Does gravity always include the Earth’s pull or can it be between other objects?
2 answers:
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Answer:
Explanation:
Given;
The operational pressure of the cooker,
area of the opening,
- We know that the gauge pressure is measured with respect to the atmospheric pressure. As the atmospheric pressure acts on all the components of the system being observed so, we only need to counter the gauge pressure.
As the petcock sits on the opening of the cross-section so, it balances the built-in pressure due to its weight.
<u>Hence:</u>
where:
mass of the petcock
acceleration due to gravity
1) First, let's calculate the value of deceleration a that the car can achieve, using the following relationship:
where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
This force will be constant, and since m is always the same, then a is the same even in the second situation.
3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping:
where S=67 m is the distance covered, vf=0 is the final velocity of the car, and vi=15 m/s is the initial velocity. From this we can find a:
2) Then, we can assume this is the value of acceleration that the car is able to reach. In fact, the force the brakes are able to apply is
This force will be constant, and since m is always the same, then a is the same even in the second situation.
3) Therefore, in the second situation we have a=-1.68 m/s^2. However, the initial velocity is different: vi=45 m/s. Using the same formula of point 1), we can calculate the distance covered by the car before stopping: