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NNADVOKAT [17]
3 years ago
6

A nanometer is a unit of mass, whereas a nanosecond is a unit of time. Question 1 options: True False

Physics
1 answer:
Alenkasestr [34]3 years ago
6 0

Answer:

True.

Explanation:

A nanometer is a unit of mass, whereas a nanosecond is a unit of time. To convert 1.3 hours to minute, you would multiply by 1 h / 60 min. Kilometer is a unit of length, whereas kilogram is a unit of mass. True.

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You are standing on a train station platform as a train goes by close to you. As the train approaches, you hear the whistle soun
adoni [48]

Answer: v = (33320 - 340f2)/( f2 + 98)

Explanation: this question refers to Doppler effect, hence,

f1 = { V / (V - v)} x f2

f1 = 98Hz, f2 = 76 Hz, V = 340m/s, v = ?, fs = fsourcs

For f1,

98 = { 340 / (340 + v)} x fs ...(i)

for f2,

f2 = { 340 / (340 - v)} x fs ... (ii)

Divide (ii) by (i)

98/f2 = [{ 340 / (340 + v)} x fs]÷ [ 340 / (340 - v)} x fs]

98/f2 = {340 / (340 + v)} x fs x (340 - v)} / 340 x fs

98/f2 = (340 + v)/ (340 - v)

Cross multiplying

98(340 - v) = f2 (340 + v)

33320 - 98v = 340f2 + vf2

Collecting like terms

vf2 +98v = 33320 - 340f2

v(f2 +98) = 33320 - 340f2

v = (33320 - 340f2)/( f2 + 98)

5 0
4 years ago
Friction always opposes an objects?<br><br> A) Power<br> B) Weight<br> C) Motion<br> D) Net force
marin [14]
Friction always opposes motion
8 0
4 years ago
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Anna [14]
Pitch is the sensation of certain frequencies to the ear. High frequency = high pitch, low frequency = low pitch. 

f = c(speed of the wave) /  <span>λ (wavelength)

1. 343m/s / 0.77955m = 439.99 Hz   
     This corresponds to pitch A 

2. 343m/s / 0.52028m = 659.26 Hz
</span>     This corresponds to pitch E 
<span>
3. 343m/s / 0.65552m = 523.349 Hz
    </span>This corresponds to pitch C

4. using f = c /  λ
  λ = c / f<span>
     = 343m/s / 587.33 = 0.583999 m = 0.584 m

</span>
3 0
3 years ago
You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.
babymother [125]

Answer:

Mass of the planet = 1.48 × 10²⁵ Kg

Mass of the star = 5.09 × 10³⁰ kg

Explanation:

Given;

Diameter = 1.8 × 10⁷ m

Therefore,

Radius = \frac{\textup{Diameter}}{\textup{2}}  = \frac{\textup{1.8}\times10^7}{\textup{2}}

or

Radius of the planet = 0.9 × 10⁷ m

Rotation period = 22.3 hours

Radius of star = 2.2 × 10¹¹ m

Orbit period = 407 earth days = 407 × 24 × 60 × 60 seconds = 35164800 s

free-fall acceleration = 12.2 m/s²

Now,

we have the relation

g = \frac{\textup{GM}}{\textup{R}^2}

g is the free fall acceleration

G is the gravitational force constant

M is the mass of the planet

on substituting the respective values, we get

12.2 = \frac{6.67\times10^{-11}\times M}{(0.9\times10^7)^2}

or

M = 1.48 × 10²⁵ Kg

From the Kepler's law we have

T² = \frac{\textup{4}\pi^2}{\textup{G}M_{star}}(R_{star})^3

on substituting the respective values, we get

35164800² = \frac{\textup{4}\pi^2}{6.67\times10^{-11}\timesM_{star}}(2.2\times10^{11})^3

or

M_{star} = 5.09 × 10³⁰ kg

6 0
3 years ago
What type of rays can pass through all matter?
Semmy [17]
Gamma rays can pass through a lot
7 0
3 years ago
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