An object is thrown directly up (positive direction) with a velocity (v o ) of 20.0 m/s and d o = 0. How high does it rise (v =
1 answer:
Answer:
h=20.38 m
Explanation:
Given that
Initial speed of object u = 20 m/s
Acceleration
We know that
Here acceleration and velocity is in opposite direction so the object will come rest after reaching at distance h.When body will reach at its highest position then velocity will become zero(v=0).
Now by putting the values
h=20.38 m
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Answer:
0.8J
Explanation:
Given parameters:
Force = 20N
Compression = 0.08m
Unknown:
Spring constant = ?
Elastic potential energy = ?
Solution:
To solve this problem, we use the expression below:
F = k e
F is the force
k is the spring constant
e is the compression
20 = k x 0.08
k = 250N/m
Elastic potential energy;
EPE = k e² =
x 250 x 0.08²
Elastic potential energy = 0.8J
Given,
The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m
The initial speed of the water, v₁=12.5 m/s
The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m
From the continuity equation,
Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.
On substituting the known values,
Thus, the flow rate of the water at the later position is 5.99 m/s