Answer: The formula of Newtons second law of motion is F=MA so therefore it would be written like this Force = Mass X Acceleration
F = 5 x 2
F = 10 N
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
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
Explanation:
In the parallel combination, the equivalent resistance is given by :

4. When three 150 ohms resistors are connected in parallel, the equivalent is given by :

5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,

Hence, this is the required solution.
The answer is B
As seen on the graph, the bus maintains a 9m/s speed for a majority of the trip to school.