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abruzzese [7]
3 years ago
9

An emission ray has a frequency of 5.10 × 10¹Ê Hz. Given that the speed of light is 2.998 × 10§ m/s, what is the wavelength of t

he ray?
Physics
1 answer:
scoundrel [369]3 years ago
6 0
The answer is 5.88 · 10⁻⁷<span> m.</span>

To calculate this we will use the light equation:

v = λ · f,
where:
v - the speed of light (units: m/s)
<span>λ - the wavelength of the ray (units: m)
</span>f - the frequency of the ray (units: Hz = 1/s <span>since Hz means cycles per second (f=1/T))
</span>
It is given:
f = 5.10 · 10¹⁴ Hz = 5.10 · 10¹⁴<span> 1/s
v = 2.998 </span>· 10⁸<span> m/s
</span><span>λ = ?
</span>
If v = λ · f, then λ = v ÷ f:
λ = 2.998 · 10⁸ m/s ÷ 5.10 · 10¹⁴ 1/s
   = 0.588 · 10⁸⁻¹⁴ · m
   = 0.588 · 10⁻⁶ m
   = 5.88 · 10⁻⁷ m
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A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c
satela [25.4K]

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

 Given the data in the question;

First course : a_{x} = 0.75 m/s², d_{x} = 20 m, u_{x} = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

v_{x}² = (10)² + (2 × 0.75 × 20)

v_{x}² = 100 + 30

v_{x}² = 130

v_{x} = √130

v_{x} = 11.4 m/s

for the Second Course:

u_{y} =  11.4 m/s,  a_{y} = -1.15 m/s²,  v_{y} = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) ×  d_{y} )

0 = 129.96 - 2.3d_{y}

2.3d_{y}  = 129.96

d_{y} = 129.96 / 2.3

d_{y} = 56.5 m

so;

|d| = √( d_{x}² + d_{y}² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

7 0
2 years ago
How much heat is needed to raise the temperature of 9g of water by 17oC?
andre [41]
Well the heat that is needed to raise the temperature of 10g of water by 17oC is 7
6 0
3 years ago
When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the
MaRussiya [10]

Answer:

(a) \alpha=-111.26rad/s

(b) s=4450.6in

(c) 8.66in

Explanation:

First change the units of the velocity, using these equivalents 1rev=2\pi rad and 1 min =60s

4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s

The angular acceleration \alpha the time rate of change of the angular speed \omega according to:

\alpha=\frac{\Delta \omega}{\Delta t}

\Delta  \omega=\omega_i-\omega_f

Where \omega_i is the original velocity, in the case the velocity before starting the deceleration, and \omega_f is the final velocity, equal to zero because it has stopped.

\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s

b) To find the distance traveled in radians use the formula:

\theta = \omega_i t + \frac{1}{2} \alpha t^2

\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad

To change this result to inches, solve the angular displacement \theta for the distance traveled s (r is the radius).

\theta=\frac{s}{r} \\s=\theta r

s=890.12(5)=4450.6in

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

\frac{890.12}{2\pi}=141.6667

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance  between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle \gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120 is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which  is also the net displacement):

c^2=a^2+b^2-2abcos(\gamma)

c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

4 0
3 years ago
A. How many atoms of helium gas fill a spherical balloon of diameter 29.6 cm at 19.0°C and 1.00 atm? b. What is the average kine
Korolek [52]

Answer:

a) 3.39 × 10²³ atoms

b) 6.04 × 10⁻²¹ J

c) 1349.35 m/s

Explanation:

Given:

Diameter of the balloon, d = 29.6 cm = 0.296 m

Temperature, T = 19.0° C = 19 + 273 = 292 K

Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa

Volume of the balloon = \frac{4}{3}\pi(\frac{d}{2})^3

or

Volume of the balloon = \frac{4}{3}\pi(\frac{0.296}{2})^3

or

Volume of the balloon, V = 0.0135 m³

Now,

From the relation,

PV = nRT

where,

n is the number of moles

R is the ideal gas constant = 8.314  kg⋅m²/s²⋅K⋅mol

on substituting the respective values, we get

1.013 × 10⁵ × 0.0135 = n × 8.314 × 292

or

n = 0.563

1 mol = 6.022 × 10²³ atoms

Thus,

0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms

b) Average kinetic energy = \frac{3}{2}\times K_BT

where,

Boltzmann constant, K_B=1.3807\times10^{-23}J/K

Average kinetic energy = \frac{3}{2}\times1.3807\times10^{-23}\times292

or

Average kinetic energy = 6.04 × 10⁻²¹ J

c) rms speed = \frac{3RT}{m}

where, m is the molar mass of the Helium = 0.004 Kg

or

rms speed = \frac{3\times8.314\times292}{0.004}

or

rms speed = 1349.35 m/s

5 0
3 years ago
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GuDViN [60]
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3 0
3 years ago
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