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1 year ago

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A train goes up a hill with a 15º incline. If the train has constant speed of 22 m/s, what are the vertical and horizontal compo

nents of the train's velocity? (Remember to use the law of sine and cosines for this problem)

1 answer:

Kobotan [32]1 year ago

8 0

Any two-dimensional vector in cartesian (x,y) coordinates can be broken down into individual horizontal and vertical components using trigonometry. If a train goes up a hill with 15 degree incline at a speed of 22 m/s, the horizontal component is 22cos(15)=21.3 m/s and the vertical component is 22sin(15)=5.5 m/s.

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A two-resistor voltage divider employing a 2-k? and a 3-k? resistor is connected to a 5-V ground-referenced power supply to prov

vesna_86 [32]

Answer:

circuit sketched in first attached image.

Second attached image is for calculating the equivalent output resistance

Explanation:

For calculating the output voltage with regarding the first image.

$Vout = Vin \frac{R_{2}}{R_{2}+R_{1}}$

$Vout = 5 \frac{2000}{5000}[/[tex][tex]Vout = 5 \frac{2000}{5000}\\Vout = 5 \frac{2}{5} = 2 V$

For the calculus of the equivalent output resistance we apply thevenin, the voltage source is short and current sources are open circuit, resulting in the second image.

so.

$R_{out} = R_{2} || R_{1}\\R_{out} = 2000||3000 = \frac{2000*3000}{2000+3000} = 1200$

Taking into account the %5 tolerance, with the minimal bound for Voltage and resistance.

if the -5% is applied to both resistors the Voltage is still 5V because the quotient has 5% / 5% so it cancels. to be more logic it applies the 5% just to one resistor, the resistor in this case we choose 2k but the essential is to show that the resistors usually don't have the same value. applying to the 2k resistor we have:

$Vout = 5 \frac{1900}{4900}\\Vout = 5 \frac{19}{49} = 1.93 V$

$Vout = 5 \frac{2100}{5100}\\Vout = 5 \frac{21}{51} = 2.05 V$

$R_{out} = R_{2} || R_{1}\\R_{out} = 1900||2850= \frac{1900*2850}{1900+2850} = 1140$

$R_{out} = R_{2} || R_{1}\\R_{out} = 2100||3150 = \frac{2100*3150 }{2100+3150 } = 1260$

so.

$V_{out} = {1.93,2.05}V\\R_{1} = {1900,2100}\\R_{2} = {2850,3150}\\R_{out} = {1140,1260}$

4 0

2 years ago

A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of wa

gavmur [86]

Answer : The correct option is, (d) $535^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of copper = $0.10cal/g^oC$

$c_2$ = specific heat of water = $1.00cal/g^oC$

$m_1$ = mass of copper = 120 g

$m_2$ = mass of water = 300 g

$T_f$ = final temperature of mixture = $35^oC$

$T_1$ = initial temperature of copper = ?

$T_2$ = initial temperature of water = $15^oC$

Now put all the given values in the above formula, we get:

$120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC$

$T_1=535^oC$

Therefore, the temperature of the kiln was, $535^oC$

7 0

2 years ago

Read 2 more answers

A 10N force pulls to the right and friction opposes 2N. If the object is 20kg,find the acceleraton.

zmey [24]

Force = mass * acceleration

10 N - 2 N = 20 kg * acceleration

8 N = 20 kg * acceleration

8 / 20 = acceleration

2/5 m/s^2 = acceleration

8 0

1 year ago

Read 2 more answers

A photon is scattered from an initially stationary electron within a metal. How does the frequency of the photon change upon sca

sammy [17]

Answer:

The frequency of the photon decreases upon scattering

Explanation:

Here we note that when a photon is scattered by a charged particle, it is referred to as Compton scattering.

Compton scattering results in a reduction of the energy of the photon and hence an increase in the wavelength (from λ to λ') of the photon known as Compton effect.

Therefore, since the wavelength increases, we have from

λf = λ'f' = c

f = c/λ

Where:

f and f' = The frequency of the motion of the photon before and after the scattering

c = Speed of light (constant)

We have that the frequency, f, is inversely proportional to the wavelength, λ as follows;

f = c/λ

As λ = increases, and c is constant, f decreases, therefore, the frequency of the photon decreases upon scattering.

5 0

2 years ago

Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another

ioda

Answer:

$6.1875\times 10^{-8}$

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies $I=\frac {k}{r^{2}}$ where r is the distance

Since intensity is given then when at 3 m

$1.1\times 10^{-7}= \frac {k}{3^{2}}$

$k=3^{2}\times 1.1\times 10^{-7}= 9.9\times 10^{-7}$

Since we have the constant then at 4m

Intensity, $I= \frac {9.9\times 10^{-7}}{4^{2}}=6.1875\times 10^{-8}$

8 0

1 year ago