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3 years ago

All known frequencies of the visible spectrum are..

Physics

1 answer:

crimeas [40]3 years ago

7 0

Explanation:

$b. \: light$

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pantera1 [17]

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3 years ago

Calculate the inhomogeneity of a 1.5 t magnet

Ludmilka [50]

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3 years ago

Which number is 0.0069 expressed in scientific notation? 6.9 × 10 4 6.9 × 10 –6 6.9 × 10 –4 6.9 × 10 –3

Komok [63]

Given: The number = 0.0069 that has to be expressed in scientific notation

Concept: When we express any number in scientific notation then we shall consider two points.

(i) If we shift the decimal point from left to right after first (non-zero) digit then we count the number of shifted place of decimal and write them in terms of the negative power of 10. For example, 0.004789 = 4.789 ×10⁻³

(ii) If we shift the decimal point from right to left after first (non-zero) digit from the left end then we count the number of shifted place and write then in terms of the positive power of 10. For example, 4789.24 = 4.78924 ×10⁺³ = 4.78924 ×10³ Now, we shall convert the given number 0.0069 in scientific notation

0.0069 = 6.9 ×10⁻³ because we have shifted the decimal from left to right for three places that is after digit 6.

Hence, the last option 6.9 ×10⁻³ will be the correct option.

5 0

3 years ago

A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric

jeka94

Answer:

a) F₃₁ = 63.0 μN

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

$F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}} = 63.0 \mu N (1)$

Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

$F_{32} = F_{3} - F_{31} = 49.0\mu N - 63.0\mu N = -14.0 \mu N (2)$

Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

$q_{2} = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC (3)$

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3 years ago

How much watt is called 1 H. P

agasfer [191]

Answer:

746 watts

Explanation:

7 0

2 years ago