Answer:
d = 2021.6 km
Explanation:
We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them
Airplane 1
Height y₁ = 800m
Angle θ = 25°
cos 25 = x / r
sin 25 = z / r
x₁ = r cos 20
z₁ = r sin 25
x₁ = 18 103 cos 25 = 16,314 103 m
= 16314 m
z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m
2 plane
Height y₂ = 1100 m
Angle θ = 20°
x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m
z₂ = 20 103 without 25 = 8.452 103 m = 8452 m
The distance between the planes using the Pythagorean Theorem is
d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2
Let's calculate
d² = (18126-16314)² + (1100-800)² + (8452-7607)²
d² = 3,283 106 +9 104 + 7,140 105
d² = (328.3 + 9 + 71.40) 10⁴
d = √(408.7 10⁴)
d = 20,216 10² m
d = 2021.6 km
Answer:
Unknown
Explanation:
By definition, we can't observe what's inside there, because no light – no information of any kind – can escape a black hole. But astrophysical theories suggest that, at the core of a black hole, all the black hole's mass is concentrated into a tiny point of infinite density. This point is known as a singularity.
Answer:
Solution
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Given:
Mass of body = 30 kg
gravitational acceleration on the moon = 1.62 m/s
2
Weight of the body on the moon = Mass of the body×gravitational acceleration on the moon=30×1.62=48 N
Answer:
Only 2,3,4 are true
Explanation:
Bosons Particles are particles that condense to the same state. Bosons particle have integral spin like 0 , 
, 
, 
, etc. Bosons particles always have asymmetric wave function and there is exchange of particles.
1) It does not obey Fermi_ Dirac statistics
2) It obeys Bose-Einstein statistics
3) The object can have intrinsic spin
4) Yes the Bosons particle is always symmetric with exchange of particles
5) No Bosons particle are symmetric and not asymmetric
Answer:
i. Cv =3R/2
ii. Cp = 5R/2
Explanation:
i. Cv = Molar heat capacity at constant volume
Since the internal energy of the ideal monoatomic gas is U = 3/2RT and Cv = dU/dT
Differentiating U with respect to T, we have
= d(3/2RT)/dT
= 3R/2
ii. Cp - Molar heat capacity at constant pressure
Cp = Cv + R
substituting Cv into the equation, we have
Cp = 3R/2 + R
taking L.C.M
Cp = (3R + 2R)/2
Cp = 5R/2
