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4 years ago

A base is a substance that's the chemical opposite of an acid. True or false

Physics

1 answer:

anygoal [31]4 years ago

6 0

True

Explanation:

A base is a substance that is often used as the chemical opposite of an acid.

Both behaves in opposite way to one another.

They can be said to complementary or conjugate chemicals.

According to Bronsted-Lowry theory, an acid is a proton donor and a base is a proton acceptor.
The lewis theory states that an acid is an electron pair acceptor while a base is an electron pair donor.
Acids turn blue litmus paper red and bases turns red litmus paper blue.

learn more :

Acid brainly.com/question/11062486

#learnwithbrainly

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a new planet is found with a density one third as much at earth and a radius twice that of earth. what is the acceleration due t

IrinaVladis [17]

The acceleration due to gravity on the new planet would be one third that of Earth's, or about 9.8 m/s^2.

What is planet?
This question might seem to have an easy solution, but it doesn't. The planets Earth, Mars, and Jupiter are well known. But before recent discoveries sparked a passionate debate about how to best characterise them, both Pluto and Ceres have been thought of as planets. This debate is still going strong today. The International Astronomical Union adopted the most recent definition of the a planet in 2006.

It specifies three requirements for a planet:

-It must move about a star (in our cosmic neighborhood, the Sun).
-It must be large enough for gravity to pull it in a spherical direction.
-It must have been large enough that any nearby objects of the a similar size were removed by its gravitational pull.

Given that the mass of the new planet is one third that of Earth's and its radius is twice that of Earth's, we can say that its volume is eight times that of Earth's. This means that its average density is one third that of Earth. Therefore, the acceleration due to gravity on the new planet would be one third that of Earth's, or about 9.8 m/s^2.

To learn more about planet
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1 year ago

A container of hydrogen gas has a volume of 2.3m³. What is the mass of hydrogen contained? The density of hydrogen = 90kg/m³.

umka2103 [35]

Answer:

container of hydrogen gas has a volume of 2.3m³. What is the mass of hydrogen

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3 years ago

It is possible a tree could fall down, even if no one did anything to it. A. True B. False

hichkok12 [17]

Answer:

This is TRUE

Explanation:

The tree can fall down even though no one did anything to it...

A hard breeze can blow and it can fall down or something else can cause it to topple...

Not only humans can make trees topple over...

PLEASE DO MARK ME AS BRAINLIEST UWU 

Bonne journée ;) 

8 0

3 years ago

Suppose a soccer player kicks the ball from a distance 29 m toward the goal. find the initial speed of the ball if it just passe

Rudik [331]

The ball's vertical velocity at the time it just passes over the goal is 0 m/s. Its initial vertical velocity is unknown and we denote it by $v\sin39^\circ$ , where $v$ here is the ball's initial speed. Vertically, the only force acting on the ball is gravity, which attributes a downward acceleration of 9.8 m/s^2. We expect the maximum height achieved by the ball to be 2.4 m, so we can find the initial speed by solving

$\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(v\sin39^\circ\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm m)$

$\implies v=11\,\dfrac{\mathrm m}{\mathrm s}$

6 0

3 years ago

A) In the figure below, a cylinder is compressed by means of a wedge against an elastic constant spring = 12 /. If = 500 , deter

Radda [10]

Explanation:

A) Draw free body diagrams of both blocks.

Force P is pushing right on block A, which will cause it to move right along the incline. Therefore, friction forces will oppose the motion and point to the left.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force N pushing up and left 10° from the vertical,

Friction force Nμ pushing down and left 10° from the horizontal,

Reaction force Fab pushing down,

and friction force Fab μ pushing left.

There are 2 forces acting on block B:

Reaction force Fab pushing up,

And elastic force kx pushing down.

(There are also horizontal forces on B, but I am ignoring them.)

Sum of forces on A in the x direction:

∑F = ma

P − N sin 10° − Nμ cos 10° − Fab μ = 0

Solve for N:

P − Fab μ = N sin 10° + Nμ cos 10°

P − Fab μ = N (sin 10° + μ cos 10°)

N = (P − Fab μ) / (sin 10° + μ cos 10°)

Sum of forces on A in the y direction:

N cos 10° − Nμ sin 10° − Fab = 0

Solve for N:

N cos 10° − Nμ sin 10° = Fab

N (cos 10° − μ sin 10°) = Fab

N = Fab / (cos 10° − μ sin 10°)

Set the expressions equal:

(P − Fab μ) / (sin 10° + μ cos 10°) = Fab / (cos 10° − μ sin 10°)

Cross multiply:

(P − Fab μ) (cos 10° − μ sin 10°) = Fab (sin 10° + μ cos 10°)

Distribute and solve for Fab:

P (cos 10° − μ sin 10°) − Fab (μ cos 10° − μ² sin 10°) = Fab (sin 10° + μ cos 10°)

P (cos 10° − μ sin 10°) = Fab (sin 10° + 2μ cos 10° − μ² sin 10°)

Fab = P (cos 10° − μ sin 10°) / (sin 10° + 2μ cos 10° − μ² sin 10°)

Sum of forces on B in the y direction:

∑F = ma

Fab − kx = 0

kx = Fab

x = Fab / k

x = P (cos 10° − μ sin 10°) / (k (sin 10° + 2μ cos 10° − μ² sin 10°))

Plug in values and solve.

x = 500 N (cos 10° − 0.4 sin 10°) / (12000 (sin 10° + 0.8 cos 10° − 0.16 sin 10°))

x = 0.0408 m

x = 4.08 cm

B) Draw free body diagrams of both blocks.

Force P is pushing block A to the right relative to the ground C, so friction force points to the left.

Block A moves right relative to block B, so friction force on A will point left. Block B moves left relative to block A, so friction force on B will point right (opposite and equal).

Block B moves up relative to the wall D, so friction force on B will point down.

There are 5 forces acting on block A:

Applied force P pushing to the right,

Normal force Fc pushing up,

Friction force Fc μ₁ pushing left,

Reaction force Fab pushing down and left 15° from the vertical,

and friction force Fab μ₂ pushing up and left 15° from the horizontal.

There are 5 forces acting on block B:

Weight force 750 n pushing down,

Normal force Fd pushing left,

Friction force Fd μ₁ pushing down,

Reaction force Fab pushing up and right 15° from the vertical,

and friction force Fab μ₂ pushing down and right 15° from the horizontal.

Sum of forces on B in the x direction:

∑F = ma

Fab μ₂ cos 15° + Fab sin 10° − Fd = 0

Fd = Fab μ₂ cos 15° + Fab sin 15°

Sum of forces on B in the y direction:

∑F = ma

-Fab μ₂ sin 15° + Fab cos 10° − 750 − Fd μ₁ = 0

Fd μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Substitute:

(Fab μ₂ cos 15° + Fab sin 15°) μ₁ = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab μ₁ μ₂ cos 15° + Fab μ₁ sin 15° = -Fab μ₂ sin 15° + Fab cos 15° − 750

Fab (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°) = -750

Fab = -750 / (μ₁ μ₂ cos 15° + μ₁ sin 15° + μ₂ sin 15° − cos 15°)

Sum of forces on A in the y direction:

∑F = ma

Fc + Fab μ₂ sin 15° − Fab cos 15° = 0

Fc = Fab cos 15° − Fab μ₂ sin 15°

Sum of forces on A in the x direction:

∑F = ma

P − Fab sin 15° − Fab μ₂ cos 15° − Fc μ₁ = 0

P = Fab sin 15° + Fab μ₂ cos 15° + Fc μ₁

Substitute:

P = Fab sin 15° + Fab μ₂ cos 15° + (Fab cos 15° − Fab μ₂ sin 15°) μ₁

P = Fab sin 15° + Fab μ₂ cos 15° + Fab μ₁ cos 15° − Fab μ₁ μ₂ sin 15°

P = Fab (sin 15° + (μ₁ + μ₂) cos 15° − μ₁ μ₂ sin 15°)

First, find Fab using the given values.

Fab = -750 / (0.25 × 0.5 cos 15° + 0.25 sin 15° + 0.5 sin 15° − cos 15°)

Fab = 1151.9 N

Now, find P.

P = 1151.9 N (sin 15° + (0.25 + 0.5) cos 15° − 0.25 × 0.5 sin 15°)

P = 1095.4 N

6 0

3 years ago