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3 years ago

10

The highest free-standing tower in North America is the CN Tower in Toronto, Canada. From a distance of 1km from its base, the a

ngle of elevation to the top of the tower is 28.81 degrees. Find the height of the tower, rounded to two decimal places.

1 answer:

Whitepunk [10]3 years ago

4 0

The height of the tower and the distance from its base form a right angle triangle. Thus,
tan(∅) = height / base
tan(28.81) = height / 1,000 m
height = 549.98 meters

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Explain why most foods and beverages<br> are mixtures.

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Answer:

Most food and drinks are mixtures as they are not chemically bonded and also beacuse they are made up of two or more substances

Explanation:

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2 years ago

Photovoltaic cells: a) have become more economical to produce and use over the past 25 years. b) are the most efficient means of

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Photovoltaic cells are the most efficient means of converting solar energy to electricity. Option b is correct.

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2 years ago

A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary

Snezhnost [94]

Explanation:

Let $N_p\ and\ N_s$ are the number of turns in primary and secondary coil of the transformer such that,

$\dfrac{N_p}{N_s}=\dfrac{1}{3}$

A resistor R connected to the secondary dissipates a power $P_s=100\ W$

For a transformer, $\dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}$

$V_s=(\dfrac{N_s}{N_p})V_p$

$V_s=3V_p$ ...............(1)

The power dissipated through the secondary coil is :

$P_s=\dfrac{V_s^2}{R}$

$100\ W=\dfrac{V_s^2}{R}$

$V_p^2=\dfrac{100R}{9}$ .............(2)

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$\dfrac{N_p'}{N_s'}=\dfrac{1}{24}$

New voltage is :

$V_s'=(\dfrac{N_s'}{N_p'})V_p'$

$V_s'=24V_p$ ...............(3)

So, new power dissipated is $P_s'$

$P_s'=\dfrac{V_s'^2}{R}$

$P_s'=\dfrac{(24V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(V_p)^2}{R}$

$P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}$

$P_s'=6400\ Watts$

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

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3 years ago

Your friend says that if Newton’s third law is correct, no object would ever start moving. Here is his argument: You pull a sled

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Answer:

Explanation:

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3 years ago

In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this

soldier1979 [14.2K]

Answer:

1). $v = - 2960526m/s$

2). Toward us

3). $v = - 493421m/s$

4). Toward us

5). $v = 1480263m/s$

6). Away from us

7). $v = 3207236m/s$

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm ( $\lambda_{0} = 121.6nm$ )

$Redshift: \lambda_{measured} > \lambda_{0}$

$Blueshift: \lambda_{measured} < \lambda_{0}$

Then, for this particular case it is gotten:

Star 1: $\lambda_{measured} = 120.4nm$

Star 2: $\lambda_{measured} = 121.4nm$

Star 3: $\lambda_{measured} = 122.2nm$

Star 4: $\lambda_{measured} = 122.9nm$

Star 1:

$Blueshift: 120.4nm < 121.6nm$

Toward us

Star 2:

$Blueshift: 121.4nm < 121.6nm$

Toward us

Star 3:

$Redshift: 122.2nm > 121.6nm$

Away from us

Star 4:

$Redshift: 122.9nm > 121.6nm$

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

$v = c\frac{\Delta \lambda}{\lambda_{0}}$ (1)

Where $\Delta \lambda$ is the wavelength shift, $\lambda_{0}$ is the wavelength at rest, v is the velocity of the source and c is the speed of light.

$v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}})$ (2)

<em>Case for star 1 $\lambda_{measured} = 120.4 nm$ :</em>

<em></em>

$v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})$

$v = - 2960526m/s$

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 $\lambda_{measured} = 121.4 nm$ :</em>

$v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})$

$v = - 493421m/s$

<em>Case for star 3 $\lambda_{measured} = 122.2 nm$ :</em>

$v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})$

$v = 1480263m/s$

<em>Case for star 4 $\lambda_{measured} = 122.9 nm$ :</em>

$v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})$

$v = 3207236m/s$

4 0

3 years ago