Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
Answer:
v₂ = 0.56 m / s
Explanation:
This exercise can be done using Bernoulli's equation
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂
Where points 1 and 2 are on the surface of the glass and the top of the straw
The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of the straw the velocity of the surface of the vessel is almost zero v₁ = 0
The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m
We substitute in the equation
+ ρ g y₁ = + ½ ρ v₂² + ρ g y₂
½ v₂² = g (y₂-y₁)
v₂ = √ 2 g (y₂-y₁)
Let's calculate
v₂ = √ (2 9.8 1.6 10⁻²)
v₂ = 0.56 m / s
Answer:
3x10⁴v
Explanation:
Using
Wavelength= h/ √(2m.Ke)
880nm = 6.6E-34/√ 2.9.1E-31 x me
Ke= 6.6E-34/880nm x 18.2E -31.
5.6E-27/18.2E-31
= 3 x 10⁴ Volts
Answer:
1÷60 h
time equals distance upon speed
