Newton's first law states that an object in motion tends to stay in motion unless if acted upon by a net force. This means that if friction is not being accounted for, there is no net force required to keep an object moving if it's in motion. A net force is only required to change an object's motion. The 500 kg object is moving at a constant velocity; therefore, there is no net force (0 Newtons) acting on the object.

Newton's first law of motion refers to objects with no external forces acting on them. Objects with no external forces will maintain the same velocity, meaning that

(1) if they are not moving, they will continue not moving

(2) if they are moving, they will keep moving with the same speed and direction

The answer "Dropping a box causes it to accelerate downwards" refers to a box with a force acting upon it—the force of gravity. Also, the box is accelerating, unlike objects referred to by Newton's first law of motion, which have constant velocities.

In a vacuum, there is no friction due to air resistance. Newton's first law states that an object in motion stays in motion unless acted upon by a net force. Thus the spaceship will travel at the constant speed (zero acceleration) of $\displaystyle 500\frac{m}{s}$ indefinitely and the net force on the spaceship must be zero. This can also be shown mathematically:

$\displaystyle F=ma$

$\displaystyle F=(1000kg)(0\frac{m}{s^2}) = 0N$

Since both you and your friend are traveling at a constant speed, the acceleration of you and your friend is zero. Thus, neither you nor your friend experiences any acceleration. This can be shown mathematically using the equation for acceleration:

$\displaystyle a=\frac{\Delta v}{\Delta t}$

Since there is no change in velocity over time acceleration is zero. Also note that when acceleration is zero, so is the net force.

Because an object at rest tends to stay at rest, when your car is hit your body/neck will 'want' to stay where it was. This will cause your body and neck to 'whip' as it will take time for it to speed up from being hit. 

The acceleration of the person in the elevator is determined by the net forces and his/her mass. The net force is calculated to be 20N upwards. To find the mass of the passenger, use the following formula:

$\displaystyle F=mg$

$\displaystyle m=\frac{100N}{10\frac{m}{s^2}}=10kg$

Then, to find the net acceleration, use Newton's second law.

$\displaystyle F=ma$

$\displaystyle a=\frac{20N}{10kg}=2\frac{m}{s^2}$

Recall Newton's first law of motion: an object will remain in its state of uniform motion unless acted upon by an external force. The car's motion is described as having a constant velocity which is a uniform state so there are no external forces.

If the box is moving at a constant velocity, we know that the sum of the forces acting up and down the inclined plane must add to zero. If the only forces acting are gravity and friction, we can show:

$\displaystyle \sum F_{x'}=0=mg \sin \theta - F_f \Rightarrow mg \ sin \theta = \mu _k N$

We can gain information about the normal force $\displaystyle N$ by looking at the forces acting perpendicular to the plane, shown by: 

$\displaystyle \sum F_{y'}= 0 = N - mg \cos \theta \Rightarrow N = mg \cos \theta$

Putting all of this together lets us write the following expression for the forces in acting along the plane, and finally, our answer:

$\displaystyle mg \ sin \theta = \mu _k mg \cos \theta\Rightarrow \mu _k = \tan \theta = \tan 10^{\circ}= 0.176$

Use conservation of momentum:

$\displaystyle P_i=P_f$

$\displaystyle (m_1+m_2)(v_i)=m_1*v_{1f}+m_2*v_{2f}$

Plug in values.

$\displaystyle 0=(550*10^5)*300+(430*10^5)*v$

Solve for $\displaystyle v$:

$\displaystyle v=\frac{-(550*10^5)*300}{(430*10^5)}$

$\displaystyle v=-384\frac{m}{s}$

$\displaystyle |v|=384\frac{m}{s}$

If the diver has reached terminal velocity, her acceleration is $\displaystyle 0$.

Thus, according to Newton's second law:

$\displaystyle F=ma$

Her net force is equal to zero.

$\displaystyle F_{net}=0$

The only forces acting on her are gravity and wind resistance, which must add up to zero.

$\displaystyle F_{net}=0=mg+F_{wind}$

Where $\displaystyle g$ is pointing down and thus negative

Plug in values:

$\displaystyle F_{net}=0=92*-9.8+F_{wind}$

Solve for $\displaystyle F_{wind}$

$\displaystyle F_{wind}=901.6N$