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3 years ago

If Brandy ran for 3 seconds and ended up 6 meters away, what was her speed?

Physics

2 answers:

aliya0001 [1]3 years ago

8 0

2 meters per second.

Zarrin [17]3 years ago

6 0

Distance Traveled/Time= Speed
6/3= 2m/s
I hope that helped. Remember the formula for next time.

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Why are galaxies visible

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Because a galaxy is a large collection of many stars, and almost every star radiates some visible light.

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3 years ago

Suppose earth's mass increased but earth's diameter didn't change. Describe how the gravitational force between Earth and the ob

zheka24 [161]

Answer:

It increases proportionally

Explanation:

The gravitational force between the Earth and an object on its surface is given by

$F=G\frac{Mm}{R^2}$

where

G is the gravitational constant

M is the Earth's mass

m is the mass of the object

R is the Earth's radius

In this problem, the Earth's mass is increased, while the diameter (and therefore, the radius) doesn't change. From the equation, we see that the gravitational force is directly proportional to the Earth's mass: therefore, if the mass is increased, the force will increase as well by the same proportion (for example, if the mass is doubled, the force will double as well)

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3 years ago

A private aviation helicopter's main rotor blades rotate at approximately

Arisa [49]

Answer: 7.5 rev/s

Explanation:

We are given the angular velocity $\omega$ a helicopter's main rotor blades:

$\omega=450 rpm=450 \frac{rev}{min}$

However, we are asked to express this $\omega$ in the International Systrm (SI) units. In this sense, the SI unit for time is second ( $s$ ):

$\omega=450 \frac{rev}{min} \frac{1 min}{60 s}$

$\omega=7.5 \frac{rev}{s}$

4 0

3 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )