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2 years ago

14

A passenger on a stopped bus notices that rain is falling vertically just outside the window. When the bus moves with constant v

elocity, the passenger observes that the falling raindrops are now making an angle of 18 ∘ with respect to the vertical.1. What is the ratio of the speed of the raindrops to the speed of the bus?2. Find the speed of the raindrops, given that the bus is moving with a speed of 19 {m/s}.

1 answer:

nekit [7.7K]2 years ago

5 0

Answer:

1)0.325

2) $6.17\ \rm m/s$

Explanation:

<u>Given:</u>

The angle that falling raindrops make with the vertical= $18^\circ$

Let $V_R$ be the velocity of the raindrops and $V_B$ be the velocity of the bus.

1)

$\dfrac{V_R}{V_B}=\tan 18^\circ\\\dfrac{V_R}{V_B}=0.315\\$

2)Speed of the raindrops $=0.315\times 19$

$=6.17\ \rm m/s$

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In what setting do all frequencies of electromagnetic waves travel at the same speed?

Irina-Kira [14]

All electromagnetic waves travel in vacuum (space) at the speed of light (3 * 10^8 m/s). Radio waves is just a member of the electromagnetic spectrum. All electromagnetic waves follow the wave equation: speed = frequency * wavelength. With all electromagnetic waves, the speed in space is the same.

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2 years ago

Water at room temperature is discharged from a pipe at a rate of 1000 gallons per minute (gpm). Express this flow rate in cubic

marshall27 [118]

Answer

given,

discharge rate from pipe = 1000 gallons/minutes

now,

flow rate in cubic meters per second

1 gallon = 0.00378541 m³

1 min = 60 s

Q = $1000\times \dfrac{0.00378541\ m^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}$

Q = 0.063 m³/s

flow rate in liters per minute

1 gallon = 3.78541 L

Q = $1000\times \dfrac{3.78541\ m^3}{1\ gallon}$

Q = 3785.41 m³/min

flow rate in cubic feet per second

1 gallon = 0.133681 ft³

1 min = 60 s

Q = $1000\times \dfrac{0.133681\ ft^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}$

Q = 2.23 ft³/s

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2 years ago

Part of one row of the periodic table is shown. 4 blue boxes in a row. The first has S n at the center and 50 above with tin and

nydimaria [60]

Answer:

Atoms of tellurium (Te) have the greatest average number of neutrons equal to 76.

Explanation:

In the periodic table, Elements are represented with their respected symbols. Above the symbol is the elements atomic number which is equal to the number of protons in each atom. Below the symbol is the mass number of that element which is roughly equal to the sum of neutrons and protons of that atom.

To calculate the number of neutrons we can take the difference of Atomic number and mass number:

Number of neutrons = mass number - atomic number

<u>- Tin:</u>

Atomic number = 50

Mass number = 119

Number of neutrons = mass number - atomic number = 119 - 50

Number of neutrons = 69

<u>- Antimony(Sb):</u>

Atomic number = 51

Mass number = 122

Number of neutrons = mass number - atomic number = 122 - 51

Number of neutrons = 71

<u>- Tellurium(Te):</u>

Atomic number = 52

Mass number = 128

Number of neutrons = mass number - atomic number = 128 - 52

Number of neutrons = <u>76</u>

<u>- Iodine(I):</u>

Atomic number = 53

Mass number = 127

Number of neutrons = mass number - atomic number = 127 - 53

Number of neutrons = 74

Here, the greatest number of neutrons is for the atoms of Tellurium(Te).

7 0

3 years ago

Read 2 more answers

Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to

jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x $10^{-8}$ J.

(b) The wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency of the photon is 2.47 x $10^{25}$ Hz.

Explanation:

Let;

$E_{1}$ = -13.60 ev

$E_{2}$ = -3.40 ev

(a) Energy of the emitted photon can be determined as;

$E_{2}$ - $E_{1}$ = -3.40 - (-13.60)

= -3.40 + 13.60

= 10.20 eV

= 10.20(1.6 x $10^{-9}$ )

$E_{2}$ - $E_{1}$ = 1.632 x $10^{-8}$ Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x $10^{-8}$ Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x $10^{-34}$ Js), c is the speed of light (3 x $10^{8}$ m/s) and λ is the wavelength.

10.20(1.6 x $10^{-9}$ ) = (6.6 x $10^{-34}$ * 3 x $10^{8}$ )/ λ

λ = $\frac{1.98*10^{-25} }{1.632*10^{-8} }$

= 1.213 x $10^{-17}$

Wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x $10^{-8}$ = 6.6 x $10^{-34}$ x f

f = $\frac{1.632*10^{-8} }{6.6*10^{-34} }$

= 2.47 x $10^{25}$ Hz

Frequency of the emitted photon is 2.47 x $10^{25}$ Hz.

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2 years ago

At the _______________ point, the particles in an object have not kinetic energy

Brums [2.3K]

The freezing point ..... :)

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3 years ago