A ball is thrown directly upward into the air. The graph below shows the vertical position of the ball with respect to time

Why is it difficult to see colors outside at night?

Stella [2.4K]

I choose D. but I'm not sure if it's right yet. Let me know if it was :)

7 0

3 years ago

A cart moves with negligible friction or air resistance along a roller coaster track. The cart starts from rest at the top of a

lina2011 [118]

Answer:

hinit = 17.5 m

Explanation:

Assuming no friction present, the mechanical energy must be conserved, which means that at any point of the trajectory, the sum of the gravitational potential energy and the kinetic energy must keep the same.
At the top of the hill, since it starts from rest, all the energy must be potential, and we can express it as follows:

$E_{o} = U_{o} = m*g*h_{init} (1)$

When the car arrives to the top of the second hill, as we know that it is lower than the first one, the energy of the car, must be part gravitational potential energy, and part kinetic energy.
We can express this final energy as follows:

$E_{f} = U_{f} + K_{f} = m*g* h_{2} + \frac{1}{2} *m*v_{f} ^{2} (2)$

In order to find hinit, we need to make (1) equal to (2), and solve for it.
In (2) we have the value of h₂ (10 m), but we still need the value of the speed at the top of the second hill, vf.
Now, when the car is at the top of the hill, there are two forces acting on it, in opposite directions: the normal force (upward) and the weight (downward).
We know also that there is a force that keeps the car along the circular track, which is the centripetal force.
This force is just the net downward force acting on the car (it's vertical at the top), and is just the difference between the weight and the normal force.
If the cart just barely loses contact with the track at the top of the second hill, this means that at that point the normal force becomes zero.
So, the centripetal force must be equal to the weight.
The centripetal force can be expressed as follows:

$F_{c} = m*\frac{v_{f} ^{2}}{R} (3)$

We have just said that (3) must be equal to the weight:

$F_{c} = m*\frac{v_{f} ^{2}}{R} = m*g (4)$

Simplifying, and rearranging, we can solve for vf², as follows:

$v_{f}^{2} = R*g (5)$

Replacing (5) in (2), simplifying and rearranging in (1) and (2) we finally have:

$h_{init} = h_{2} + \frac{1}{2} R = 10m + 7.5 m = 17.5 m (6)$

7 0

3 years ago

A circular test track for cars has a circumference of 3.5 km . A car travels around the track from the southernmost point to the

Marta_Voda [28]

<span>The car would have traveled exactly one-half of the circumference of the track, since it would have gone from one extreme point to its opposite extreme point. This would be equal to (3.5 / 2), or 1.75 km. The northernmost point would be 1.75km away from the southernmost point.</span>

8 0

4 years ago

Using parallax, what is the distance to the<br> farthest stars you can measure?

aev [14]

Parallax angles of less than 0.01 arcsec are very difficult to measure from Earth because of the effects of the Earth's atmosphere. This limits Earth based telescopes to measuring the distances to stars about 1/0.01 or 100 parsecs away.

6 0

3 years ago

If you drive through water, your brakes may become slippery and ineffective. To dry the brakes off, __________.

GenaCL600 [577]

Answer:

C

Explanation:

Water at normal temperature is a good lubricant and will hinder the brake from working properly when the road is wet. To dry the brakes off, apply brakes gently as you accelerate. This will generate some heat that will evaporate the water with the help of friction.

8 0

4 years ago