Should people eat animals?

Suzie Spacewalker hovers in space beside a rotating space station in outer space. Both she and the center of mass of the space s

Inga [223]

Answer:

is in the earths orbit

Explanation:

for Suzie to hover in space beside the rotating space station, she and the center of mass of the space station are at relative rest which happens when space station is in Earth orbit, hence she is in the earths orbit.

3 0

3 years ago

Bullets from two revolvers are fired with the same velocity. The bullet from gun #1 is twice as heavy as the bullet from gun #2.

IrinaVladis [17]

Answer:

option C

Explanation:

Let mass of the bullet be m and velocity be v

mass of gun be M and bullet be V

now,

using conservation of momentum for gun 1

(M+m) V' = 2 mv + 3 MV

V' = 0

3 M V = - 2 mv

momentum of gun 1 =- 2 mv---------(1)

now for gun 2

(M+m) V' = mv + MV

V' = 0

M V = - mv

momentum of gun 1 = -mv-----------(2)

dividing equation (1) by (2)

$\dfrac{P_m1}{P_m2} = \dfrac{- 2mv}{-mv}$

$\dfrac{P_m1}{P_m2} = \dfrac{2}{1}$

the correct answer is option C

7 0

3 years ago

Although these quantities vary from one type of cell to another, a cell can be 1.9 μm in diameter with a cell wall 60 nm thick.

DaniilM [7]

Answer: $2(10)^{-9} mg$

Explanation:

We know the total diameter of the cell (assumed spherical) is:

$d=1.9\mu m=1.9(10)^{-6} m$

Then its total radius $r=\frac{d}{2}=\frac{1.9(10)^{-6} m}{2}=9.5(10)^{-7} m$

On the other hand, we know the thickness of the cell wall is $r_{t}=60 nm= 60(10)^{-9} m$ and its density is the same as water ( $\rho=997 kg/m^{3}$ ).

Since density is the relation between the mass $m$ and the volume $V$ :

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

The mass is: $m=\rho V$ (1)

Now if we are talking about this cell as a thin spherical shell, its volume will be:

$V=\frac{4}{3}\pi R^{3}$ (2)

Where $R=r-r_{w}=9.5(10)^{-7} m - 60(10)^{-9} m$

Then:

$V=\frac{4}{3}\pi (9.5(10)^{-7} m - 60(10)^{-9} m)^{3}$ (3)

$V=2.952(10)^{-18} m^{3}$ (4)

Substituting (4) in (1):

$m=(997 kg/m^{3})(2.952(10)^{-18} m^{3})$ (5)

$m=2.94(10)^{-15} kg$ (6)

Knowing $1 kg=1000 g$ and $1 mg=0.001 g$ :

$m=2.94(10)^{-15} kg=2(10)^{-9} mg$

7 0

4 years ago

Three uniform spheres of radius 2R, R, and 3R are placed in a line, in the order given, so their centers are lined up and the sp

kolezko [41]

Answer:

x = 2.33 R from the center of mass of the smallest sphere.

Explanation:

Due to the symmetry of the spheres, the center of mass of any of them, is located just in the center of the sphere.

If we align the centers of the spheres with the x-axis, the center of mass of any of them will have only coordinates on the x-axis, so the center of mass of the system will have coordinates on the x-axis only also.

By definition, the x-coordinate of the center of mass of a set of discrete masses m₁, m₂, m₃, can be calculated as follows:

$Xcm = \frac{m1*x1+m2*x2+m3*x3}{m1+m2+3}$

In this case, we need to get the coordinates of the center of mass of each sphere:

If we place the spheres in such a way that the center of the first sphere has the x-coordinate equal to its radius (so it is just touching the origin), we will have:

x₁ = 2*R

For the second sphere, the center will be located at a distance equal to the diameter of the first sphere plus its own radius, as follows:

x₂ = 4*R + R = 5*R

Finally, for the third sphere, the center will be located at a distance equal to the diameter of the first sphere, plus the diameter of the second sphere, plus its own radius, as follows:

x₃ = 4*R + 2*R + 3*R = 9*R

We can calculate the mass of each sphere (assuming that all are from the same material, with a constant density), as the product of the density and the volume:

m = ρ*V

For a sphere, the volume can be calculated as follows:

$\frac{4}{3} *\pi *(r)^{3}$

So, we can calculate the masses of the spheres, as follows:

m₁ = ρ* $\frac{4}{3} *\pi *(2r)^{3}$

m₂ = ρ* $\frac{4}{3} *\pi *(r)^{3}$

m₃ = ρ* $\frac{4}{3} *\pi *(3r)^{3}$

The total mass can be calculated as follows:

M= ρ* $\frac{4}{3} *\pi$ * (8*r³ + r³ + 27*r³) =ρ* $\frac{4}{3} *\pi$ * 36*r³

Replacing by the values, and simplifying common terms, we can calculate the x-coordinate of the center of mass of the system as follows:

$Xcm = \frac{m1*x1+m2*x2+m3*x3}{m1+m2+3}$

$Xcm = \frac{(8*R^{3} *2*R)+(R^{3}*(5*R))+27*R^{3}*(9*R))}{36*R^{3} }=\frac{264*R^{4} x}{36*R^{3}} = 7.33 R$

As the x-coordinate of the center fof mass of the entire system is located at 7.33*R from the origin, and the center of mass of the smallest sphere is located at 5*R from the origin, the center of mass of the system is located at a distance d:

d = 7.33*R - 5*R = 2.33 R

4 0

3 years ago

State two reasons why the rod is not in equilibrium

Lubov Fominskaja [6]

Answer:

For equilibrium the net torque on the rod must be zero. So the 100N forces should on either side of the pivot point i.e. one on the left and one the right, and in same direction.

5 0

3 years ago