Answer:
v = 384km/min
Explanation:
In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.
You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

hence, the speed of the Hubble is approximately 384km/min
False. That description fits the wave's 'frequency'.
It has nothing to do with refraction.
Answer:
the distance from charge A to C is r₁₃= 1.216 m
Explanation:
following Coulomb's law , the force exerted by 2 point charges between themselves is:
F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant
since C ( denoted as 3) is at equilibrium
F₁₃=F₂₃
k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²
q₁/r₁₃²=q₂/r₂₃²
r₁₃²/q₁=r₂₃²/q₂
r₂₃=r₁₃*√(q₂/q₁)
since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have
r₁₃+r₂₃=d=r₁₂
r₁₃+r₁₃*√(q₂/q₁)=d
r₁₃*(1+√(q₂/q₁))=d
r₁₃=d/(1+√(q₂/q₁))
replacing values
r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m
thus the distance from charge A to C is r₁₃= 1.216 m
Answer:
Explanation:
Vm = Δs/Δt
700km/h = Δs/1.5h
700 = Δs/1.5

S = 700 x 1.5
S = 1050 Km
*(S = Δs)
Answer: The rocket will have traveled 1050 Km
Hope this help ☺
Answer:

Explanation:
When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

Here, n = 10
Spring constant of each spring = k
Thus,



