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3 years ago

If a town was 90 miles away and you travel at 45 mph how long would it take to get there

Physics

1 answer:

notka56 [123]3 years ago

5 0

Answer:

t = 2 hours

Explanation:

Given that,

Distance of the town, d = 90 miles

Speed, v = 45 mph

We need to find the time to get there. The speed of an object is given by :

$v=\dfrac{d}{t}$

Where

t is time

$t=\dfrac{d}{v}\\\\t=\dfrac{90}{45}\\\\t=2\ h$

So, the required time is 2 hours.

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When a neutrally charged atom loses an electron to another atom, the result is the creation of

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That would make a cation. If it gained one instead, then it would make an anion.

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3 years ago

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more t

olga2289 [7]

The force result in stretching the spring 10.0 centimeters is 2.5N.

<h3>What is Hooke's law?</h3>

If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.

F = kx

where k is the proportionality constant called the spring constant or force constant.

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.

Let the spring constant be very low 0.04N/m

The force applied is

F = 10 cm / 0.04

F = 0.1 m / 0.04

F = 2.5 N

Thus, the force result in stretching the spring 10cm is 2.5 N.

Learn more about hooke's law.

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2 years ago

A thin, uniform rod is bent into a square of side length a. If the total mass is M, find the moment of inertia about an axis thr

Papessa [141]

Answer:

The moment of inertia about an axis through the center and perpendicular to the plane of the square is

$I_s = \frac{Ma^2}{3}$

Explanation:

From the question we are told that

The length of one side of the square is $a$

The total mass of the square is $M$

Generally the mass of one size of the square is mathematically evaluated as

$m_1 = \frac{M}{4}$

Generally the moment of inertia of one side of the square is mathematically represented as

$I_g = \frac{1}{12} * m_1 * a^2$

Generally given that $m_1 = m_2 = m_3 = m_4 = m$ it means that this moment inertia evaluated above apply to every side of the square

Now substituting for $m_1$

$I _g= \frac{1}{12} * \frac{M}{4} * a^2$

Now according to parallel-axis theorem the moment of inertia of one side of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

$I_a = I_g + m [\frac{q}{2} ]^2$

=> $I_a = I_g + {\frac{M}{4} }* [\frac{q}{2} ]^2$

substituting for $I_g$

=> $I_a = \frac{1}{12} * \frac{M}{4} * a^2 + {\frac{M}{4} }* [\frac{q}{2} ]^2$

=> $I_a = \frac{Ma^2}{48} + \frac{Ma^2}{16}$

=> $I_a = \frac{Ma^2}{12}$

Generally the moment of inertia of the square about an axis through the center and perpendicular to the plane of the square is mathematically represented as

$I_s = 4 * I_a$