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3 years ago

Gold has a density of 19300 kg/m3 calculate the mass of 0.02m3 of gold in kilograms

Physics

1 answer:

aev [14]3 years ago

5 0

Answer:

The mass of 0.02 m³ of gold is 386 kilograms

Explanation:

Given:

The density of the gold = 19300 kg/m³.

The volume of gold = 0.02 m³

To Find:

The mass of gold = ?

Solution:

We know that density is mass divided per unit volume.

Thus mathematically

Density = \frac{mass}{volume}Density=

volume

mass

Rewriting in terms of mass ,

Mass = density * volume

On substituting the known values

Mass = 19300 kg/m³ * 0.02 m³

Mass = 386 kilograms

Learn more about Mass and Density:

Mass=?,volume=190,density=4

Mass 350 kg volume 175 density ans

This is not my answer I copied it but hope it helps:)

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Electromagnetic waves, which include light, consist of vibrations of electric and magnetic fields, and they all travel at the sp

Alekssandra [29.7K]

Answer:

2.92682 m

$1.5\times 10^{18}\ Hz$

250000000000 Hz

$2.88462\times 10^{-8}\ m$

0.21126 m

0.12244 m

Explanation:

c = Speed of light = $3\times 10^8\ m/s$

$\lambda$ = Wavelength

f = Frequency

Wavelength is given by

$\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{102.5\times 10^6}\\\Rightarrow \lambda=2.92682\ m$

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Frequency is given by

$f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{0.2\times 10^{-9}}\\\Rightarrow f=1.5\times 10^{18}\ Hz$

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The wavelength is $2.88462\times 10^{-8}\ m$

$\lambda=\dfrac{c}{f}\\\Rightarrow \lambda=\dfrac{3\times 10^8}{1.42\times 10^{9}}\\\Rightarrow \lambda=0.21126\ m$

The wavelength is 0.21126 m

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3 years ago

You place a 3.0-m-long board symmetrically across a 0.5-m-wide chair to seat three physics students at a party at your house. If

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Answer:

Komila should sit 0.33m from the middle of the board towards tahreen.

Explanation:

We are told to treat each student as point-like objects. So i have attached a rigid body diagram to depict this.

From the diagram,

F_d is force exerted by dan

F_t is force exerted by tahreen

F_k is force exerted by komila

F_b is force of board at the mid point.

x1 is distance of dan from the centre of the chair

x2 is distance of komila from the centre of the chair

x3 is distance of tahreen from komila

We are given;

Mass of Dan;m_d = 62 kg

Mass of tahreen;m_t = 50 kg

Mass of komila;m_k = 54 kg

Now, taking moments about the centre of the chair, we have;

(F_d*x1) - (F_k*x2) - (F_t(x2 + x3)) = 0

Now,F_d = m_d*g ; F_t = m_t*g ; F_k = m_k*g

We are told that the board is 3m long. So, if we assume that the fulcrum position of the chair coincides with the midpoint of boards length, we'll have;

x1 = (x2 + x3) = 1.5

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(F_d*1.5) - (F_k*x2) - (F_t*1.5) = 0

F_d = m_d*g = 62 * 9.8 = 607.6 N

F_t = m_t * g = 50 x 9.8 = 490 N

F_k = m_k * g = 54 x 9.8 = 529.2 N

So plugging in these values, we have;

(607.6 * 1.5) - (529.2 * x2) - (490 * 1.5) = 0

911.4 - 735 = 529.2 x2

529.2 x2 = 176.4

x2 = 176.4/529.2

x2 = 0.33m

Komila should sit 0.24m from the middle of the board towards tahreen

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3 years ago

Gold has a density of 19300 kg/m3 calculate the mass of 0.02m3 of gold in kilograms​

Gold has a density of 19300 kg/m3 calculate the mass of 0.02m3 of gold in kilograms