Answer:
A system that includes the stone and the earth.
Explanation:
If the system of being dropped from the height of the cliff consists of just the stone alone, then it means that its momentum will certainly undergo changes as it falls freely. However, If the system is now expanded to include not only the stone but also the Earth, then it implies that the momentum of the stone which is in the downward direction will be equal and opposite to the momentum of the Earth in the upwards direction towards the stone. Therefore, the momentum will cancel out and net momentum will be zero.
Answer:
B. A solenoid carries an electric current.
C. An electromagnet is a solenoid with a metal core.
E. A solenoid induces a magnetic field in a metal core.
Explanation:
An electroiman is a large number of wire turns (solenoid) that are wrapped around a magnetic core made of a ferromagnetic or ferrimagnetic material, such as iron. The metal core is composed of "tiny imans", when a current is passed through the solenoid, its magnetic field penetrates the iron, and causes the "tiny imans" to aligning parallel to the magnetic field, creating a large magnetic field.
There mass is a different size (weight)
<span />
Explanation:
Given that,
Weight of the friend, W = 600 N
When the friend sits on the metal frame it bends downward 4 cm, we can say that the compression in the it is 4 cm or 0.04 m
To find,
Spring constant for this chair or k
Solve :
The weight of an object is equal to the force exerted by the gravitational force, F = 600 N
According to Hooke's law, the force exerted by the spring is given by :
F = kx
k is the spring constant
![k=\dfrac{F}{x}](https://tex.z-dn.net/?f=k%3D%5Cdfrac%7BF%7D%7Bx%7D)
![k=\dfrac{600\ N}{0.04\ m}](https://tex.z-dn.net/?f=k%3D%5Cdfrac%7B600%5C%20N%7D%7B0.04%5C%20m%7D)
k = 15000 N/m
Therefore, the spring constant of the spring is 15000 N/m.
Explanation:
Area of ring ![\ 2{\pi} a d a](https://tex.z-dn.net/?f=%5C%202%7B%5Cpi%7D%20a%20d%20a)
Charge of on ring ![d q=-(\ 2{\pi} a d a)](https://tex.z-dn.net/?f=d%20q%3D-%28%5C%202%7B%5Cpi%7D%20a%20d%20a%29)
Charge on disk
![Q=-\left(\pi R^{2}\right)](https://tex.z-dn.net/?f=Q%3D-%5Cleft%28%5Cpi%20R%5E%7B2%7D%5Cright%29)
![\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