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3 years ago

An object in projectile motion will follow wich path

1 answer:

6 0

It will tend to follow a trajectory path. When a projectile is being flung or flown away by a certain type of force, the trajectory is what it follows through the means of both force and function of time itself. It's also called as flight path.

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the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity

NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

$\int\limits^s_0 {a} \, ds = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\$

Remember that

$v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2$

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0

3 years ago

What is the closest distance the electrodes used in an NCV test can be placed on a nerve in order to measure the voltage change

sammy [17]

Answer:

0.1 m

Explanation:

The closest distance the electrodes used in an NCV test in oerder to measure

the voltage change as a response to the stimulus is 0.1 m.

This is because the shortest observable time period is not less than the action-potential time response of 1 mili second the length traveled by the sensation during this time is 1 m sec x 100 m / s =0.1 m, which is the shortest distance the electrodes could be positioned on the nerve.

6 0

3 years ago

Does research lessen its value for its being cyclical? Enumerate Two Possible advantages of a cyclical research

Tanya [424]

Answer:

Research usually is a cyclical process because it starts with a problem and ends with a problem.

This is bad or lessens the value of the research?

No, because there are problems that in nature are cyclical, (for example the related ones to action research).

And it also may be a good thing, solving one "problem" leads to another problem, but in the process of solving the first one we may win a lot of knowledge, and the same happens with the next one, and so on.

Two possible advantages of cyclical research are:

Knowing beforehand that the research will be cyclical, will allow us to estimate better the amount of time and money needed because we already know (more or less) where to aim.

It also may lead to a better end product, as we already know that we must focus in solving one thing and then we can focus in the next one.

Another possible advantage may be that we know that after the work, there will be a new thing to research, and it is fun, so if you are curious enough this may be a good thing (especially in scientific areas, physics, chemistry, etc).

7 0

3 years ago

You are holding a shopping basket at the grocery store with two 0.55-kg cartons of cereal at the left end of the basket. the bas

stepan [7]

Refer to the diagram shown below.

The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.

The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N

The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N

The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.

For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m

Answer: 0.63 m from the left end.

5 0

3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

$P_{2} > P_{1}$ .

6 0

3 years ago