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3 years ago

(4. A bus is moving at 25 m/s when the driver steps on the brakes and brings the bus to a stop in

Physics

1 answer:

zlopas [31]3 years ago

6 0

Answer:

(a) the average acceleration of the bus while braking is 8.333 m/s

(b) if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a. [¹/₂ (8.333 m/s) = 4.16 s]

Explanation:

Given;

initial velocity of the bus, v = 25 m/s

time of the motion, t = 3 s

(a) the average acceleration of the bus while braking

a = dv/dt

where;

a is the bus acceleration

dv is change in velocity

dt is change in time

a = 25 / 3

a = 8.333 m/s

(b) If the bus took twice as long to stop, the duration = 2 x 3s

a = 25 / (2 x 3s)

a = ¹/₂ x (25 / 3)

a = ¹/₂ (8.333 m/s) = 4.16 s

Thus, if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a.

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If an object left earth which direction would it travel

IceJOKER [234]

Answer:

In a straight line

Explanation:

Why, because of the power of inerta

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3 years ago

Which of the following will cause an increase in gas pressure in a closed container?

Ludmilka [50]

C. Both A. and B.

Explanation:

Statement A. Reducing the volume is true because of Boyle's law, which states that for a gas at fixed temperature, the pressure p and the volume V are inversely proportional:

$pV=const.$

Therefore, when the volume V is reduced, the pressure p increases.

Statement B. Adding more gas is also true: in fact, if we add gas into the container, we will have more molecules of the gas hitting the wall of the container. But the pressure of a gas is exactly given by this: by the collision of the molecules against the wall of the container, so the more the molecules of gas, the greater the pressure.

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4 years ago

Read 2 more answers

Eli states that sodium phosphate is a mixture because it is composed of both sodium ions and phosphate ions.

galina1969 [7]

The question is incomplete, the complete question is;

Eli states that sodium phosphate is a mixture because it is composed of both sodium ions and phosphate ions. Which is the best analysis of Eli’s statement? It is correct because each ion is a pure substance, so sodium phosphate is made up of two pure substances. It is correct because the composition of sodium phosphate changes depending on the sample. It is incorrect because sodium phosphate is a compound that has a single composition. It is incorrect because the two types of ions in sodium phosphate cannot be seen.

Answer:

It is incorrect because sodium phosphate is a compound that has a single composition

Explanation:

A compound is a neutral substance made up of two or more atoms which are chemically combined together.

Ionic substances are made up of ions. These ions are not separate entities, they are part of the compound.

Hence, Eli's statement is incorrect because sodium phosphate is a compound that has a single composition.

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3 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .