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Murljashka [212]

3 years ago

7

Please someone answer this ASAP❗️❗️‼️‼️

1 answer:

kogti [31]3 years ago

5 0

Starting making jokes and rapping

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The density of gold is 19.3 g cm³. What is the mass of a bar of gold in Kg that measures 6 cm x 4 cmx to 2 cm ?

klemol [59]

Answer: 0.9264 kg

Explanation: [I'll use "cc" for cubic centimeter, instead of cm^3.

The volume is 6cm*4cm*2cm = 48 cm^3 (cc).

Density of Au is 19.3 g/cc

Mass of gold = (48 cc)*(9.3 g/cc) = 926.4 grams Au

1 kg = 1,000 g

(926.4 grams Au)*(1 kg/1,000 g) = 0.9264 kg, 0.93 kg to 2 sig figs

At gold's current price of $57,500/kg, this bar is worth $53,268. Keep it hidden from your lab partner (and instructor).

3 0

3 years ago

By what factor must the amplitude of a sound wave be decreased in order to decrease the intensity by a factor of 3?

s344n2d4d5 [400]

Answer:

Amplitude is decreased by a factor of $\sqrt3$ if intensity is decreased by a factor of 3.

Explanation:

Intensity of a sound wave is directly proportional to the square of its amplitude.

Therefore, if intensity is $I$ and amplitude is $A$ , then

$I=kA^2$ , where, $k$ is constant of proportionality.

Now, if intensity of sound wave is decreased by a factor of 3. So,

New intensity is, $I_{new}=\frac{I}{3}$

$I_{new}=kA_{new}^2\\\frac{I}{3}=kA_{new}^2$

Plug in $kA^2$ for $I$ . This gives,

$\frac{kA^2}{3}=kA_{new}^2\\A_{new}^2=\frac{A^2}{3}\\A_{new}=\sqrt{\frac{A^2}{3}}=\frac{A}{\sqrt{3}}$

Therefore, amplitude is decreased by a factor of $\sqrt3$ .

4 0

3 years ago

A 100 kg box is suspended from two ropes. The "left rope makes an angle of 20" degrees with the vertical, and the right rope mak

GarryVolchara [31]

Explanation:

It is given that,

Mass of the box, m = 100 kg

Left rope makes an angle of 20 degrees with the vertical, and the right rope makes an angle of 40 degrees.

From the attached figure, the x and y component of forces is given by :

$T_{1x} =-T_1 cos (20)$

$T_{2x} = T_2 cos (40)$

$mg_x = 0$

$T_{1y} = T_1 sin (20)$

$T_{2y} = T_2 sin (40)$

$mg_y= -mg$

Let $R_x$ and $R_y$ is the resultant in x and y direction.

$R_x=-T_1 cos (20)+T_2 cos (40)+0$

$R_y=T_1 sin(20)+T_2 sin(40)-mg$

As the system is balanced the net force acting on it is 0. So,

$-T_1 cos (20)+T_2 cos (40)+0=0$ .............(1)

$T_1 sin(20)+T_2 sin(40)-100\times 9.8=0$ ..................(2)

On solving equation (1) and (2) we get:

$T_1=866.86\ N$ (tension on the left rope)

$T_2=1063.36\ N$ (tension on the right rope)

So, the tension on the right rope is 1063.36 N. Hence, this is the required solution.

7 0

3 years ago

A jetliner, traveling northward, is landing with a speed of 71.9 m/s. Once the jet touches down, it has 675 m of runway in which

Leviafan [203]

Answer:

The value is $a = -3.7 \ m/s^2$

Explanation:

From the question we are told that

The landing speed is $u = 71.9 \ m/s$

The distance traveled is $d = 675 \ m$

The velocity it is reduced to is $v = 11.3 \ m/s$

Generally the average acceleration is mathematically represented as

$a = \frac{ v^2 - u^2 }{ 2 * d }$

=> $a = \frac{ 11.3^2 - 71.9^2 }{ 2 * 675 }$

=> $a = -3.7 \ m/s^2$

5 0

3 years ago

A tank with a flat bottom is filled with water to a height of 4 meters. What is the pressure at any point at the bottom of the t

Mariulka [41]

<u>Answer</u>

C.39.2 kPa

<u>Explanation</u>

The pressure in liquids depend on the depth or what we call the liquid column(h), the density of the liquid(σ) and the the acceleration due to gravity(g).

h = 4.00 m

σ = 1000 Kg/m³

g = 9.8m/s²

Pressure Is given by;

P = σgh

= 1000 × 9.8 × 4

= 39200 Pa

=39.2 KPa

8 0

3 years ago

Read 2 more answers