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Ask QuestionPosted by Alisha Saini 6 years, 10 months ago
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Posted by Omkar Raushan 6 years, 10 months ago
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Posted by Prinçe Çharming 6 years, 10 months ago
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Suhani Rana? 6 years, 10 months ago
Posted by Shriya Gautam 6 years, 10 months ago
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Ram Kushwah 6 years, 10 months ago
SLOPE हिंदी में 'झुकाब '
{tex}\begin{array}{l}\mathrm{In}\;\triangle\;\mathrm{OPQ}\\\tan\;\mathrm\alpha\;=\frac{\mathrm{OQ}}{\mathrm{OP}}=(\mathrm{this}\;\mathrm{value}\;\mathrm{is}\;0\;\mathrm{to}\;\;\mathrm{infinite}\;\mathrm{as}\;\mathrm{per}\;\mathrm{value}\;\mathrm{of}\;\mathrm{angle})\\\end{array}{/tex}
in physics and maths it is denoted by m
like y=mx +c
Posted by Swetha Swetha 6 years, 10 months ago
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Posted by Mahesh Prajapati 6 years, 10 months ago
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Anuj Sharma 6 years, 10 months ago
Posted by Himanshu Mamgain 6 years, 10 months ago
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Posted by Devendra Singh Yadav 6 years, 10 months ago
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Posted by Khushboo Kumari 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
a)
By = 12 x 10-8 sin [1.20 x 107 z + 3.60 x 1015t]
Now Bo = 12 x 10-8 T and c = 3 x 10 8 m/s
Since Eo = Bo c = 12 x 10-8 x 3 x 10 8 = 36 V/m
Energy Density = (1/2) ε0Eo2 = (1/2)(8.85 x 10-12) (36x36)
Energy Density = 5.74 x 10-15 J/m3
b)
By = 12 x 10-8 sin [1.20 x 107 z + 3.60 x 1015t]
Now k= 1.2 x 107 and w = 3.6 x 1015
Since, λ = (2π / k) and f = (w / 2π)
V = λ x f = kw = 3 x 108 m/s (approx.)
Posted by Khushboo Kumari 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
What is the amount of work done in moving a point charge around a circular arc of radius r at the centre where another point charge is located?
<section data-topic-id="7363" id="topic"> <article data-post-id="8709" data-topic-id="7363" data-user-id="2" id="post_1">
Posted by Raj Kumar 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Energy band gap of the given photodiode, <i>E</i><i>g</i> = 2.8 eV
Wavelength, λ = 6000 nm = 6000 × 10−9 m
The energy of a signal is given by the relation:
<i>E</i> =
Where,
<i>h</i> = Planck’s constant
= 6.626 × 10−34 Js
c = Speed of light
= 3 × 108 m/s
<i>E</i>
= 3.313 × 10−20 J
But 1.6 × 10−19 J = 1 eV
∴<i>E</i> = 3.313 × 10−20 J
The energy of a signal of wavelength 6000 nm is 0.207 eV, which is less than 2.8 eV − the energy band gap of a photodiode. Hence, the photodiode cannot detect the signal.
Posted by Majid Majeed 6 years, 10 months ago
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Kuldeep Singh 6 years, 10 months ago
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Sinchana Malagond 6 years, 10 months ago
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Ranjeet Jacker 6 years, 10 months ago
Posted by Kiran Dhaliwal 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
(i) Difference: Field lines of a solenoid form continuous current loops, while in the case of an electric dipole the field lines begin from a positive charge and end on a negative charge or escape to infinity.
(ii) Two magnetic field lines cannot intersect because at the point of intersection, these will be two directions of magnetic field which is impossible.
Posted by Deepak Yadav 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Q. To which part of the electromagnetic spectrum does a wave of frequency 5 x 1019 Hz belong?
<hr />A. The part of the electromagnetic spectrum is X-rays.
Posted by Shubkarman Badesha 6 years, 10 months ago
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Tushar Goyal 6 years, 10 months ago
Posted by Kumari Chandani 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
A virtual image is one that can be seen, and a real image can be projected. Although we usually use plane mirrors for the purpose of producting virtual images, that is being able to see an object in the mirror. It possible for a plane mirror to produce a real image in conjuction with another mirror. In order for a real image to be produced the mirror must bend the incident light rays in such a way that they converge to a single point. According to the law of reflection the angle of incidence of an incident ray equals the angle of reflection of a reflected ray, so parallel light rays are reflected back at the same angle producing a virtual image of a real object.
In order for a plane mirror to produce a real image ir must be reflecting the virtual image produced by another mirror or a lens. In this context we call the virtual image a virtual object. If the light rays coming from the virtual object converge at a point behind the plane mirror the reflected rays will converge in front of the mirror, forming a real image.
Posted by Surekha Sharma 6 years, 10 months ago
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Posted by Ritu Raj 6 years, 10 months ago
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Kuldeep Singh 6 years, 10 months ago
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