Have you done dot or cross products yet? If you have the easiest way to compute the angle is through the association that
\mathbf{A}\cdot\mathbf{B}=|\mathbf{A}||\mathbf{B}|\cos{\theta}
or with cross products
|\mathbf{A}\times\mathbf{B}|=|\mathbf{A}||\mathbf{B}|\sin{\theta}
where...
First, you know that a force in the opposite direction and half the magnitude is just 3\hat{k}. So take a look at the torque;
\tau=\mathbf{r}\times\mathbf{F}=\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
a & b & 0 \\
2 & 1 & 0
\end{vmatrix}
=[a-2b]\hat{k}
So what possiblities could...
I think there is a simpler way to derive the escape velocity (if we're just talking about one surge of velocity at t=0 and not a rocket or anything). From conservation of energy we have
K_i + U_i = K_f + U_f
Two of these go to zero; the final potential energy (we're going to infinity) and the...
Think about what you're trying to find in part (a)--it's much like what the value g is on earth right? Note that g can be expressed as
\mathbf{g}=-\frac{GM}{|\mathbf{r}|^2}\hat{\mathbf{r}}=\frac{\mathbf{F}}{m}
For part (b), think of the fact that escape velocity is really something that comes...
I can't tell from the question but is the particle fired from an angle or simply a height above the target in the straight x direction? Nevertheless, any time in projectile motion you are trying to find the maximum height of something the solution is to usually just take the projectile equation...
The formula for terminal velocity would be,
V_t=\sqrt{\frac{2mg}{\rho A C_d}}
where m is the mass of your jumper in kilograms g=9.81 meters per second squared, and there are more specific things involved in the formula like the density of air (\rho) the drag coefficient (C_d) and the area...
The force of friction is defined to be \mu \mathbf{N}\text{ where } \mu \text{ is the frictional constant and } \mathbf{N} is the normal force. But there is of course no friction involved in this problem.
You've got it--I think you meant subtract the radii of the earth and the moon though, not the sun. Think about it--velocity is distance over time--so you get...
c= 2\left(\frac{3.84 \times 10^8 - (E_R+m_R)}{2.51}\right)