Using the velocity-time graph, the displacement can be calculated by the area under the velocity-time graph. At 3 seconds the total displacement is then equal to (4)(2) + (4 + 2)*1/2 = 11 m. Assuming that the starting point is at x = 0, then the particle at t=3s is at x=11 m.
I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
Answer: First option
Explanation: The higher the frequency, the higher the energy.
λν=c where λ is the wavelength, ν is the frequency and c is the speed of light. So when wavelength decreases, v increases and so does energy.
An AM radio wave has a very long wavelength. It therefore has a very low frequency and low energy.
A light wave has a very short wavelength. It therefore has a high frequency and high energy.
Answer:
Frequency = 260 Hz and wavelength = 1.31 m
Explanation:
Given that,
A hammer begins vibrating back and forth at approximately 260 cycles per second.
(a) The frequency of an object is the number of vibrations per unit time. The frequency of the sound wave is 260 Hz.
(b) The speed of sound in air is 343 m/s. So,
Hence, this is the required solution.
Answer:
Time, t = 13.34 seconds.
Explanation:
Given the following data;
Initial velocity, u = 85km/hr to meters per seconds = 85*1000/3600 = 23.61 m/s
Final velocity, v = 45km/hr to meters per seconds = 45*1000/3600 = 12.5 m/s
Acceleration, a = -3 km/hr/sec to meters per seconds square = -3*1000/3600 = -0.833m/s²
To find the time;
Acceleration = (v - u)/t
-0.833 = (12.5 - 23.61)/t
-0.833t = -11.11
t = 11.11/0.833
Time, t = 13.34 seconds.
<span>Star a is more distant and is approximately 5 times as far away as star b
Parallax is the change in angle that one must do in order to observe the same object from different locations. The further away an object is, the smaller the parallax is. As the angles approach zero, the trig functions tend to be fairly linear. And 0.1 arc seconds and 0.02 arc seconds are close enough to zero for this to hold true.
Since the parallax for star a is smaller than the parallax for star b, it is the more distant star. And since 0.1 divided by 0.02 = 5, it is approximately 5 times further away than star b.</span>
