a ball is thrown straight up with an initial speed 40m/s.a) how high does it go? b) what will be its velocity be in 7 secs?c) ho
1 answer:
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Answer:
121.3 cm^3
Explanation:
P1 = Po + 70 m water pressure (at a depth)
P2 = Po (at the surface)
T1 = 4°C = 273 + 4 = 277 K
V1 = 14 cm^3
T2 = 23 °C = 273 + 23 = 300 K
Let the volume of bubble at the surface of the lake is V2.
Density of water, d = 1000 kg/m^3
Po = atmospheric pressure = 10^5 N/m^2
P1 = 10^5 + 70 x 1000 x 10 = 8 x 10^5 N/m^2
Use the ideal gas equation
By substituting the values, we get
V2 = 121.3 cm^3
Thus, the volume of bubble at the surface of lake is 121.3 cm^3.
Answer:
Frequencia = 1.25 Hz
Explanation:
<u>Dados los siguientes datos;</u>
- Velocidad = 50 m/s
- Longitud de onda = 40 metros
Para encontrar la frecuencia de la onda;
Matemáticamente, la velocidad de una onda viene dada por la fórmula;
Haciendo de la frecuencia el tema de la fórmula, tenemos;
Sustituyendo en la fórmula, tenemos;
<em>Frequencia = 1.25 Hz</em>
Answer:
If the canoe heads upstream the speed is zero. And directly across the river is 8.48 [km/h] towards southeast
Explanation:
When the canoe moves upstream, it is moving in the opposite direction of the normal river current. Since the velocities are vector (magnitude and direction) we can sum each vector:
Vr = velocity of the river = 6[km/h}
Vc = velocity of the canoe = -6 [km/h]
We take the direction of the river as positive, therefore other velocity in the opposite direction will be negative.
Vt = Vr + Vc = 6 - 6 = 0 [km/h]
For the second question, we need to make a sketch of the canoe and we are watching this movement at a high elevation. So let's say that the canoe is located in point 0 where it is located one of the river's borders.
So we are having one movement to the right (x-direction). And the movement of the river to the south ( - y-direction).
Since the velocities are vector we can sum each vector, so using the Pythagoras theorem we have:
