A solid sphere of uniform density has a mass of 4.0×104 kg and a radius of 2.5 m. What is the gravitational force due to the sphere on a particle of mass 1.0 kg located at a distance of 3.00 m from the center of the sphere?

The gravitational force between two objects of mass $m_1$ and $m_2$ separated by a distance $r$ is given as: $F_("grav")=G(m_1*m_2)/r^2$ Where $G$ is the Universal Gravitational Constant$=6.67xx10^-11 (Nm^2)/(kg^2)$

Using Newton's gravitational force equation $F=(Gm_1m_2)/(r^2)$ and assuming that the mass of is $m_1=5.972*10^24kg$ and $m_2$ is the given mass of the moon with $G$ being $6.674*10^-11Nm^2/(kg)^2$ gives $1.989*10^20...

There is no such thing as a gravitational force. Newton described gravity in terms of a force. It is however just a good approximation. Einstein's general theory of relativity describes...

The acceleration you would experience is: $a=(GM)/r^2$ Where $M=10^11kg$ is the mass of the black hole, $G=6.674m^2kg^(-1)s^(-2)$ is the gravitational constant and $r$ is the distance from the black hole....

The radius of the Earth is approximately $4000$ miles. Gravity obeys an inverse square law, decreasing by the square of the distance from the centre. So the gravitational acceleration due...

The radius= 1.08cm = 0.43in $density=(mass)/(volume)$ $Volume = (mass)/(density)$ $Volume=(82)/(2.7)=30.37cm^3$ The volume of a sphere is given by: $V =4/3pir^3$ So $ r^3=(3V)/(4pi) =(3xx30.37)/(4xx3.142)=1.274$ $r = 1.084cm$

From definition, is the mass per unit volume. $therefore rho=m/V=m/(4/3pir^3)$ $=(1794g)/(4/3pixx4^3)$ $=6.692g//cm^3$.

The first thing to focus on here is finding the of the solution. As you know, is defined as mass per unit of volume. You know that your...

Stokes' Law Stokes' law states that the settling velocity $v$ of small spheres in a fluid is given by $v = (gd^2(ρ_p - ρ_f))/(18eta)$ where $g$ = the gravitational...

The problem tells you that lithium sulfate has a of $color(blue)("2220 g")$ $color(purple)("L"^(-1))$, which means that every $color(purple)("1 L")$ of lithium sulfate has a mass of $color(blue)("2220 g")$. In your...