Answer:
The building are not getting farther and farther away, road signs, exits, other cars
Explanation:
Calculate the first and second velocities of the car with three washers attached to the pulley, using the formulas
v1 = 0.25 m / t1, and
v2 = 0.25 m / (t2 – t1)
where t1 and t2 are the average times the car took to reach the 0.25 and the 0.50 meter marks. Record these velocities, to two decimal places, in Table E.
What is the first velocity of the car with three washers at the 0.25 meter mark?
m/s
What is the second velocity of the car with three washers at the 0.50 meter mark?
m/s
Answer:
0.19 and 0.45
Explanation:
edge
The first and second velocities of the car are [tex]0.25/t_1[/tex] and [tex]0.50/t_2[/tex]respectively.
What is velocity?The velocity of an object is defined as the product of the distance travelled in a given direction and the time required to travel that distance. The velocity is expressed mathematically as,
v = d/t
Here, d represents the travelled distance.
And t is the typical amount of time needed to go the distance.
Given,
The value of the first distance is, d1 = 0.25 m.
The value of the second distance is, d2 = 0.50 m.
Then the first velocity of the car at 0.25 m is,
[tex]v_1 = d_1/t_1\\v_1 = 0.25 / t_1[/tex]
Thus,[tex]t_1[/tex] denotes the first distance's typical time.
The second velocity of the car with three washers at the 0. 50 m mark is
[tex]v_2 = d_2/t_2\\v_2 = 0.50 /t_2[/tex]
Here, [tex]t_2[/tex] stands for the second distance's typical time.
Thus, the first and second velocities of the car are [tex]0.25/t_1[/tex] and [tex]0.50/t_2[/tex]respectively.
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A dentist uses a small concave mirror to locate a cavity in a patient's tooth. The focal length of the mirror is 2.00 cm. If the mirror is held 1.60 cm from the tooth, how many times bigger will the image be? (In other words, what is the magnification?)
Answer:
m = 4.0
Explanation:
For this exercise in geometric optics we will use the equation of the constructor
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distance to the object and image, respectively.
In the exercise indicate the focal length f = 2.00 cm, the distance to the object p = 1.60 cm, let's find the distance to the image
[tex]\frac{1}{q} = \frac{1}{f} - \frac{1}{p}[/tex]
let's calculate
[tex]\frac{1}{q} = \frac{1}{2} - \frac{1}{1.6}[/tex]1 / q = ½ - 1 / 1.6
[tex]\frac{1}{q}[/tex] = -0.125
q = -8 cm
the magnification is
m = - q / p
m = 8/2
m = 4.0
Can you describe how and why the molecules move from one side to the other?
The molecules move from one side to another across the concentration gradient by breaking weaker bonds among the atom into stronger bonds. This is done to decrease the overall kinetic energy to become a more stable molecule.
The kinetic strength of the molecules consequences in random movement, causing diffusion. In simple diffusion, this method proceeds without the useful resource of a transport protein. it is the random motion of the molecules that reasons them to move from a place of excessive attention to a place with decreased awareness.
The molecules in a gas, a liquid, or a strong are in consistent movement due to their kinetic electricity. Molecules are in steady movement and collide with each different. those collisions cause the molecules to move in random guidelines. over time, however, greater molecules may be propelled into the less concentrated place.
The majority of the molecules flow from better to decrease awareness, although there can be some that circulate from low to excessive. the general (or net) motion is consequently from high to low concentration.
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