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4 years ago

What is the mass of an object that weighs 500 newtons on earth?

Physics

1 answer:

Eva8 [605]4 years ago

6 0

The mass would be 51 kg. So b is the right answer.

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Velocity of a wave is the. (a) Wavelength x frequency (b) Wave number x frequency (c) Time period x phase (d) None

Elden [556K]

Answer:

option (a)

Explanation:

Wavelength is defined as the distance traveled by the wave in one complete oscillation.

The number of oscillations completed in one second is called frequency.

The relation for the wave velocity is given by

wave velocity = frequency x wavelength

4 0

3 years ago

The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate

Artemon [7]

Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

$L = 10log_{10}\frac{I}{I_{o}}$

$I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}$

$L' = 10log_{10}\frac{I'}{I_{o}}$

$I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}$

where

$I_{o} = 10^{- 12} W/m^{2}$

I = Intensity of sound

Now,

$I = \frac{P}{4\pi R^{2}}$

Similarly,

$I' = \frac{P}{4\pi (R + 11.0)^{2}}$

Now,

$\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}$

$\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}$

$R ^{2} + 22R + 121 = 11.64R^{2}}$

$10.64R ^{2} - 22R - 121 = 0$

Solving the above quadratic eqn, we get:

R = 4.56 m

8 0

3 years ago

Language uses_ to quickly express concepts and ideas

ludmilkaskok [199]

Answer:

i try B

Explanation:

7 0

3 years ago

Read 2 more answers

Escribir usando prefijos, en unidades del Sistema Internacional: longitud del ecuador, radios del núcleo y átomo, segundos de un

julia-pushkina [17]

la característica de los elementos

5 0

3 years ago

Two capacitors, C1 and C2, give an equivalent capacitance of 9.0 pF when connected in parallel and an equivalent capacitance of

Lilit [14]

Answer:

C₁ = 3 pF

C₂ = 6 pF

Explanation:

Let the capacitance be C₁ and C₂.

For parallel combination

C₁ +C₂ = 9

For series combination

$2=\frac{C_1\times C_2}{C_1+C_2}$

$2=\frac{C_1\times C_2}{9}$

C₁ X C₂ = 18

( 9-C₂ )X C₂ = 18

C₂²- 9C₂ +18 = 0

C₂ = 6 or 3.

C₁ = 3 or 6.

C₁ = 3 pF

C₂ = 6 pF

5 0

3 years ago