Answer:
so maximum velocity for walk on the surface of europa is 0.950999 m/s
Explanation:
Given data
legs of length r = 0.68 m
diameter = 3100 km
mass = 4.8×10^22 kg
to find out
maximum velocity for walk on the surface of europa
solution
first we calculate radius that is
radius = d/2 = 3100 /2 = 1550 km
radius = 1550 × 10³ m
so we calculate no maximum velocity that is
max velocity = √(gr) ...............1
here r is length of leg
we know g = GM/r² from universal gravitational law
so G we know 6.67 ×
N-m²/kg²
g = 6.67 ×
( 4.8×10^22 ) / ( 1550 × 10³ )
g = 1.33 m/s²
now
we put all value in equation 1
max velocity = √(1.33 × 0.68)
max velocity = 0.950999 m/s
so maximum velocity for walk on the surface of europa is 0.950999 m/s
Answer:
I think the answers March 21
Answer:
(a) 42.28°
(b) 37.08°
Explanation:
From the principle of refraction of light, when light wave travels from one medium to another medium, we have:
= sinθ
/sinθ
In the given problem, we are given the refractive indices of light which are parallel and perpendicular to the axis of the optical lens as 1.4864 and 1.6584 respectively.
For critical angle θ
= θ
, θ
= 90°; 
(a) 
= sinθ
/sin90°
0.6728 = sinθ![_{c}θ[tex]_{c} = sin^(-1) 0.6728 = 42.28°(b) [tex]n_{a} = 1.6584](https://tex.z-dn.net/?f=_%7Bc%7D%3C%2Fp%3E%3Cp%3E%CE%B8%5Btex%5D_%7Bc%7D%20%3D%20sin%5E%28-1%29%200.6728%20%3D%2042.28%C2%B0%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3E%28b%29%20%5Btex%5Dn_%7Ba%7D%20%3D%201.6584)
= sinθ
/sin90°
0.60299 = sinθ[tex]_{c}
θ[tex]_{c} = sin^(-1) 0.60299 = 37.08°
Answer:
32 m/s
Explanation:
The speed of a bus is 30 m/s due East wrt the passenger
He also sees a passenger on the bus walking to the back at 2 m/s.
We need to find the passenger's velocity relative to the bus. As the observer sees that the bus and the passenger are moving in opposite direction. Let v is the relative velocity. So,
v = 30 m/s + 2 m/s
v = 32 m/s
Hence, the passenger's velocity relative to the bus is 32 m/s.