Answer:
La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.
Explanation:
El movimiento rectilíneo uniforme (MRU) es el movimiento que describe un cuerpo o partícula a través de una línea recta a velocidad constante.
La distancia recorrida, x
, por un móvil que tiene un MRU con un velocidad v durante el intervalo de tiempo t es:
x= x0 + v*t
donde x0 es la posición inicial.
En este caso:
Reemplazando:
x= 22 m + 5 m/s* 30 s
Resolviendo:
x= 22 m + 150 m
x= 172 m
<u><em>La posición en la que se encuentra el móvil en el instante t = 30 s es 172 m.</em></u>
Hello Again! I think the Answer might be 220 m! ( 1/2) ( 21 m/s + 0 m/s) (21 s) = 220 m
An important aspect of fission reactions is that they produce free neutrons, which causes chain reactions.
A rubber ball and a stone of the same size are examples which will have more inertia and is therefore denoted as option A.
<h3>What is Inertia?</h3>
This is referred to as the property exhibited by a body in which it has the tendency to remain at rest or in uniform motion.This property is dependent on the mass of the substance as we can deduce that the greater the mass, the greater the inertia and vice versa.
The size of a rubber ball and stone will have different masses in which that of the stone will be greater. This is as a result of the difference in the nature of the substances which are used to make both items mentioned above.
This is therefore the reason why a rubber ball and a stone of the same size as having more inertia(mass) where chosen as the most appropriate choice in this scenario.
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Possible beat frequencies with tuning forks of frequencies 255, 258, and 260 Hz are 2, 3 and 5 Hz respectively.
The beat frequency refers to the rate at which the volume is heard to be oscillating from high to low volume. For example, if two complete cycles of high and low volumes are heard every second, the beat frequency is 2 Hz. The beat frequency is always equal to the difference in frequency of the two notes that interfere to produce the beats. So if two sound waves with frequencies of 256 Hz and 254 Hz are played simultaneously, a beat frequency of 2 Hz will be detected. A common physics demonstration involves producing beats using two tuning forks with very similar frequencies. If a tine on one of two identical tuning forks is wrapped with a rubber band, then that tuning forks frequency will be lowered. If both tuning forks are vibrated together, then they produce sounds with slightly different frequencies. These sounds will interfere to produce detectable beats. The human ear is capable of detecting beats with frequencies of 7 Hz and below.
A piano tuner frequently utilizes the phenomenon of beats to tune a piano string. She will pluck the string and tap a tuning fork at the same time. If the two sound sources - the piano string and the tuning fork - produce detectable beats then their frequencies are not identical. She will then adjust the tension of the piano string and repeat the process until the beats can no longer be heard. As the piano string becomes more in tune with the tuning fork, the beat frequency will be reduced and approach 0 Hz. When beats are no longer heard, the piano string is tuned to the tuning fork; that is, they play the same frequency. The process allows a piano tuner to match the strings' frequency to the frequency of a standardized set of tuning forks.
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