Explanation:
Given that,
The mass of rock, m = 2.35-kg
It was released from rest at a height of 21.4 m.
(a) The kinetic energy is given by : 
As the rock was at rest initially, it means, its kinetic energy is equal to 0.
(b) The gravitational potential energy is given by : 
It can be calculated as :
(c) The mechanical energy is equal to the sum of kinetic and potential energy such that,
M = 0 J + 492.84 J
M = 492.84 J
Hence, this is the required solution.
There are many factors that determine if an aircraft can operate from a given airport. Of course the availability of certain services, such as fuel, access to air stairs and maintenance are all necessary. But before considering anything else, one must determine if the plane can physically land at an airport, and equally as important, take off.
What is the minimum runway length that will serve?
Looking at aerial views of runways can lead some to the assumption that they are all uniform, big and appropriate for any plane to land. This couldn’t be further from the truth.
A given aircraft type has its own individual set of requirements in regards to these dimensions. The classic 150’ wide runway that can handle a wide-body plane for a large group charter flight isn’t a guarantee at every airport. Knowing the width of available runways is important for a variety of reasons including runway illusion and crosswind condition.
Runways also have different approach categories based on width, and have universal threshold markings that indicate the actual width.
To learn more about runway
brainly.com/question/11553726
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Calculate its average speed in meters per second
Answer:
5.77 m/s
Explanation:
Speed= Distance/Time
Distance= 40+ half of 40= 40+20= 60 m
Time= 8.8+1.6=10.4 s
Average speed= 60/10.4=5.769230769 m/s
Approximately, the average speed is 5.77 m/s
<span>The formulas are,
v1d1² = v2d2² ........ (1)
h = (v2²-v1²)/2g ...... (2)
Given that,
v1 = 1.71 m/s
we assume that the stream has decreased by a factor
d2 =0.805d1
then,
v1d1² = v2 (0.805d1)²
cancelled both side d1² then we get,
v1 = v2 (0.805)²
v1 = v2 (0.648025)
Sub v1 = 1.71,
1.71 = v2 (0.648025)
v2 = 1.71/0.648025
v2 = 2.638787083831642
v2 = 2.64 m/s
The vertical distance formula,
h = (v2²-v1²)/2g
We know that value of gravity constant is 9.8 m/s²
h = {(2.64)² - (1.71)²)/2(9.8)
h = {(6.9696) - (2.9241)}/19.6
h = (4.0455)/19.6
h = 0.2064030612244898
h = 0.21 cm
Therefore, the vertical distance h = 0.21 cm.</span>
Answer:
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.
Explanation:
From coulomb's law, F = Eq
Thus,
F = E₁q₁
F = E₂q₂
Then
E₂q₂ = E₁q₁
where;
E₂ is the external electric field due to second test charge = ?
E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C
q₁ is the first test charge = 13 mC
q₂ is the second test charge = 23 mC
Substitute in these values in the equation above and calculate E₂.
The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.
However, the direction of the external field is still to the right.
