Answer:
(a) Yes, it is possible by raising the object to a greater height without acceleration.
Explanation:
The work-energy theorem states that work done on an object is equal to the change in kinetic energy, and change in kinetic energy requires a change in velocity.
If kinetic energy will not change, then velocity will not change, this means that there will be constant velocity and an object with a constant velocity is not accelerating.
If the object is not accelerating (without acceleration) and it remains at the same height (change in height = 0, and mgh = 0).
Thus, for work to be done on the object, without changing the kinetic energy of the object, the object must be raised to a greater height without acceleration.
Correct option is " (a) Yes, it is possible by raising the object to a greater height without acceleration".
A soccer ball is released from rest at the top of a grassy incline. After 2.2 seconds, the ball travels 22 meters. One second later, the ball reaches the bottom of the incline. (Assume that the acceleration was constant.) How long was the incline
Answer:
x = 46.54m
Explanation:
In order to find the length of the incline you use the following formula:
[tex]x=v_ot+\frac{1}{2}at^2[/tex] (1)
vo: initial speed of the soccer ball = 0 m/s
t: time
a: acceleration
You first use the the fact that the ball traveled 22 m in 2.2 s. Whit this information you can calculate the acceleration a from the equation (1):
[tex]22m=\frac{1}{2}a(2.2s)^2\\\\a=9.09\frac{m}{s^2}[/tex] (2)
Next, you calculate the distance traveled by the ball for t = 3.2 s (one second later respect to t = 2.2s). The values of the distance calculated is the lenght of the incline:
[tex]x=\frac{1}{2}(9.09m/s^2)(3.2s)^2=46.54m[/tex] (3)
The length of the incline is 46.54 m
A uniformly charged sphere has a potential on its surface of 450 V. At a radial distance of 8.1 m from this surface, the potential is 150 V. What is the radius of the sphere
Answer:
The radius of the sphere is 4.05 m
Explanation:
Given;
potential at surface, [tex]V_s[/tex] = 450 V
potential at radial distance, [tex]V_r[/tex] = 150
radial distance, l = 8.1 m
Apply Coulomb's law of electrostatic force;
[tex]V = \frac{KQ}{r} \\\\V_s = \frac{KQ}{r} \\\\V_r = \frac{KQ}{r+ l}[/tex]
[tex]450 = \frac{KQ}{r} ------equation (i)\\\\150 = \frac{KQ}{r+8.1} ------equation (ii)\\\\divide \ equation (i)\ by \ equation \ (ii)\\\\\frac{450}{150} = (\frac{KQ}{r} )*(\frac{r+8.1}{KQ} )\\\\3 = \frac{r+8.1}{r} \\\\3r = r + 8.1\\\\2r = 8.1\\\\r = \frac{8.1}{2} \\\\r = 4.05 \ m[/tex]
Therefore, the radius of the sphere is 4.05 m
The smallest shift you can reliably measure on the screen is about 0.2 grid units. This shift corresponds to the precision of positions measured with the best Earth-based optical telescopes. If you cannot measure an angle smaller than this, what is the maximum distance at which a star can be located and still have a measurable parallax
Answer:
The distance is [tex]d = 1.5 *10^{15} \ km[/tex]
Explanation:
From the question we are told that
The smallest shift is [tex]d = 0.2 \ grid \ units[/tex]
Generally a grid unit is [tex]\frac{1}{10}[/tex] of an arcsec
This implies that 0.2 grid unit is [tex]k = \frac{0.2}{10} = 0.02 \ arc sec[/tex]
The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as
[tex]d = \frac{1}{k}[/tex]
substituting values
[tex]d = \frac{1}{0.02}[/tex]
[tex]d = 50 \ parsec[/tex]
Note [tex]1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km[/tex]
So [tex]d = 50 * 3.08 *10^{13}[/tex]
[tex]d = 1.5 *10^{15} \ km[/tex]