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:
83%
Explanation:
On the surface, the weight is:
W = GMm / R²
where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.
In orbit, the weight is:
w = GMm / (R+h)²
where h is the height of the shuttle above the surface of the Earth.
The ratio is:
w/W = R² / (R+h)²
w/W = (R / (R+h))²
Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:
w/W = (6.4×10⁶ / 7.03×10⁶)²
w/W = 0.83
The shuttle in orbit retains 83% of its weight on Earth.
Answer:
The answer is 26/98 how i did this is i divided them mulitiplyed well i cant really explain it but im pretty dure its right
Explanation:
Answer:
The carrier lengthen is 0.08436 m.
Explanation:
Given that,
Length = 370 m
Initial temperature = 2.0°C
Final temperature = 21°C
We need to calculate the change temperature
Using formula of change of temperature



We need to calculate the carrier lengthen
Using formula of length

Put the value into the formula


Hence, The carrier lengthen is 0.08436 m.
Answer:
26 days
Explanation:
m = 9.4×1021 kg
r= 1.5×108 m
F = 1.1×10^ 19 N
We know Fc = 
==> 1.1 ×
= (9.4 ×
×
) ÷ 1.5 × 
==> 1.1 ×
=
× 6.26×
==>
= 1.1 ×
÷ 6.26×
==>
= 0.17571885 × 
==> v= 0.419188323 ×
m/sec
==> v= 419.188322834 m/s
Putting value of r and v from above in ;
T= 2πr ÷ v
==> T= 2×3.14×1.5×
÷ 0.419188323 × 
==> T = 22.472× 100000 = 2247200 sec
but
86400 sec = 1 day
==> 2247200 sec= 2247200 ÷ 86400 = 26 days