What is 1.3 for PE=mgh and for KE=1(m)v^2 and E=PE+KE and V
1 answer:
You might be interested in
Answer:
138.46 ft
Explanation:
When the ball is dropped until the moment it hits the water, the ball moves in a uniform acceleration motion. Therefore, the equation that describes the movement of the ball is:
Where X is the distance that the ball has fallen at a time t. is the initial velocity, which is 0 ft/s as the ball was simply dropped.
is the initial position, we will say that this value is 0 in the position where the ball was dropped for simplicity, and it increases as the ball is falling. Now, we replace x with 16 feets and solves for t:
The velocity that the ball will have at the moment the ball that the ball hits the water will be:
The time that will take the ball to reach the bottom from the top of the lake will be t = 5.3s - 1s = 4.3s. And as the ball will travel with constant velocity equal to 32.2 ft/s^2, the depth of the lake will be:
Answer:
t₂=6.35min
Explanation:
t₁ = first observed time (=5.1 min)
t₂ = unknown; this is the quantity we want to find
V₁ = observer's initial speed (=0.84c)
V₂ = observer's final speed (=0.90c)
Lorentz factors for V₁ and V₂:
γ₁ = 1/√(1−(V₁/c)²)
γ₂ = 1/√(1−(V₂/c)²)
The "proper time" (the time measured by the person filling her car) is:
t′ = t₁/γ₁
The proper time is stated to be the same for both observations, so we also have:
t′ = t₂/γ₂
Combine those two equations and solve for t₂
t₂ = t₁(γ₂/γ₁)
t₂= t₁√((1−(V₁/c)²)/(1−(V₂/c)²))