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3 years ago

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Brad is working on a speed problem in physics class. The problem tells him that a girl runs from her house to the park 0.05 km a

way in 10 s. Brad calculates that her speed is 0.005 m/s. Is he correct? If not, explain the flaw or flaws in his problem solving process.

2 answers:

Dmitriy789 [7]3 years ago

7 0

It is corecct becasue it 10 s more added from 0.05

Olegator [25]3 years ago

3 0

Answer:

He is incorrect! Her speed was 5m/s.

Explanation:

For calculating the speed, first we shall remember that:

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

Where $v$ is the speed, $d$ is the distance travelled and, $t$ is the time it takes to travel distance $d$ .

So one migth think that velocity can be easely compute:

$v=\dfrac{0.05}{10}$

$v=0.005\dfrac{m}{s}$

Be carefull, he does not make a proper dimensional analisis!

Before computing the speed we must know in what dimensions our values are.

$d=0.05km$ , distances is measure in Kilometers.

$t=10s$ , time is measure in seconds.

If we want our speed to be in $m/s$ , first we need to be sure that our values are expressed in meters and seconds.

Time is already expressed in seconds, distance is not in Kilometers.

So

$0.05Km=50m$ ,

now we can compute the speed:

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

$v=\dfrac{50m}{10s}$

$v=\5dfrac{m}{s}$

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The pressure of a gas is reduced from 1200.0 mm hg to 850.0 mm hg as the volume of its container is increased by moving a piston

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From gas laws:
$\frac{PV}{T} = Constant$

Therefore,
$\frac{ P_{1} V_{1} }{ T_{1} } = \frac{ P_{2} V_{2} }{ T_{2} }$

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3 years ago

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Which of the following statements are true at some time during the course of the motion? Check all that apply. Check all that ap

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Answer:

The object can have zero velocity and, simultaneously, nonzero acceleration.

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

Explanation:

An object in simple harmonic motion has a total mechanical energy (sum of elastic potential energy and kinetic energy) that is constant:

E=U+K=1/2kx^2 + 1/2}mv^2

where,

k is equal to the spring constant

x is equal to the displacement

m is the mass

v is the speed

We can note that the force on the spring is given by Hook's law:

F=-kx

In Newton's law F = ma, this can be also be written as

ma=-kx

a=-k/mx

This implies that the acceleration is proportional to the displacement.

From the first equation, we can now states that:

When the displacement is zero, x=0, the acceleration is zero, a=0, and the velocity is maximum

When the velocity is zero, v=0, the acceleration is maximum, which occurs when the displacement is maximum

In all the other intermediate situations, both velocity and acceleration are nonzero.

So the correct answers are

The object can have zero acceleration and, simultaneously, nonzero velocity.

The object can have nonzero velocity and nonzero acceleration simultaneously.

The object can have zero velocity and, simultaneously, nonzero acceleration.

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3 years ago

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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv

cestrela7 [59]

<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":

$P=\sigma A T^{4}$ (1)

Where:

$P=300J/min=5J/s=5W$ is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing $1W=\frac{1Joule}{second}=1\frac{J}{s}$

$\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

$A=5cm^{2}=0.0005m^{2}$ is the Surface area of the body

$T=727\°C=1000.15K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

$P=\sigma A \epsilon T^{4}$ (2)

Where $\epsilon$ is the body's emissivity

(the value we want to find)

Isolating $\epsilon$ from (2):

$\epsilon=\frac{P}{\sigma A T^{4}}$ (3)

Solving:

$\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}}$ (4)

Finally:

$\epsilon=0.17$ (5) This is the body's emissivity

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3 years ago