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2 years ago

A sample of gold has a mass of 38.6 grams and a volume of 2 cm3. What is the density of gold

Physics

1 answer:

Natali5045456 [20]2 years ago

5 0

Answer:

$19.3 g/cm^3$

Explanation:

The density of a material is given by

$\rho = \frac{m}{V}$

where

m is the mass of the sample

V is its volume

for the sample of gold in this problem,

m = 38.6 g

$V=2 cm^3$

Substituting into the formula, we find the density of gold:

$\rho = \frac{38.6}{2}=19.3 g/cm^3$

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A train increases its speed steadily from 10 m/s to 20 m/s in

rewona [7]

Answer:

15m/s

Explanation:

add the two speeds and divide by 2

10+20=30

30/2=15

3 0

3 years ago

A circuit is set up such that it has a current of 8 A. What would be the new current if the resistance was increased by a factor

RUDIKE [14]

Answer:

4 A

Explanation:

The relationship between current, voltage and resistance in a circuit is given by Ohm's law:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

The equation can also be rewritten as

$I=\frac{V}{R}$

from which we see that the current is inversely proportional to the resistance, R.

In this problem, the initial current is I = 8 A. Then the resistance is doubled:

R ' = 2R

So the new current is

$I'=\frac{V}{R'}=\frac{V}{2R}=\frac{1}{2}(\frac{V}{R})=\frac{I}{2}=4 A$

so the current is halved.

7 0

3 years ago

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

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