If a comet is close to the sun how fast might it travel?
1 answer:
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The gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.
<h3>What is the gravitational force of the earth on the person?</h3>The gravitational force exerted by the earth on a person standing on the earth's surface is given below as follows:
where
G = 6.67 * 10⁻¹¹
m¹ = 62 kg
m² = 5.97 * 10²⁷ kg
r = 6.4 * 10⁶ m
Therefore, the gravitational force exerted by the earth on a person standing on the earth's surface is 602.74 N.
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Answer:
λ = 2042 nm
Explanation:
given data
screen distance d = 11 m
spot s = 4.5 cm = 4.5 × m
separation L = 0.5 mm = 0.5 × m
to find out
what is λ
solution
we will find first angle between first max and central bright
that is tan θ = s/d
tan θ = 4.5 × / 11
θ = 0.234
and we know diffraction grating for max
L sinθ = mλ
here we know m = 1 so put all value and find λ
L sinθ = mλ
0.5 × sin(0.234) = 1 λ
λ = 2042.02 × m
λ = 2042 nm
Answer:
The question is incomplete, below is the complete question
"The displacement of a wave traveling in the negative y-direction is D(y,t) = ( 4.60cm ) sin ( 6.20 y+ 60.0 t ), where y is in m and t is in s.
A) What is the frequency of this wave?
B) What is the wavelength of this wave?
C) What is the speed of this wave?"
Answers:
a.
b.
c.
Explanation:
The equation of a wave is represented as
Where A=amplitude
w=angular frequency=2πf
K=wave numbers =2π/λ
since we re giving he equation D(y,t) = ( 4.60cm ) sin ( 6.20 y+ 60.0 t ),
we can compare and get the value for the wave number and angular frequency.
By comparing we have
w=60rads/s
k=6.20
a. to determine the frequency, from the expression fr angular wave frequency we have
w=2πf hence
f=w/2π
if we substitute we arrive at
b. to determine the wave length, we use
c. the wave speed v is express as the product of the frequency and the wavelength. Hence