If the diver stays at a constant depth, we know that all the forces on him must cancel out:

$\displaystyle F_{net}=0$

We can write out all of the forces on the diver, using force of the flippers to denote the vertical force he exerts to stay at constant depth:

$\displaystyle F_{gravity}+F_{buoyancy}+F_{flippers} = 0$

The force of gravity and bouyancy counteract each other, so they will have opposite signs. If we assume that the force of flippers will be down, we can write:

$\displaystyle F_{flippers} = F_{buoyancy}-F_{gravity}$

The bouyancy force is simply the weight of the water displaced by the diver. Plugging in expressions for the latter two forces:

$\displaystyle F_{flippers} = V_{diver}\rho_{water}g - mg$

We have values for all of these, allowing us to solve:

$\displaystyle F_{flippers} = (0.3m^3)(1,000\frac{kg}{m^3})(10\frac{m}{s^2}) - (150kg)(10\frac{m}{s^2})$

$\displaystyle F_{flippers} = 1500 N$

Since the spring is initially in equilibrium, we can write:

$\displaystyle F_{net}=0$

There are three forces in play: spring force, gravitational force, and the additional force resulting from the acceleration of the box. If we say that any forces pointing downward are positive, we can write:

$\displaystyle F_{spring}+F_{g}-F_{acceleration} = 0$

The force resulting from the additional acceleration must be subtracted since all forces cancel out and both of the other two forces are postive. This means that the resulting force from acceleration is pointing upward; Thus, the box is accelerating downward (think about riding in an elevator).

Substituting expressions for each force, we get:
$\displaystyle kx + mg - ma = 0$

Rearrange to solve for the acceleration of the box:
$\displaystyle a = \frac{kx+mg}{m}= \frac{(0.2m)(25\frac{N}{m})+(2kg)(10\frac{m}{s^2})}{2kg}$

$\displaystyle a = \frac{25N}{2kg} = 12.5\frac{m}{s^2}$

There are three relevant forces in play in this scenario: friction, tension, and gravity.

We can use Newton's second law to create an expression for the net force:

$\displaystyle F_{net}=0$

$\displaystyle F_g -F_f - T = 0$

Substituting expressions in for the forces, we get:

$\displaystyle mgsin(\theta)-\mu_s mgcos(\theta)-T = 0$

Rearrange to solve for the mass:

$\displaystyle m(gsin(\theta)-\mu_sgcos(\theta))=T$

$\displaystyle m = \frac{T}{gsin(\theta)-\mu_s gcos(\theta)}$

Use our given values to solve:

$\displaystyle m = \frac{10N}{(10\frac{m}{s^2})sin(28^{\circ})-(0.40)(10\frac{m}{s^2})cos(28^{\circ})}$

$\displaystyle m = 8.6kg$

The correct answer is normal force. Normal force will occur when the astronaut is standing on a surface, which will push back on him. Gravity is present since the astronaut is in orbit (gravity provides the centripetal force). The feeling of weightlessness he experiences is due to the absence of a surface to push back on him/her.

To solve this problem, we need to look at all of the forces acting on the sphere, this includes the force of gravity, the normal force provided by the scale, and the buoyancy force.  If the object is in equilibrium, the net force acting on the sphere must be zero.

$\displaystyle N+F_B=mg\Rightarrow N=mg-F_B=mg-\rho g V_{dis}=g \left( m-\rho \frac{1}{2}\frac{4}{3}\pi r^3 \right)$

$\displaystyle N=9.81 \frac{m}{s^2}\left( 20 kg-1000\frac{kg}{m^3} \frac{2}{3}\pi(0.1 m)^3\right )=175.7 N$

If a person were to use their hand to push a book across the table, the type of force that their hand is applying to the book is classified as an electromagnetic force. Even though this may not seem intuitive, the reason is that the electrostatic repulsion of the atoms in the person's hand repel the atoms in the book. Due to this electrostatic repulsion between the atoms in the person's hand and the atoms in the book, the force is electromagnetic.

Gravitational forces occur due to gravity, that is, from an attraction between any two objects with mass. The strong and weak forces are both nuclear forces that act only across very, very short distances. Generally, these forces are only significant at the sub-atomic level in the atom's nucleus, and thus do not have a role at greater distances (such as between the person's hand and the book).

The normal force will be equal to the magnitude of the force of gravity pushing the buggy into the ground.

 $\displaystyle F_N=|-G\frac{m_1m_2}{r^2}|$

Using definition of frictional force and work equation:

$\displaystyle .5mv_i^2-F_N*d*\mu=.5mv^2_f$

Combining equations:

$\displaystyle .5mv_i^2-G\frac{m_1m_2}{r^2}*d*\mu=.5mv^2_f$

Canceling out the mass of the buggy and plugging in values:

$\displaystyle .5(25)^2-6.67*10^{-11}\frac{6.39*10^{23}}{(3.39*10^6)^2}*250*\mu=0$

Solving for $\displaystyle \mu$

$\displaystyle \mu=84$

When a diver reaches terminal velocity, we know that the net force on the diver is equal to 0. There are two forces acting on the diver, gravity and drag, so we can say that:

$\displaystyle F_g = F_d$

Plugging in expressions, we get:

$\displaystyle mg = \frac{1}{2}\rho v_t^2C_DA$

Rearranging for density, we get:

$\displaystyle \rho = \frac{2mg}{v_t^2C_DA}$

We have all of these values, so we can plug and chug:

$\displaystyle \rho = \frac{2(50kg)\left ( 10\frac{m}{s^2}\right )}{\left ( 25\frac{m}{s}\right )^2(1.1)(1.2m^2)}$

$\displaystyle \rho = 1.21\frac{kg}{m^3}$

This question tests your understanding of the concept of centripetal force. Centripetal force is a force that maintains a body's circular travel along a curved surface and points inward towards the center of the curve from each point that the body travels along. Centripetal force is calculated as follows:

$\displaystyle F_{centripetal} = mass * acceleration$

$\displaystyle acceleration_{centripetal} = \frac{velocity^{2}}{radius}$

$\displaystyle F_{c} = \frac{mass*velocity^{2}}{radius}$

$\displaystyle F_{C} = \frac{500 kg*(50 \frac{m}{s})^{2}}{10 m}$

$\displaystyle F_{C} = 125,000 N$

Therefore the centripetal force on the car is $\displaystyle 125,000 N$.

Let's use the midpoint as our point of reference. With that said, $\displaystyle A$ is $\displaystyle 10 cm$ from the center, and $\displaystyle B$ is $\displaystyle 5 cm$ from the center. However, each of these are on different sides of the midpoint. When we set up our center of gravity equation, we must determine a (+) and (-) side in order to denote under which side our mobile string will fall. It may be helpful to look at the mobile as a number line, with the left being negative, and the right side positive, but you can really use which ever is more comfortable.

Our main equation is this:

$\displaystyle X_{CG} = \frac{w_{A}x_{A}+w_{B}x_{B}}{w_{A}+w_{B}}$

Where $\displaystyle w$ is our weight in newtons and $\displaystyle x$ is our distance in meters. Typically, we would want to convert everything to SI units, so let's go ahead and do that (ex: $\displaystyle cm\rightarrow m$)

Now let's plug in our numbers, remember about our negative/positive sides!

$\displaystyle {X_{CG}} = \frac{120N\cdot (-.10m)+40N\cdot (.05m)}{120N+40N}$

This should give us $\displaystyle -0.0625m = -6.25cm$

Now, because this number is negative, we know it's to the left of our midpoint (if you chose to set up your -/+ sides opposite of how we did it, your answer will be positive). Regardless of the outcome, this measurement is meant to be taken from object A to the midpoint. Well, do any of our answers have either (+) or (-) $\displaystyle 6.25cm$? No. But if we read them carefully, we can determine that the answer is $\displaystyle 8.75cm$ from the left of the dowel by using simple subtraction.

One side is $\displaystyle 15cm - 6.25cm = 8.75 cm$

Think about it logically too (use your pencil). The heavier end of a pencil usually has a bulky eraser on it (just like the bulkier object on the left). Try holding the pencil at the tip's end, and then gradually try balancing it in the same manner as you move closer to the eraser. 

ALSO NOTE: the question states the weight of the dowel as well. But look we didn't even need it. Sometimes the AP exam will give you some aspects in the question to distract you.