a solid sphere and a hollow sphere with equal mass are rotated about an axis through their centers. both spheres experience equa
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Answer:
v’= 279.66 m / s
Explanation:
We work this exercise using the conservation of the moment. For this we define the system formed by the two blocks, therefore the forces during the collision are internal of the action and reaction type.
Initial instant. Before the crash
p₀ = m v₀ + 0
Final moment. After the crash
p_f = m v + M v ’
how the tidal wave is preserved
p₀ = p_f
m v₀ = m v + M v ’
v =
let's calculate
v ’=
v ’=
v ’= 279.66 m / s
Answer:
2023857702.507m
Explanation:
recall from newton's law of gravitation
G=gravitational constant
mshew=50g
melephant=5*10^3kg
rearth=radius of the earth 6400km or 6400000m
mearth= masss of the earth
Gm(shrew)m(earth)/r(earth)^2 = Gm(elephant)m(earth)/r^2
strike out the left hand side and right hand side variables
m(shrew)/r(earth)^2 = m(elephant)/r^2
r^2 = m(elephant).r(earth)^2 / m(shrew) .........make r^2 the subject of the equation
r^2=
r^2=40960000000000
r=2023857702.507m
Formula of range is given by
here given that
θ=14.1∘
now by above equation
R = 66.7 m
so range will be 66.1 m