In calculating the energy of a photon of light, we need the relationship for energy and the frequency which is expressed as:
E=hv
where h is the Planck's constant (6.626 x 10-34 J s)and v is the frequency.
E = 6.626 x 10-34 J s (<span>7.33 x 10^14 /s) = 4.857 x 10^-19 J</span>
Answer:
Tension in the chains - In a chain drive, technically, you have a closed-chain (which has no end) going around 2 pulley or gears; looking closely you have 2 parallel chains going in opposite direction. If kept in horizontal direction, the one below the other is the slack side and the other the tight side. The tension on the upper or tight side is more than the slack side. So you need to keep in mind to keep your chain drive tight so that there is no loss or rotation or lags.
Sizes of the pulley/gear - The chain will be warped around a pair of pulley or gear. The sizes of these pulley/gear will also determine the efficiency of the chain drive (consider one big and one small)
Number of pulley/gear - If the number of pulley/gear is more and chain wrapped on it with little complexity will result in decrease in efficiency because of extra tension.
Length of the chain drive - You cannot have much too long chain drive. It will make your slack side more heavy because the end are further away. You have to apply more power and possibilities of lag increases decreasing efficiency. In an ideal situation, this won't happen, but this world isn't ideal.
Friction between chains & pulley/gear - If you have studied gears (involving its teeth), you will come to know that there is friction offered on the two meeting surfaces.
Angle of contact - This would have been explained better with a diagram. Although, if you are familiar with the terms you won't have difficulty understanding. Angle of contact is the angle the chain forms with the pulley/gear at the point of contact with the center of the pulley. The angle of contact should not be too small, or else the things will be slippery.
Explanation:
I can think of two possible and logical questions for the problem given. First, you can calculate for the maximum height reached by the blue ball. Second, you can compute the length of time for the two balls to be at the same height. If so, the solution are as follows:
When the object is thrown upwards or when the object is dropped from a height, the only force acting upon it is the gravitational force. Because of this, it simplifies equations of motion.
1. For the maximum height, the equation is
H = v₀²/2g
where
v₀ is the initial speed
g is the acceleration due to gravity equal to 9.81 m/s²
For the blue ball, v₀ = 21.8 m/s. Substituting the values:
H = (21.8 m/s)²/2(9.81m/s²)
H = 24.22 m
The maximum height reached by the blue ball is 24.22 m + 0.9 = 25.12 m.
2. For this, you equate the y values of both balls:
y for red ball = y for blue ball
v₀t + 0.5gt² = v₀t + 0.5gt²
(10.4 m/s)t + 0.5(9.81 m/s²)(t²) + 26.6 m = (21.8 m/s)t + 0.5(9.81 m/s²)(t²) + 0.9 m
Solving for t,
t = 2.25 seconds
Thus, the two balls would be at the same height after 2.25 seconds.
Answer:
4.245s
Explanation:
Given that,
Hypothetical value of speed of light in a vacuum is 18 m/s
Speed of the car, 14 m/s
Time given is 6.76 s, and we're asked to find the observed time, T
The relationship between the two times can be given as
T = t / √[1 - (v²/c²)]
The missing variable were looking for is t, and we can find it if we rearrange the formula and make t the subject
t = T / √[1 - (v²/c²)]
And now, we substitute the values and insert into the equation
t = 6.76 * √[1 - (14²/18²)]
t = 6.76 * √[1 - (196/324)]
t = 6.76 * √(1 - 0.605)
t = 6.76 * √0.395
t = 6.76 * 0.628
t = 4.245 s
Therefore, the time the driver measures for the trip is 4.245s
I think the correct answer is C
