) distance to get to the window;
using the second equation equation of motion;
s = ut + 1/2 at²
we substitute
2.8/2 = 0×t + 1/2 × 0.85 × t²
1.4 = 0.425t²
t = √( 1.4 / 0.425 )
t = 1.81497 sec
now, at acceleration 0.10 m/s²
the dog has to cover the distance;
s = ut + 1/2 at²
s = ( 1.4 × 1.81497) - 1/2 × 0.10 × 1.81497²
s = 2.540958 - 0.1647
s = 2.3762 m
The cant in front of the dog as it leaps through the window;
distance = 2.8 m - 2.3762 m
distance = 0.4238 m
Therefore, the cat is 0.4238 m in front of the dog as it leaps through the window
Answer:
a)

b)

Explanation:
L = inductance of the Inductor = 3.14 mH = 0.00314 H
C = capacitance of the capacitor = 5.08 x 10⁻⁶ F
a)
f = frequency = 55.7 Hz
Impedance is given as



b)
f = frequency = 11000 Hz
Impedance is given as



There are two types of electric charges; positive and negative
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