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3 years ago

If a data point is way off the trend line, what will not resolve the problem?

Physics

1 answer:

Elden [556K]3 years ago

4 0

Going out to a movie will not resolve the problem.

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Why do scientists believe that light is made of streams of particles?

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3 years ago

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10. A boy weighs 475 N. What is his mass? (acceleration due to gravity on Earth is 9.8m/s2 = g)

Darya [45]

Answer: mass = 48.47 kg.

Explanation:

Formula : Weight = mg , where m = mass of body , g= acceleration due to gravity .

Given: Weight = 475 N

$g= 9.8\ m/s^2$

Substitute all values in formula , we get

$475= m \times9.8\\\\\Rightarrow\ m = \dfrac{475}{9.8}\\\\\Rightarrow\ m \approx 48.47\ kg$

Hence, his mass = 48.47 kg.

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3 years ago

For the past few days, the weather in Miami has been cool. When the temperature did warm up, the meteorologists noticed that it

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3 years ago

A ball is dropped off the side of a bridge,<br> After 1.55 S, how far has it fallen?<br> (Unit=m)

kow [346]

Answer:

Distance S = 11.77 m (Approx.)

Explanation:

Given:

Time t = 1.55 Second

Gravity acceleration = 9.8 m/s²

Find:

Distance S

Computation:

S = ut + (1/2)(g)(t)²

S = (0)(1.55) + (1/2)(9.8)(1.55)²

Distance S = 11.77 m (Approx.)

7 0

3 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

8 0

3 years ago