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3 years ago

Which formula can be used to calculate the horizontal displacement not of a horizontally launched projectile

1 answer:

8 0

<span>If you are looking to get an object up the highest, shoot it straight up. If you want to go for a specific horizontal displacement, use the range equation. R = v2sin(twice the launch angle)/ g. g is the gravitaional constant, 9.8 meters per second. Use degrees for the angle. v is the launch velocity. R is the horizontal displacement. This formula only works if your start altitude and end altitude are the same, i.e. you must shoot over a level field. it depends on the gravitational force of attraction of earth and air resistance. if we are neglecting air resistance, the max.horizontal distance is according to this formulae, V0/2 * sin (2theta) where V0 is the initial velocity theta is the angle with x axis and the projection. There are a number of ways that you could find a horizontally displaced object. You could for example just look.</span>

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An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar

defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency $\zeta$ = 3%

where;

$\zeta = \dfrac{W_{out}}{Q_{supplied }}$

$Q_{supplied } = \dfrac{2}{0.03} \ MW$

$Q_{supplied } = 66.66 \ MW$

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

$LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}$

Also;

$\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K$

$\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K$

$LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}$

$LMTD = \dfrac{8}{In (5)}$

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

$Q_H = UA (LMTD)$

where;

U = overall heat coefficient given as 1200 W/m².K

$66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}$

The mass flow rate:

$Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}$

3 0

3 years ago

Each orbit is limited to a maximum of four electrons true or false

Naya [18.7K]

This question would be false

7 0

3 years ago

Why do you think this appeal system is necessary in the legal system?

MatroZZZ [7]

Appeal system enables the higher court to review the case once again. Possible situation for appeal arises when the defendant consider punishment very tough.

<u>Explanation:</u>

The Appeal system is very important in the legal system. Appeal means a process which helps in reviewing the cases. In case of the appeal, a request is made to a higher court to look into the case once again.

This process helps those who feel that the earlier judgment in case of the case was not correct.

The Possible situation where someone needs to appeal is when the defendant is found guilty and we feel that the punishment given is very harsh. This is possible mostly in criminal cases.

6 0

3 years ago

An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17

Softa [21]

Answer:

distance cover is = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

$t_1 = 3.32 sec$