Ask question

Ask question

All categories

3 years ago

8

Consider these scenarios.

1 answer:

never [62]3 years ago

8 0

Answer:

Scenario 2

Explanation:

because literally theres no point in scenario one and i did the quiz and got it right

You might be interested in

Fine grains of beach sand are assumed to be spheres of radius 64.8 µm. These grains are made of silicon dioxide which has a dens

Nutka1998 [239]

Answer:

1) Mass of the grain is $2.9632\times 10^{-9} kg$ .

2) 0.08901 kg of sand would have surface area equal the surface area of the cube.

Explanation:

1) Radius of the grain,r = 64.8 µm = $6.48\times 10^{-5}m$

Volume of the sphere = $\frac{4}{3}\pi r^3$

Volume of the grain of a sand:

$V=\frac{4}{3}\times \pi r^3=\frac{4}{3}\times 3.14\times (6.48\times 10^{-5} m)^3$

$V=1.1397\times 10^{-12} m^3$

Density of a grain of sand = $d=2600 kg/m^3$

Mass of a grain of a sand = M

$d=2600 kg/m^3=\frac{M}{1.1397\times 10^{-12} m^3}$

$M=2.9632\times 10^{-9} kg$

Mass of the grain is $2.9632\times 10^{-9} kg$ .

2) Surface are of sphere: $4\pi r^2$

Surface area of a grain:

$A=4\times 3.14\times (6.48\times 10^{-5}m)^2$

$A=5.2766\times 10^{-8} m^2$

Length of the cube = a = 0.514 m

Total surface area of cube ,A'= $6a^2$

$A'=6\times (0.514 m)^2=1.5851 m^2$

let the number grains with area equal to total surface area of cube be x.

$A'=A\times x$

$x=\frac{1.5851 m^2}{5.2766\times 10^{-8} m^2}=3.003\times 10^7$

Volume of x number of grains :V'

$V'=V\times x$

$V= 1.1397\times 10^{-12} m^3\times 3.003\times 10^7$

$V'=3.4236\times 10^{-5} m^3$

Mass of $3.4236\times 10^{-5} m^3$ of sandL:

= $3.4236\times 10^{-5} m^3\times 2600 kg/m^3=0.08901 kg$

0.08901 kg of sand would have surface area equal the surface area of the cube.

8 0

3 years ago

A 22 µF capacitor charged to 0.7 kV and a second 115 µF capacitor charged to 5.5 kV are connected to each other, with the positi

vesna_86 [32]

Answer:

0.099C

Explanation:

First, we need to get the common potential voltage using the formula

$V=\frac {C_2V_2-C_1V_1}{C_1+C_2}$

Where V is the common voltage, C and V represent capacitance and charge respectively. Subscripts 1 and 2 to represent the the first and second respectively. Substituting the above with the following given values then

$C_1=22\times 10^{-6} F\\ C_2=115\times 10^{-6} F\\ V_1= 0.7\times 10^{3}\\V_2=5.5\times 10^{3}$

Therefore

$V=\frac {115\times 10^{-6}\times 5.5\times 10^{3}-22\times 10^{6}\times 0.7\times 10^{3}}{22\times 10^{-6}+115\times 10^{-6}}=4504.3795620437$

Charge, Q is given by CV hence for the first capacitor charge will be $Q_1=C_1V$

Here, $Q_1=22\times 10^{-6}\times 4504.3795620437=0.0990963503649C\approx 0.099C$

8 0

3 years ago

A box measures 20cm by 10cm by 5cm

Levart [38]

length × width × height

20 cm × 10 cm × 5 cm = 1,000 cm

5 0

3 years ago

Read 2 more answers

A positive symptom of schizophrenia would be

labwork [276]

Answer:

i will like to know the answer but idk

Explanation:

7 0

3 years ago

Read 2 more answers

Radioactive decay of 40k atoms in an igneous rock has resulted in a radio of 25 percent 40k atoms to 75 percent 40Ar and 40Ca at

Ivanshal [37]

In order to answer this, we would need to know
the half-life of ⁴⁰K. Perhaps when you look that up,
you'll be able to answer this on your own.

6 0

3 years ago