Answer: the conclusion is that
Explanation:
Answer:
Hello your question is incomplete below is the complete question
Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000
answer : V = 1.624* 10^-5 m/s
Explanation:
First we have to calculate the value of a
a = 93 * 10^6 mile/m * 1609.344 m
= 149.668 * 10^8 m
next we will express the distance between the earth and the sun
--------- (1)
a = 149.668 * 10^8
E (eccentricity ) = ( 1/60 )^2
= 90°
input the given values into equation 1 above
r = 149.626 * 10^9 m
next calculate the Earths velocity of approach towards the sun using this equation
------ (2)
Note :
Rc = 149.626 * 10^9 m
equation 2 becomes
(
therefore : V = 1.624* 10^-5 m/s
Answer:
I know someone anwsered but it would be 400M
Explanation:
i initial velocity (u)=10m/s
acceleration (a)=0
time taken (t) =40s
then distance (s)=u t +1/2 a t^2
s= u t +0 (as a is 0)
s= 10 x 40
s= 400M
Answer:
= 9.8°
Explanation:
Width of one slit (a₁ ) = 1 / 1000 mm=0.001 mm = 10⁻⁶ m.
width of one slit in case 2 (a₂ ) = 1/500 =2 x 10⁻⁶ m
angular position of fringe, Sinθ = n λ /a
n is order of fringe , λ is wave length of light and a is slit aperture
So Sinθ ∝ 1 / a
Sin θ₁ /Sin θ₂ = a₂/a₁ ;
Sin20°/sinθ₂ = 2 / 1
sinθ₂ = Sin 20° / 2 = .342/2 = .171
θ₂ = 9.8 °
because the building are weaker and they are most likley to fall