Answer:
v=0.94 m/s
Explanation:
Given that
M= 5.67 kg
k= 150 N/m
m=1 kg
μ = 0.45
The maximum acceleration of upper block can be μ g.
a= μ g ( g = 10 m/s²)
The maximum acceleration of system will ω²X.
ω = natural frequency
X=maximum displacement
For top stop slipping
μ g =ω²X
We know for spring mass system natural frequency given as

By putting the values

ω = 4.47 rad/s
μ g =ω²X
By putting the values
0.45 x 10 = 4.47² X
X = 0.2 m
From energy conservation


150 x 0.2²=6.67 v²
v=0.94 m/s
This is the maximum speed of system.
Explanation:
given,
mass of one planet (m1)=2*10^23 kg
mass of another planet (m2)=5*10^22kg
distance between them(d)=3*10^16m
gravitational constant(G)=6.67*10^-11Nm^2kg^-2
gravitational force between them(F)=?
we know,
F=Gm1m2/d^2
or, F=6.67*10^-11*2*10^23*5*10^22/(3*10^16)^2
or, F=6.67*2*5*10^-11+23+22/3*3*10^32
or, F=66.7*10^34/9*10^32
or, F=7.41*10^34-32
•°• F=7.41*10^2
thus, the gravitational force between them is 7.14*10^2
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Answer:
3.2075*10^16
Explanation:
Q=P/V just search up a converter and youll get 30V and so you do 15/30 which is a half and a single coulomb is 6.415*10^16 so you half it. I belive this is correct if you dont belive me wait for someone else smarter to answer and compare.