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sertanlavr [38]
2 years ago
15

Dharna is said to be concentration it is true or false​

Physics
1 answer:
Lelechka [254]2 years ago
5 0

dharna is said to be concentration.. it is true

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Compare and contrast a series and parallel circuit. Give at least one way that they are alike and one way that they are differen
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A parallel circuit has multiple paths from the energy source to the resistors (e.x., lightbulbs). If one bulb was to burn out, the rest would stay on, unlike a series circuit. If one bulb was to burn out, the circuit is no longer fully connected because there is only one path from the energy source to the resistors. But, despite these differences, they are similar in the way that they do have paths from the energy source to the resistors.
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Whats the standard unit for measuring mass
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Kilograms


The kilogram is the SI base unit of mass and is equal to the mass of the international prototype of the kilogram, a platinum-iridium standard that is kept at the International Bureau of Weights and Measures (BIPM)

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1. A 46000N helicopter feels a net force of 9200N. What litt force is exerted by the air on the propellers?
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Answer:

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3 years ago
Write about Archimedes principle​
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The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr
MA_775_DIABLO [31]

Answer:

q=49Q/64

and

x =16R/15

Explanation:

See  attached figure.

E_{Q}= E due to sphere

E_{q}= E due to particule

E_{total}=E_{Q}-E_{q}=0  (1)

according to the law of gauss and superposition Law:

E_{Q}=E_{1}+E_{2}=E_{2} ; electric field due to the small sphere with r1=R/4

E_{Q}=kq_{2}/(r_{1}^{2})=

q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}

then: E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})  (2)

on the other hand, for the particule:

E_{q}=kq/(r_{p}^{2})

r_{p}=2R-R/4=7R/4   ⇒    E_{q}=16kq/(49R^{2})   (3)

We replace (2) y (3) in (1):

E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})

q=49Q/64

--------------------

if R<x<2R   AND E_{total}=E_{Q}-E_{q}=0

E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})

remember that  q=49Q/64

then:

Q(2R-x^{2})=49/64*x^{2}

solving:

x_{1} =16R/15

x_{2} =16R

but: R<x<2R  

so : x =16R/15

7 0
3 years ago
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