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3 years ago

X+1 +1 =2x+1_____________x+6+x=x+6

1 answer:

6 0

I got no solution! but I believe the answer is zero.

I added like terms so

X+ 2=2x+1
—————
x^2 +6 =x +6

Then

X+1=2x
————
x^2+0=x

Then

3x=-1
———
x^3

So

-3
—
-X^2 = 1

So no solution :/

But if you do the equations separate the answer is 0 because
The top equation will equal 1 and the bottom will equal zero.

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what acceleration does the force of earth's gravity produceit varies depending on how long it has been falling.it changes direct

iren2701 [21]

If gravity is the ONLY force acting on an object, then the force of
Earth's gravity produces acceleration of 9.8 meters per second²
downward. That number doesn't change, and it doesn't depend
on the mass or weight of the object.

If there are any other forces acting on the object at the same time,
like air resistance or some other kind of friction for example, then
all bets are off, and we can't say what the object's acceleration will
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3 years ago

A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat

UNO [17]

<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

Force = Spring constant x Elongation

F = kx

Here force is weight of mass

F = W = mg = 0.54 x 9.81 = 5.3 N

Elongation, x = 36 cm = 0.36 m

Substituting

F = kx

5.3 = k x 0.36

k = 14.72 N/m

Spring constant is 14.72 N/m

6 0

3 years ago

Un objeto de 0.5kg de masa se desplaza a lo largo de una trayectoria rectilínea con aceleración constante de 0.3m/s2. Si partió

den301095 [7]

Answer:

p = 1.2 kg-m/s

Explanation:

The question is, "An object of mass 0.5kg is moving along a rectilinear path with constant acceleration of 0.3m / s2. If it started from rest and the magnitude of its momentum in kg * m / s after 8s is".

Mass of the object, m = 0.5 kg

Acceleration of the object, a = 0.3 m/s²

We need to find the momentum after 8 seconds.

We know that,

$p=F\times t$

i.e.

p = mat

So,

$p=0.5\times 0.3\times 8\\\\p=1.2\ kg-m/s$

So, the momentum of the object is 1.2 kg-m/s.

6 0

3 years ago

A student standing on a stationary skateboard tosses a textbook with a mass of mb = 1.25 kg to a friend standing in front of him

juin [17]

Answer:

The velocity of the student has after throwing the book is 0.0345 m/s.

Explanation:

Given that,

Mass of book =1.25 kg

Combined mass = 112 kg

Velocity of book = 3.61 m/s

Angle = 31°

We need to calculate the magnitude of the velocity of the student has after throwing the book

Using conservation of momentum along horizontal direction

$m_{b}v_{b}\cos\theta= m_{c}v_{c}$

$v_{s}=\dfrac{m_{b}v_{b}\cos\theta}{m_{c}}$

Put the value into the formula

$v_{c}=\dfrac{1.25\times3.61\times\cos31}{112}$

$v_{c}=0.0345\ m/s$

Hence, The velocity of the student has after throwing the book is 0.0345 m/s.

3 0

3 years ago

The part of a longitudinal wave where the particles are spread apart is called a______?

lilavasa [31]

Answer:

In a longitudinal wave, particles of the medium vibrate in a direction that is parallel to the direction that the wave travels. Places where particles of the medium crowd closer together are called compressions. Places where particles of the medium spread farther apart are called rarefactions.

Explanation:

3 0

3 years ago

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