Answer:
2.28 s
Explanation:
Let g = 9.8 m/s2 and neglect air resistance. The box of sand with an initial velocity of 4.2m/s in free fall would yield the following equation of motion
t = 2.28 or t = -3.14
Since t can only be positive we will pick t = 2.28
s
Answer:
a) t = 0.75 s, b) t = 2.25 s
Explanation:
The speed of sound is constant in a material medium
v = 340 m / s
we can use the relations of uniform motion to find the time
v = x / t
t = x / v
In the exercise, the observer's distance to the wall is indicated d = 510 m, it also indicates that the shot is fired at the midpoint
x = d / 2
a) direct sound the distance from the observer to the screen is
x = 510/2 = 255 m
t = 255/340
t = 0.75 s
b) echo sound.
In this case the sound reaches the wall bounces, the distance is
x = d / 2 + d
x = 3/2 d
x = 3/2 510
x = 765 m
the time is
t = x / v
t = 765/340
t = 2.25 s
Responder:
39200 J
Explicación:
Dado:
Masa de Tamara (m) = 50 kg
Altura a la que se encuentra Tamara (h) = 80 m
Aceleración debido a la gravedad (g) = 9.8 m / s²
La energía potencial de un objeto de masa 'm' ubicada a una altura 'h' sobre el suelo se da como:
Ahora, conecte los valores dados y resuelva la energía potencial. Esto da,
Por lo tanto, la energía potencial de Tamara ubicada a una altura de 80 m es 39200 J.
Answer:
hope this helps!
Explanation:
Volume of the air bubble, V1=1.0cm3=1.0×10−6m3
Bubble rises to height, d=40m
Temperature at a depth of 40 m, T1=12oC=285K
Temperature at the surface of the lake, T2=35oC=308K
The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa
The pressure at the depth of 40 m: P1=1atm+dρg
Where,
ρ is the density of water =103kg/m3
g is the acceleration due to gravity =9.8m/s2
∴P1=1.103×105+40×103×9.8=493300Pa
We have T1P1V1=T2P2V2
Where, V2 is the volume of the air bubble when it reaches the surface.
V2=
