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mrs_skeptik [129]

3 years ago

13

Arthur and Betty start walking toward each other when they are 100 m apart. Arthur has a speed of 3.0 m/s and Betty has a speed

of 2.0 m/s. Their dog, Spot, starts by Arthur's side at the same time and runs back and forth between them at 5.0 m/s. By the time Arthur and Betty meet, what distance has Spot run?

2 answers:

elixir [45]3 years ago

6 0

Let t be the time they both meet. When they meet they both have covered the distance between them.

Arthur with 3 m/s would have covered 3t m
Betty with 2 m/s would have covered 2t m

3t + 2t = 100
5t = 100
t = 100/5
t = 20 seconds.

Spot who is helplessly runing back and forth at 5 m/s would have done this for 20 seconds.

This distance covered by spot = 5m/s * 20s = 100m

solniwko [45]3 years ago

5 0

The distance Spot has traveled is 100 m

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

distance = d = 100 m

Arthur's speed = v₁ = 3.0 m/s

Betty's speed = v₂ = 2.0 m/s

Spot's speed = v = 5.0 m/s

<u>Unknown:</u>

Spot's distance = s = ?

<u>Solution:</u>

At first we will find time when the two of them meet.

$\text {Arthur's distance + Betty's Distance} = 100 ~ m$

$v_1t + v_2t = 100$

$3t + 2t = 100$

$5t = 100$

$t = 100 \div 5$

$\boxed {t = 20 ~ seconds}$

Because Spot runs back and forth until these two people meet, then:

$\text {Spot's Distance} = vt$

$\text {Spot's Distance} = 5(20)$

$\large {\boxed {\text {Spot's Distance} = 100 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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Explanation:

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The answer would be Newton’s Second Law

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The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

The given parameters;

<em>distance between the two black holes, r = 10 AU = 1.5 x 10¹² m</em>
<em>gravitational force between the two black holes, F = 6.9 x 10²⁵ N.</em>
<em>combined mass of the two black holes = 5.20 x 10³⁰ kg</em>

The product of the two masses is calculated from Newton's law of universal gravitational as follows;

$F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2$

The sum of the two masses is given as;

m₁ + m₂ = 5.2 x 10³⁰ kg

m₂ = 5.2 x 10³⁰ kg - m₁

The first mass is calculated as follows;

m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰

5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰

m₁² - 5.2 x 10³⁰m₁ + 2.328 x 10⁶⁰ = 0

<em>solve the quadratic equation using formula method</em>;

a = 1, b =- 5.2 x 10³⁰, c = 2.328 x 10⁶⁰

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m₂ = 5.2 x 10³⁰ kg - 4.9 x 10²⁹ kg

m₂ = 4.7 x 10³⁰ kg

Thus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.

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Answers: (1) a. weight, (2)b. Force changes by 2/9, (3)b. movement, (4)a. 40,000 Joules, (5)c. the soil will be 5°C.

<h2>Answer 1: a. weight</h2>

Mass and weight are very different concepts.

Mass is the amount of matter that exists in a body, which only depends on the quantity and type of particles within it. This means mass is an intrinsic property of each body and remains the same regardless of where the body is located.

On the other hand, weight is a measure of the gravitational force acting on an object and is directly proportional to the product of the mass $m$ of the body by the acceleration of gravity $g$ :

$W=m.g$

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<h2>Answer 2: b. Force changes by 2/9</h2>

According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:

$F=G\frac{m_{1}m_{2}}{r^2}$ (1)

Where:

$F$ is the module of the force exerted between both bodies

$G$ is the universal gravitation constant

$m_{1}$ and $m_{2}$ are the masses of both bodies.

$r$ is the distance between both bodies

If we double the mass of one object (for example $2m_{1}$ ) and triple the distance between both (for example $3r$ ). The equation (1) will be rewritten as:

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$F=\frac{2}{9}G\frac{m_{1}m_{2}}{r^2}$ (3)

If we compare (1) and (2) we will be able to see the force changes by 2/9.

<h2>Answer 3: b. movement</h2>

The Work $W$ done by a Force $F$ refers to the release of potential energy from a body that is <u>moved</u> by the application of that force to overcome a resistance along a path.

When the applied force is constant and <u>the direction of the force and the direction of the movement are parallel,</u> the equation to calculate it is:

$W=(F)(d)$

Now, <u>when they are not parallel, both directions form an angle</u>, let's call it $\alpha$ . In that case the expression to calculate the Work is:

$W=Fdcos{\alpha}$

Therefore, pushing on a rock accomplishes no work unless there is movement (independently of the fact that movement is parallel to the applied force or not).

<h2>Answer 4: a. 40,000 Joules</h2>

The Kinetic Energy is given by:

$K=\frac{1}{2}mV^{2}$ (4)

Where $m$ is the mass of the body and $V$ its velocity

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$K_{1}=\frac{1}{2}mV_{1}^{2}$ (5)

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$m=\frac{2K_{1}}{V_{1}^{2}}$ (6)

$m=\frac{2(10000J)}{(13.4112m/s)^{2}}$ (7)

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For the second case (unknown kinetic energy $K_{2}$ for a car with the same mass at $V_{2}=60 mph=26.8224m/s$ ):

$K_{2}=\frac{1}{2}mV_{2}^{2}$ (9)

$K_{2}=\frac{1}{2}(111.197kg)(26.8224m/s)^{2}$ (10)

$K_{2}=40000J$ (11)

<h2>Answer 5: c. the soil will be 5°C</h2>

The formula to calculate the amount of calories $Q$ is:

$Q=m. c. \Delta T$ (12)

Where:

$m$ is the mass

$c$ is the specific heat of the element. For water is $c_{w}=1 kcal/g\°C$ and for soil is $c_{s}=0.20 kcal/g\°C$

$\Delta T$ is the variation in temperature (the amount we want to find for both elements)

This means we have to clear $\Delta T$ from (12) :

$\Delta T=\frac{Q}{m.c}$ (13)

For Water:

$\Delta T_{w}=\frac{Q_{w}}{m_{w}.c_{w}}$ (14)

$\Delta T_{w}=\frac{1kcal}{(1kg)(1 kcal/g\°C)}$ (15)

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For Soil:

$\Delta T_{s}=\frac{Q_{s}}{m_{s.c_{s}}$ (17)

$\Delta T_{s}=\frac{1kcal}{(1kg)(0.20 kcal/g\°C)}$ (18)

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Hence the correct option is c.

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So what we can do is apply the<span> Hooke's law wich states that
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