Answer:
Definimos:
Rapidez media es igual al cociente entre la distancia recorrida y el tiempo que se tarda en recorrer esa distancia.
En este caso la distancia recorrida es 400m, y el tiempo que se tarda es 30s, entonces la rapidez media va a ser:
RM = 400m/30s = 13.33 m/s
La velocidad media por otro lado, es igual al cociente entre el desplazamiento y el tiempo necesario para desplazarse.
El desplazamiento es igual a la distancia entre la posición final y la posición inicial, que en este caso eso 40m, y el tiempo necesario sigue siendo 30s, entonces la velocidad media va a ser:
VM = 40m/30s = 1.33 m/s
This would happen later at night or early in the morning.
The reason being land becomes warm and cold quicker than the water because of the heat capacity. So during the day water warms up because of sunlight but at night the land becomes a lot cooler as compared to the water which is still war. So the air over water is significantly warmer than the air over land.
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
Here it is given that initial speed of the package will be same as speed of the helicopter
displacement of the package as it is dropped on ground
acceleration is due to gravity
now by kinematics
by solving above equation we have
so it will take 5.2 s to reach the ground
Answer:
The thrown rock strike 2.42 seconds earlier.
Explanation:
This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:
So now we have an equation and unkown value.
for the thrown rock
for the dropped rock
solving both equation with the quadratic formula:
we have:
the thrown rock arrives on t=5.4 sec
the dropped rock arrives on t=7.82 sec
so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)
