Answer:
28 cm and 32 cm
Explanation:
1. The spring pendulum hangs vertically, oscillates harmonic with amplitude 2cm and angular frequency 20 rad/s. The natural length of
a spring is 30cm. What is the minimum and maximum length of the spring during the oscillation? Take g = 10m/s2.
As the amplitude is 2 cm and the natural length is 30 cm. So, it oscillates between 30 -2 = 28 cm to 30 + 2 = 32 cm.
So, the minimum length is 28 cm and the maximum length is 32 cm.
The eroded rock and soil materials that are transported downstream by a river are called its load. A river transports, or carries, its load in three different ways: in solution, in suspension, and in its bed load.
Mineral matter that has been dissolved from bedrock is carried in solution. Common minerals carried in solution by rivers include dissolved calcium, magnesium, and bicarbonate. Most of a river’s solution load comes from groundwater seeping into the river. Before it reaches the stream,thegroundwaterhastraveledthroughfracturesinthebedrock, chemically eroding rock along the way.
When river water looks muddy, it is carrying rock material in suspension. Suspended material includes clay, silt, and fine sand. Although these suspended materials are heavier than water, the turbulence of the stream flow stirs them up and keeps them from sinking. Turbulence includes swirls and eddies that form in water as a result of friction between the stream and its channel. The faster a stream flows, the more turbulent and muddy it becomes. A rough or irregular channel also increases turbulence.
A river may also transport rock materials in its bed load. The bed load consists of sand, pebbles, and boulders that are too heavy to be carried in suspension. These heavier materials are moved along the streambed, especially during floods. Boulders and pebbles roll or slide along the river bed. Large sand grains are pushed along the bottom in a series of jumps and bounces.
The relative amounts of a river’s load that are carried in solution, in suspension, and in the bed load depend on the nature of the river, the climate, the type of bedrock, and the season of the year. As a general rule, most of the load carried by the world’s streams and rivers is carried in suspension. The size of a river’s suspended load increases with human land use. Road and building construction and removal of vegetation make it easier for rain to wash sediment into streams and rivers.
Answer:
electric potential, V = -q(a²- b²)/8π∈₀r³
Explanation:
Question (in proper order)
Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings
<em>consider the attached diagram below</em>
the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below
Va = q/4π∈₀ [1/(a² + b²)¹/²]

Also
the electric potential at point p, distance r from the center of the inner charged ring with radius b is

Sum of the potential at point p is
V = Va + Vb
that is


![V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%7D%20%2A%20%5B%5Cfrac%7B1%7D%7B%28a%5E%7B2%7D%20%2B%20r%5E%7B2%7D%20%29%5E%7B1%2F2%7D%20%7D%20-%20%5Cfrac%7B1%7D%7B%28b%5E%7B2%7D%20%2B%20r%5E%7B2%7D%20%29%5E%7B1%2F2%7D%20%7D%5D)
the expression below can be written as the equivalent

likewise,

hence,
![V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%7D%20%2A%20%5B%5Cfrac%7B1%7D%7B%7Br%281%5E%7B2%7D%20%2B%20%5Cfrac%7Ba%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B1%2F2%7D%20%7D%20-%20%5Cfrac%7B1%7D%7B%7Br%281%5E%7B2%7D%20%2B%20%5Cfrac%7Bb%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B1%2F2%7D%20%7D%5D)
1/r is common to both equation
hence, we have it out and joined to the 4π∈₀ denominator that is outside
![V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B%5Cfrac%7B1%7D%7B%7B%281%5E%7B2%7D%20%2B%20%5Cfrac%7Ba%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B1%2F2%7D%20%7D%20-%20%5Cfrac%7B1%7D%7B%7B%281%5E%7B2%7D%20%2B%20%5Cfrac%7Bb%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B1%2F2%7D%20%7D%5D)
by reciprocal rule
1/a² = a⁻²
![V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B%7B%281%5E%7B2%7D%20%2B%20%5Cfrac%7Ba%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B-1%2F2%7D%20-%20%7B%281%5E%7B2%7D%20%2B%20%5Cfrac%7Bb%5E%7B2%7D%20%7D%7Br%5E%7B2%7D%20%7D%20%29%7D%5E%7B-1%2F2%7D%5D)
by binomial expansion of fractional powers
where 
if we expand the expression we have the equivalent as shown

also,

the above equation becomes
![V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B%28%281-%5Cfrac%7Ba%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%20%29%20-%20%281-%5Cfrac%7Bb%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%20%29%5D)
![V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B1-%5Cfrac%7Ba%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%20-%201%2B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%5D)
![V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B-%5Cfrac%7Ba%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%20%2B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%5D%5C%5C%5C%5CV%20%3D%20%5Cfrac%7Bq%7D%7B4%5Cpi%20e0%20r%7D%20%2A%20%5B%5Cfrac%7Bb%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%20-%5Cfrac%7Ba%5E%7B2%7D%20%7D%7B2r%5E%7B2%7D%20%7D%5D)


Answer

OR

'The normal is a line perpendicular to the surface of the mirror'.This is the correct statement that corrects an error on the site.
<h3>What is the law of reflection?</h3>
The law of reflection specifies that upon reflection from a downy surface, the slope of the reflected ray is similar to the slope of the incident ray.
The reflected ray is consistently in the plane determined by the incident ray and perpendicular to the surface at the point of reference of the incident ray.
When the light rays descend on the smooth surface, the angle of reflection is similar to the angle of incidence, also the incident ray, the reflected ray, and the normal to the surface all lie in a similar plane.
Hence 'The normal is a line perpendicular to the surface of the mirror'.This is the correct statement that corrects an error on the site
To learn more about the law of reflection refer to the link;
brainly.com/question/12029226
Answer:
123 J transfer into the gas
Explanation:
Here we know that 123 J work is done by the gas on its surrounding
So here gas is doing work against external forces
Now for cyclic process we know that

so from 1st law of thermodynamics we have


so work done is same as the heat supplied to the system
So correct answer is
123 J transfer into the gas