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:
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Answer:
4.5m/s
Explanation:
Linear speed (v) = 42.5m/s
Distance(x) = 16.5m
θ= 49.0 rad
radius (r) = 3.67 cm
= 0.0367m
The time taken to travel = t
Recall that speed = distance / time
Time = distance / speed
t = x/v
t = 16.5/42.5
t = 0.4 secs
tangential velocity is proportional to the radius and angular velocity ω
Vt = rω
Angular velocity (ω) = θ/t
ω = 49/0.4
ω = 122.5 rad/s
Vt = rω
Vt = 0.0367 * 122.5
Vt =4.5 m/s