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3 years ago

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Newton's First Law of Motion states that an object will remain at rest or in uniform motion in a straight line unless acted upon

by an external force change the motion. This is sometimes called the law of

1 answer:

atroni [7]3 years ago

3 0

This is sometimes called the Law of Inertia

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Which of the following will help you stay motivated to achieve your goals?

Dimas [21]

Answer:

C. Post your goals in an obvious place

Explanation:

Being able to look at your goals and check in on the progress of those goals in a place you know where they are so there's no worrying if you might have forgotten one or two and or havent done them is a good way to stay active and keep motivated with your goals. Making notes or specifications on those goals and trying to think over goals is also something you should do. Ignoring or forgetting your goals because you didnt write them down or put them in a secure place can ruin motivation and make you tired cause keeping track of those goals become a chore.

7 0

2 years ago

025 10.0 points

mamaluj [8]

Answer:

$m_b=278.73\ kg$

Explanation:

<u>Center of Gravity </u>

It refers to a point where all the forces of gravity of a body make a zero total torque. To find the solution we use the fact that the net force acting on the system boat-man is in every moment equal to zero. It's assured by the first Newton’s law, the center of gravity is at rest or in uniform motion in both moments. From an external viewer's point of view, the center of gravity remains unchanged. The formula to compute it is shown below

$\displaystyle x_c=\frac{\sum x_im_i}{\sum m_i}$

Originally, the man sits on the stern of the boat. His weight is applied at a distance xm=4.9 m from the pier (assumed as x=0). The boat is assumed to have a uniformly distributed mass applied at its center, i.e. at xb = 4.9 / 2 = 2.45 m. The center of gravity is located originally at

$\displaystyle x_c=\frac{(4.9)(90.4)+(2.45)(m_b)}{90.4+m_b}$

$\displaystyle x_c=\frac{442.96+2.45m_b}{90.4+m_b}$

When the man walks to the prow, the boat moves x = 1.2 m from the pier, so its center is located at a distance

$x_b=1.2+2.45=3.65\ m$

The man is located at

$x_m=1.2\ m$

The center of gravity is computed now as

$\displaystyle x_c=\frac{(1.2)(90.4)+(3.65)(m_b)}{90.4+m_b}$

$\displaystyle x_c=\frac{108.48+3.65m_b}{90.4+m_b}$

Both centers of gravity are equal, thus

$\displaystyle \frac{442.96+2.45m_b}{90.4+m_b}=\frac{108.48+3.65m_b}{90.4+m_b}$

Simplifying

$442.96+2.45m_b=108.48+3.65m_b$

Rearranging

$1.2m_b=334.48$

Thus

$\boxed{m_b=278.73\ kg}$

6 0

3 years ago

A light-rail train going from one station to the next on a straight section of track accelerates from rest at 1.1 m/s^2 for 20s.

svet-max [94.6K]

Answer:

A) The distance between the stations is 1430m

B) The time it takes the train to go between the stations is 80s

Explanation:

First we will calculate the distance covered for the first 20s.

From one the equations of kinematics for linear motion

$S = ut + \frac{1}{2}at^{2} \\$

Where $S$ is distance traveled

$u$ is the initial velocity

$t$ is time

and $a$ is acceleration

Since the train starts from rest, $u$ = 0 m/s

Hence, for the first 20s

$a =$ 1.1 m/s²; $t =$ 20s, $u$ = 0 m/s

∴ $S = ut + \frac{1}{2}at^{2} \\$ gives

$S = (0)(20) + \frac{1}{2}(1.1)(20)^{2}$

$S = \frac{1}{2}(1.1)(20)^{2}$

$S =$ 220m

This is the distance covered in the first 20s.

The train then proceeds at constant speed for 1100m.

Now, we will calculate the speed attained here

From

$v = u +at$

Where $v$ is the final velocity

Hence,

$v = 0 + 1.1(20)$

$v = 1.1(20)$

$v =$ 22 m/s

This is the constant speed attained when it proceeds for 1100m

The train then slows down at a rate of 2.2 m/s² until it stops

We can calculate the distance covered while slowing down from

$v^{2} = u^{2} + 2as$

The initial velocity, $u$ here will be the final velocity before it started slowing down

∴ $u$ = 22 m/s

The final velocity will be 0, since it came to a stop.

∴ $v$ = 0 m/s

$a = -$ 2.2 m/s² ( - indicates deceleration)

Hence,

$v^{2} = u^{2} + 2as$ gives

$0^{2} =22^{2} +2(-2.2)s$

$0=22^{2} - (4.4)s\\4.4s = 484\\s = \frac{484}{4.4} \\s = 110m$

This is the distance traveled while slowing down.

A) The distance between the stations is

220m + 1100m + 110m

= 1430m

Hence, the distance between the stations is 1430m

B) The time it takes the train to go between the stations

The time spent while accelerating at 1.1 m/s² is 20s

We will calculate the time spent when it proceeds at a constant speed of 22 m/s for 1100m,

From,

$Speed =\frac{Distance}{Time}\\$

Then,

$Time = \frac{Distance}{Speed}$

$Time = \frac{1100}{22}$

Time = 50 s

And then, the time spent while decelerating (that is, while slowing down)

From,

$v = u + at\\0 = 22 +(-2.2)t\\2.2t = 22\\t = \frac{22}{2.2} \\t= 10 s$

This is the time spent while slowing down until it stops at the station.

Hence, The time it takes the train to go between the stations is

20s + 50s + 10s = 80s

The time it takes the train to go between the stations is 80s

6 0

3 years ago

In which direction does the gravity of the earth act?

jolli1 [7]

Answer:

Satellites don't fall from the sky because they are orbiting Earth. Even when satellites are thousands of miles away, Earth's gravity still tugs on them. Gravity--combined with the satellite's momentum from its launch into space--cause the satellite go into orbit above Earth, instead of falling back down to the groundv

Explanation:

5 0

3 years ago

What 3 things can you learn about energy in a roller coaster?

Alika [10]

<u>Answer:</u>

The following are the things we can learn about energy in roller coaster.

All roller coasters work using the principle of gravitational energy.
All roller coaster utilise the Potential energy.
Roller coasters convert the potential energy into kinetic energy.
Friction is reduced in roller coasters as it decreases the speed of roller coaster.

Roller coasters do not have engines whereas Gravity pulls them and potential energy gets converted into kinetic energy providing its motion. Due to the friction and air resistance, the energy gets reduced as the ride proceeds.

4 0

3 years ago