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4 years ago

Im Stuck In This Thing Called ''Life" Can You Help Me.

Physics

2 answers:

lora16 [44]4 years ago

8 0

I'll be willing to help

pantera1 [17]4 years ago

4 0

im still trying to find someone who can help me

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The molecules of a substance become more closely packed and move more quickly.

andrey2020 [161]

When the substance are moved close together and they move more quickly they get compressed.

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3 years ago

A vector must always have both size and ..

DochEvi [55]

Both magnitude and DIRECTION
For example,
• 12m East
• -2 miles
•9 meter north
• 8 miles up

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3 years ago

Two carts have a compressed spring between them and are initially at rest. One of the carts has total mass, including its conten

ladessa [460]

Answer:

A) - 1.8 m/s

Explanation:

As we know that whole system is initially at rest and there is no external force on this system

So total momentum of the system must be conserved

so we will have

$m_1v_1 + m_2v_2 = 0$

now plug in all data into above equation

$5(v) + 3(3)$

$5v = -9$

$v = -1.8 m/s$

so correct answer is

A) - 1.8 m/s

3 0

4 years ago

You move a 2.5 kg book from a shelf that is 1.2 m above the ground to a shelf that is 2.6 m above the ground. What is the change

Sophie [7]

The change in potential energy of an object is given by
$U=mg \Delta h$
where
m is the mass of the object
g is the gravitational acceleration
$\delta h$ is the increase in altitude of the object

In our problem, $m=2.5 kg$ is the mass of the book, $g=9.81 m/s^2$ and
$\Delta h=2.6 m -1.2 m=1.4 m$ is the increase in altitude of the book, so its variation of potential energy is
$U=mg\Delta h=(2.5 kg)(9.81 m/s^2)(1.4 m)=34.3 J$

8 0

4 years ago

In a novel from 1866 the author describes a spaceship that is blasted out of a cannon with a speed of about 11.000 m/s. The spac

Elan Coil [88]

Answer:

$a=0.284\ m/s^2$

Explanation:

Given that,

Initially, the spaceship was at rest, u = 0

Final velocity of the spaceship, v = 11 m/s

Distance accelerated by the spaceship, d = 213 m

We need to find the acceleration experienced by the occupants of the spaceship during the launch. It is a concept based on the equation of kinematics. Using the third equation of motion to find acceleration.

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(11)^2-(0)^2}{2\times 213}\\\\a=0.284\ m/s^2$

So, the acceleration experienced by the occupants of the spaceship is $0.284\ m/s^2$ .

5 0

3 years ago