They are marsupials that are considered "specialist eaters" since they only eat a certain type of leaf. While generalist feeders aren't as picky as the koala bears. I hope this helps!
<h3><u>Answer;</u></h3>
40 light bulbs
<h3><u>Explanation</u>;</h3>
The total resistance of components or bulbs in series is given as the sum of resistance of all the components.
Thus; if there are bulbs in series each with a resistance of 1.5 Ω, the the total resistance will be; 1.5nΩ
From the ohms law;
V = IR , where V is the voltage, I is the current and R is the resistor.
Thus; R = V/i
R = 120/2
= 60 Ω
But, there are n bulbs each with 1.5 Ω; thus there are;
n = 60/1.5
<u> = 40 Bulbs </u>
The electrical force between these two charges remains the
same. In coulomb’s law, it states that the magnitude of two charges (product of
two charges) is inversely proportional to the square of the distance. Since both
the magnitude and the distance are halved, therefore, the change in both quantities
will have no effect in the value of electrical force.
Answer:
About 7.67 m/s.
Explanation:
Mechanical energy is always conserved. Hence:
Where <em>U</em> is potential energy and <em>K</em> is kinetic energy.
Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:
Substitute and solve for final velocity:
In conclusion, the child's speed at the bottom of the slide is about 7.67 m/s.
Explanation:
Area of ring 
Charge of on ring 
Charge on disk
![\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dd%20v%20%26%3D%5Cfrac%7Bk%20d%20q%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5C%26%3D2%20%5Cpi-k%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5Cv%281%29%20%26%3D2%20%5Cpi%20c%20k%20%5Cint_%7B0%7D%5E%7BR%7D%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5Ccdot_%7B2%20%5Cvarepsilon_%7B0%7D%7D%5E%7B2%7D%20R%20%5C%5C%26%3D2%20%5Cpi%20%5Csigma%20k%5B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%26%3D%5Cfrac%7B2%20%5Cpi%20%5Csigma%7D%7B4%20%5Cpi%20%5Cvarepsilon_%7B0%7D%7D%5B%5Csqrt%7Bz%5E%7B2%7D%2BR%5E%7B2%7D%7D-%2821%29%5D%20%5C%5C%26%3D%5Cfrac%7B%5Csigma%7D%7B2%7D%28%5Csqrt%7B2%5E%7B2%7D%2BR%5E%7B2%7D%7D-2%29%5Cend%7Baligned%7D)
Note: Refer the image attached
