IMPORTANT ANSWER ALL 3 PLEASE!

The mean diameters of planets A and B are 8.1 × 103 km and 1.4 × 104 km, respectively. The ratio of the mass of planet A to that

lisabon 2012 [21]

Answer:

dA/dB = 4.955

Approximately, the ratio is 5/1

(Where dA is mean density for planet A while dB is mean density for planet B)

Explanation:

Mass of A = mA

Mass of B = mB

mA/mB = 0.96

Mean radius for A = mA = (8.1 × 10^3)/2 = 4.05 × 10^3 km

Mean radius for B = mB = (1.4 × 10^4)/2

= 7×10^3km

Density = mass/volume

Volume of a sphere = 4/3Πr3

Mean volume for A = (4/3) × Π × (4.05 × 10^3)^3

= 2.784 × 10^11 km3

Mean volume for B = 4/3×Π×(7×10^3)^3

= 1.437 × 10^12km3

Since m/v = d ( where m = mass, v = volume and d = density)

mA = 2.784 × 10^11 km3 × dA ...equation 1

mB= 1.437 × 10^12km3 × dB... equation 2

but mA/mB= 0.96

mA = 0.96 × mB

substitute for mA in equation 1

0.96 × mB = 2.784 × 10^11 x dA equation 3

Substitute for mB in equation 3..

(refer to equation 2)

0.96×1.437×10^12 × dB = 2.784 × 10^11 × dA .....equation 4

divide through by the coefficient of dA

dA = (0.96×1.437×10^12×dB)/(2.784 × 10^11)

divide through by dB

dA/dB = 4.955

therefore, the ratio of dA to dB is 5/1

Therefore, the mean density of A is almost five times that of B

7 0

3 years ago

A parallel-plate air capacitor is connected to a constant-voltage battery. If the separation between the capacitor plates is dou

Oliga [24]

Answer:

Drop to half of the previous value

Explanation:

Energy stored in capacitor is inversly propotional to the distance between the plates.

3 0

4 years ago

Work is measured in _____.

DaniilM [7]

Answer:

work is measured in joule.

8 0

3 years ago

Read 2 more answers

Suppose a soup can is made from a sheet of steel19 which is .13 mm thick. If the can is 11 cm high and 6 cm in diameter, use dif

Flura [38]

Answer:

The can mass is 0,00359 kg or 3,59 g

Explanation:

1. Relevant Data:

Steel thickness= 0.13 mm or 0.013 mm

h=11 cm

d=6 cm

ρ=800 kg/m^3

2. Calculate mass from densisty equation:

$\rho=\frac{m}{v}$ , then $m=\rho .v$

We need to estimate the volume of the can to calculate the mass.

3. Estimate volume using differentials:

Cylinder volume equation is:

$V=\frac{1}{4}\pi d^{2}h$

Considering that the can is an object with a hole inside, then we need to estimate the real volume of the sheet of steel.

Using differentials we have:

$dV=\frac{1}{2}\pi Dh (dD)$

Then, we could say that $dD=0.013 cm$

Replacing the values of d, h and dD, we obtain:

$dV=\frac{1}{6}\pi (6 cm)(11 cm)(0,013 cm)$

$dV=0,4492 cm^3$

4. Calculate the mass

Convert volume unit into $m^3$

$0,4492 cm^3x\frac{1 m^3}{1x10^6 cm^3} =0,4492 x 10^-6 m^3$

Calculate mass

$m=\rho .v$

$m=8000 \frac{kg}{m^3}.0,4492 x10^-6 m^3$

$m=0,00359 kg =3,59 g$

5 0

4 years ago

A particle undergoes two displacements. The first has a magnitude of 11 m and makes an angle of 82 ◦ with the positive x axis. T

Luden [163]

Answer:

$\theta = 211.7 degree$

Explanation:

First displacement of the particle is given as

$r_1$ = 11 m at 82 degree with positive X axis

so we can say

$\vec r_1 = 11 cos82 \hat i + 11 sin82 \hat j$

$\vec r_1 = 1.53\hat i + 10.9 \hat j$

resultant displacement of the particle after second displacement is given as

r = 8.7 m at 135 degree with positive X axis

so we can say

$r = 8.7 cos135\hat i + 8.7 sin135\hat j$

$r = -6.15 \hat i + 6.15 \hat j$

now we know that

$r = r_1 + r_2$

now we have

$r_2 = r - r_1$

so we will have

$r_2 = (-6.15 \hat i + 6.15 \hat j) - (1.53\hat i + 10.9 \hat j)$

$r_2 = -7.68 \hat i - 4.75 \hat j$

so angle of the second displacement is given as

$tan\theta = \frac{r_y}{r_x}$

$tan\theta = \frac{-4.75}{-7.68}$

$\theta = 211.7 degree$

8 0

3 years ago