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2 years ago

HOW DOES A GRAPH HELP YOU SEE ANOMALOUS DATA?

Physics

1 answer:

Ghella [55]2 years ago

8 0

Answer:

Anomalies consist of one or more modifications, insertions or deletions. As was described in Section 3.1, there are only three types of changes that can be made to a graph. Therefore, anomalies that consist of structural changes to a graph must consist of one of these types. Assumption 4.

Explanation:

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What do you need to know to be able to determine how far a projectile travels horizontally?

Effectus [21]

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D. The horizontal velocity of the projectile and <span>B. The length of time before it lands
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3 years ago

NEED ANSWER ASAP WILL MARK BRAINLIEST

lukranit [14]

Speed=30 m/s - 1.5 m/s = 28.5 m/s forward

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3 years ago

A basketball with a mass of 0.5 kilograms is accelerated at 2

Paul [167]

Answer: 1N

Explanation: its not 0N.

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2 years ago

If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g

EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408 × 10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M_1, M_2$ are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

$\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}$

$\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2$

Since $M_2 = 2m_2$ and $r = R/3$

$\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5$

So gravity would have been decreased by a factor of 4.5

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3 years ago

A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis thr

Papessa [141]

Answer:

The moment of inertia about an axis through the center and perpendicular to the plane of the square is

$I_s = \frac{Ma^2}{3}$

Explanation:

From the question we are told that

The length of one side of the square is $a$

The total mass of the square is $M$

Generally the mass of one size of the square is mathematically evaluated as

$m_1 = \frac{M}{4}$

Generally the moment of inertia of one side of the square is mathematically represented as

$I_g = \frac{1}{12} * m_1 * a^2$

Generally given that $m_1 = m_2 = m_3 = m_4 = m$ it means that this moment inertia evaluated above apply to every side of the square

Now substituting for $m_1$

$I _g= \frac{1}{12} * \frac{M}{4} * a^2$

Now according to parallel-axis theorem the moment of inertia of one side of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

$I_a = I_g + m [\frac{q}{2} ]^2$

=> $I_a = I_g + {\frac{M}{4} }* [\frac{q}{2} ]^2$

substituting for $I_g$

=> $I_a = \frac{1}{12} * \frac{M}{4} * a^2 + {\frac{M}{4} }* [\frac{q}{2} ]^2$

=> $I_a = \frac{Ma^2}{48} + \frac{Ma^2}{16}$

=> $I_a = \frac{Ma^2}{12}$

Generally the moment of inertia of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

$I_s = 4 * I_a$

=> $I_s = 4 * \frac{Ma^2}{12}$

=> $I_s = \frac{Ma^2}{3}$

8 0

3 years ago