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3 years ago

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If Team 4 wants to increase the horizontal distance that the skittles are projected through the air but keep the height as close

to the original trail as possible, How should the term revise the catapult design?

1 answer:

photoshop1234 [79]3 years ago

3 0

Answer:

The speed of the catapult arm needs to be increased.

Explanation:

If the height needs to stay the same, they can not change the length of the catapult arm because this is the variable that determines the height of the thrown object. To change the horizontal distance the object travels, in other words the object's speed, the rope or counterweigth that pulls the catapult arm can be changed to increase the speed that the object is thrown.

I hope this answer helps.

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Tamsen is interested in history, and read that because of its regular period, the pendulum constituted the basis of the most acc

geniusboy [140]

We will apply the concept of period in a pendulum, defined as the product between 2 $\pi$ by the square root of the length over gravity, this is mathematically

$T = 2\pi \sqrt{\frac{L}{g}}$

Here,

T = Period

L = Length

g = Acceleration due to gravity

For the period to be 1 second, then we must look for the necessary length for such a requirement so

$1 = 2\pi \sqrt{\frac{L}{9.8}}$

$(\frac{1}{2\pi})^2 = \frac{L}{9.8}$

$L = 9.8(\frac{1}{2\pi})^2$

$L = 0.2482m$

The meter's length would be slight less than one-fourth of its current length. Also, the number of significant digits depends only on how precisely we know g, because the time has been defined to be exactly 1s.

Therefore the correct answer is C.

3 0

3 years ago

The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.

S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

$\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }$

Where:

$\omega$ - Angular frequency, measured in radians per second.

$m$ - Mass of the physical pendulum, measured in kilograms.

$g$ - Gravitational constant, measured in meters per square second.

$d$ - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

$I_{O}$ - Moment of inertia with respect to pivot point, measured in $kg\cdot m^{2}$ .

In addition, frequency and angular frequency are both related by the following formula:

$\omega =2\pi\cdot f$

Where:

$f$ - Frequency, measured in hertz.

If $f = 0.658\,hz$ , then angular frequency of the physical pendulum is:

$\omega = 2\pi \cdot (0.658\,hz)$

$\omega = 4.134\,\frac{rad}{s}$

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

$\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}$

$I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}$

Given that $m = 1.15\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $d = 0.425\,m$ and $\omega = 4.134\,\frac{rad}{s}$ , the moment of inertia associated with the physical pendulum is:

$I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}$

$I_{o} = 0.280\,kg\cdot m^{2}$

The rotational inertia of the pendulum around its pivot point is $0.280\,kg\cdot m^{2}$ .

8 0

3 years ago

As the wave interacts with a wall, which kind of wave interaction is shown?

kow [346]

The answer would be a reflection. This is because, t<span>he color of an object is actually the wavelengths of the light reflected while all other wavelengths are absorbed. Color, in this case, refers to the different wavelengths of light in the </span>visible light spectrum<span>perceived by our eyes. The physical and chemical composition of matter determines which wavelength (or color) is reflected.</span>

5 0

3 years ago

Read 2 more answers

Simple diffusion and facilitated diffusion are related in that both

kaheart [24]

Answer:

allow the downward movement of the concentration gradient by passive transport

Explanation:

Passive transport is a process of substance transport, which is carried out spontaneously, without energy expenditure and in favor of the concentration gradient, that is, from a medium where the molecules are more concentrated towards a medium where their concentration is lower.

Three types of passive transport are distinguished: osmosis, simple diffusion and facilitated diffusion

<u>Simple diffusion</u>

It is the passage, through the plasma membrane, of small molecules without charge soluble in the lipid bilayer, such as some gases (oxygen and carbon dioxide). For a molecule to diffuse through the membrane it is necessary that there is a difference in concentration between the external and the internal environment.

<u>Diffusion facilitated </u>

There are molecules such as amino acids, glucose and small ions that, due to their chemical and size characteristics, cannot diffuse through the lipid bilayer and require transport proteins for diffusion.

The transport proteins are immersed in the plasma membrane and can be of two types: protein channels, formed by proteins that generate a channel in the membrane, and permeases, which are proteins that, when joined to the molecule to be transported, change their shape by carrying them into the cell.

4 0

3 years ago

Sirius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. An

Vikentia [17]

Answer: Sirius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Another bright star, Regulus, has a parallax of 0.042 arcseconds. Then, the distance in parsecs will be,23.46.

Explanation: To find the answer, we have to know more about the relation between the distance in parsecs and the parallax.

<h3>What is the relation between the distance in parsecs and the parallax?</h3>

Let's consider a star in the sky, is d parsec distance from the earth, and which has some parallax of P amount.
Then, the equation connecting parallax and the distance in parsec can be written as,

$d=\frac{1}{P}$

We can say that,

$dP=constant.\\thus,\\d_1P_1=d_2P_2$

<h3>How to solve the problem?</h3>

We have given that,

$d_1=2.6 parsecs.\\P_1=0.379arcseconds.\\P_2=0.042 arcseconds.\\d_2=?$

Thus, we can find the distance in parsecs as,

$d_2=\frac{d_1P_1}{P_2} =23.46 parsecs$

Thus, we can conclude that, the distance in parsecs will be, 23.46.

Learn more about the relation connecting distance in parsecs and the parallax here: brainly.com/question/28044776

#SPJ4

6 0

2 years ago