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2 years ago

At a sports event, 1, 045 spectators will be

Physics

2 answers:

LiRa [457]2 years ago

8 0

Explanation:

Total number of buses =9178181

devlian [24]2 years ago

8 0

85n >/= 1,045

85 people times n buses has to equal or exceed 1,045 people

85n is greater than or equal to 1,045

85n >/= 1, 045

n >/= 12.29

So, since n has to be greater than or equal to 12.29, and there is no 29/100ths of a bus, they would need 13 buses (in case there's a follow-up question)

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How many miles away is the sun to the earth?

geniusboy [140]

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3 years ago

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Describe two experiments to determine the speed of propagation of a transverse wave on a rope. You have the following tools to u

AnnZ [28]

Answer:

#See solution for details.

Explanation:

Tools: $stopwatch, \ meter \ stick, \ mass \ measuring \ scale , \ force \ measuring \ device$ .

$Experiment \ 1$ :Calculate the speed of the wave using the time, $t$ it takes to travel along the rope. Rope's length, $L$ is measured using the meter stick.

-Attach one end of rope to a wall or post, shake from the unfixed end to generate a pulse. Measure the the time it takes for the pulse to reach the wall once it starts traveling using the stopwatch.

-Speed of the pulse can then be obtained as:

$v=\frac{L}{t}$

$Experiment \ 2$ : Apply force of known value to the rope then use the following relation equation to find the speed of a pulse that travels on the rope.

$v=\sqrt{\frac{F}{\mu}}\ ,\mu=\frac{m}{L}$

-Use the measuring stick and measuring scale to determine $L,m$ values of the rope then obtain $\mu$ .

-Use the force measuring constant to determine $F$ . These values can the be substituted in $Experiment \ 1$ to obtain $v.$

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2 years ago

Which answer best explains why the contents of a warm can of carbonated drink might rush out of the can faster than the contents

Flura [38]

Answer:

its A

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3 years ago

In the Bronsted-Lowry Theory of acids and bases, an acid is this.

madreJ [45]

Answer: An acid is a substance that donates a proton and produces a conjugate base.

Explanation:

According to Bronsted-Lowry theory, an acid is a substance that donates a proton and produces a conjugate base while a base is a molecule or ion which accepts the proton.

An example of Bronsted-Lowry acid and base is Ethanoic acid, CH3COOH and hydroxide ion, OH- respectively as shown in the reaction below

CH3COOH(aq) + OH-(aq) <---> CH3COO-(aq) + H2O(l)

Thus, ethanoic acid acts as an acid by donating a proton to the hydroxide ion which accepts it, thus producing ethanoate ion, CH3COO- as a conjugate base.

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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:

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

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3 years ago