If the mass of a materials is 44grams and the volume of the material is 11cm^3 what would the density of the material be

How many questions are on cumulative tests? on edgenutiy

Evgesh-ka [11]

Roughly 50 for me i dont know about anyone else

8 0

3 years ago

Two particles of equal mass m are at the vertices of the base of an equilateral triangle. The triangle’s center of mass is midwa

Helga [31]

Answer:Twice of given mass

Explanation:

Given

Two Particles of Equal mass placed at the base of an equilateral Triangle

let mass of two equal masses be m and third mass be m'

Taking one of the masses at origin

Therefore co-ordinates of first mass be (0,0)

Co-ordinates of other equal mass is (a,0)

if a is the length of triangle

co-ordinates of final mass $(\frac{a}{2},\frac{\sqrt{3}a}{2})$

Given its center of mass is at midway between base and third vertex therefore

$x_{cm},y_{cm}=\frac{a}{2},\frac{\sqrt{3}a}{4}$

$y_{cm}=\frac{m_1y_1+m_2y_2+m_3y_3}{m_1+m_2+m_3}$

$\frac{\sqrt{3}a}{4}=\frac{m\cdot 0+m\cdot 0+m'\cdot \frac{\sqrt{3}a}{2}}{m+m+m'}$

$2m+m'=4\times (\frac{m'}{2})$

$2m+m'=2m'$

$m'=2m$

8 0

3 years ago

Please help guyss!!!

Savatey [412]

Explanation:

The distance traveled north ward = 2 miles

distance traveled south ward = 3miles

Displacement is the distance traveled in a specific direction. So, it depend on the start and finish point.

The displacement = 3 miles - 2miles = 1mile southward

Distance is the total length of path;

Distance = 3 miles + 2 miles = 5miles

5 0

3 years ago

What are density conversions

shutvik [7]

$\rho= \frac{kg}{m^3} = \frac{g}{cm^3} = \frac{g}{dm^3} \ and \ many \ others.$

8 0

3 years ago

An object in space has altitude of 2210 km, velocity of 7000 m/s and flight path angle of 20 degrees. Find the eccentricity, sem

Dmitry [639]

Answer:

Eccentricity = 0.0557

Semi-major axis = 9,095 km

Angular momentum/mass = 60,116 $\frac{km^{2} }{Sec}$

Kinetic energies/mass = 24,500 KJ

Potential energies/mass = 21,673 KJ

Explanation:

Eccentricity

To find the eccentricity use the following formula

Eccentricity = [ Altitude from the earth x ( $Velocity^{2}$ / Gravitational parameter for the Earth ) ] - 1

Where

Altitude from the earth radius = Radius of the earth + Altitude of the earth from radius = 6,378 km + 2,210 km = 8,588 km

Velocity = 7,000 m/s

Gravitational parameter for the Earth = 3.986004418 × $10^{14}$

Eccentricity = ?

Placing values in the formula

Eccentricity = [ ( 8,588 km x 1000 ) x ( $7000^{2}$ / 3.986004418 × $10^{14}$ ) ] - 1

Eccentricity = 0.0557

Semi-major axis

Total Distance = Semi-major axis x ( 1 - Eccentricity )

Where

Total Distance = 8,588 km

Eccentricity = 0.0557

8,588 x 1,000 m = Semi-major axis x ( 1 - 0.0557 )

8,588,000 m = Semi-major axis x 0.9443

Semi-major axis = 8,588,000 m / 0.9443

Semi-major axis = 9,094,567.40 m

Semi-major axis = 9,094.56740 km

Semi-major axis = 9,095 km

Angular momentum/mass

L = MVR

L/M = VR

Where

V = Velocity = $\frac{7,000 m/s}{1000}$ = 7 km/s

R = Total Radius = Radius of Earth + Altitude = 6,378 km + 2,210 km = 8,588 km

Placing values in the formula

L/M = 7 km/s x 8,588 km = 60,116 $\frac{km^{2} }{Sec}$

Kinetic energies/mass

Ke = I $W^{2}$

Ke = $\frac{1}{2} mr^{2}$ $W^{2}$ ( Where I = $mr^{2}$ )

$\frac{ke}{m}$ = $\frac{r^2}{2}$ $W^{2}$

$\frac{ke}{m}$ = $\frac{r^2}{2}$ $\frac{V^2}{r^2}$ ( $W^{2}$ = $\frac{V^2}{r^2}$ )

$\frac{ke}{m}$ = $\frac{V^2}{2}$

$\frac{ke}{m}$ = $\frac{7000^2}{2}$

$\frac{ke}{m}$ = 24,500,000 J

$\frac{ke}{m}$ = 24,500 KJ

Potential energies/mass

PE = mgh

$\frac{PE}{m}$ = gh

Where

g = 9.807 $\frac{m}{s^2}$

h = 2,210 km x 1,000 = 2,210,000 m

Placing values in the formula

$\frac{PE}{m}$ = 9.807 $\frac{m}{s^2}$ x 2,210,000 m

$\frac{PE}{m}$ = 21,673,470 J

$\frac{PE}{m}$ = 21,673.470 KJ

$\frac{PE}{m}$ = 21,673 KJ

8 0

3 years ago