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2 years ago

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Compare a wave that has a period of 0.03 second with a second wave that has a period of 1⁄4 second. Which wave has the greater f

requency? Be sure to show the steps for your work.

2 answers:

yaroslaw [1]2 years ago

4 0

Answer:

The wave with 0.03 time period have the greater frequency

Explanation:

Given in the question,

Period T $_{1}$ of one wave = 0.03 seconds

Period T $_{2}$ of another wave = 1/4 seconds

We know that

T = 1 / f

So,

f $_{1}$ = 33.3 hz

f $_{2}$ = 4 hz

The wave with 0.03 time period have the greater frequency.

o-na [289]2 years ago

4 0

Answer:

The first wave has a frequency of 33.3 Hz:

f1 = 1⁄T1

f1 = 1⁄0.03

f1 = 33. 3 Hz

The second wave has a frequency of 4 Hz. f2 = 1⁄T2

f2 = 1⁄1⁄4

f2 = 1 ÷ 1⁄4

f2 = 1 × 4⁄1

f2 = 1⁄1 × 4⁄1

f2 = 4 Hz

Therefore, the first wave has a higher frequency.

Explanation:

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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

<u>Step 4:</u> calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

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2 years ago

Janice's mother often lets her 6-month-old baby sit in front of the television, watching episodes of Sesame Street. What is Jani

brilliants [131]

B.

The child is too old to be gaining something from the screen time.

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2 years ago

Person lifting a chair convert from what energy to another

vagabundo [1.1K]

A person lifting a chair is converting chemical energy to mechanical energy.

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2 years ago

Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe

matrenka [14]

<span> (a) does an electric field exert a force on a stationary charged object?
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span> $F=qE$
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with charge q and speed v is
</span> $F=qvB \sin \theta$
where $\theta$ is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object?
Yes, The intensity of the electric force is still
</span> $F=qE$
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
$F=qvB \sin \theta$
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so?
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span> $F=qE$
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span> $F=qvB \sin \theta$
<span>is different from zero because v is different from zero.</span>

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2 years ago

A weather balloon is rising vertically from a launching pad on the ground. A technician standing 300 feet from the launching pad

avanturin [10]

Answer:

$\dfrac{dh}{dt} =5\ ft/s$

Explanation:

Let

h = height of balloon (in feet).

θ = angle made with line of sight and ground (in radians).

h = 300 tanθ

$\dfrac{dh}{d\theta } = 300 sec^2\theta$

now $\dfrac{dh}{dt}$ can be written as

$\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}$

$\dfrac{d\theta }{dt} = \dfrac{1}{120}\at \ \theta =\dfrac{\pi}{4}$

When θ = π/4,

$\dfrac{dh}{d\theta } = 300 sec^2\theta$

$\dfrac{dh}{d\theta } = 600$

$\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}$

$\dfrac{dh}{dt} =600\times \dfrac{1}{120}$

$\dfrac{dh}{dt} =5\ ft/s$

5 0

2 years ago