Answer:
The distance is 
=
7
m
 
Explanation:
Apply the equation of motion
s
(
t
)
=
u
t
+
1
2
a
t
2
 
The initial velocity is 
u
=
0
m
s
−
1
 
The acceleration is 
a
=
2
m
s
−
2
 
Therefore, when 
t
=
3
s
 , we get
s
(
3
)
=
0
+
1
2
⋅
2
⋅
3
2
=
9
m
 
and when 
t
=
4
s
 
s
(
4
)
=
0
+
1
2
⋅
2
⋅
4
2
=
16
m
 
Therefore,
The distance travelled in the fourth second is
d
=
s
(
4
)
−
s
(
3
)
=
16
−
9
=
7
m
 
        
             
        
        
        
Answer:
For vector u, x component = 10.558 and  y component =12.808
unit vector = 0.636 i+ 0.7716 j
For vector v, x component = 23.6316 and y component = -6.464
unit vector = 0.9645 i-0.2638 j
Explanation:
Let the vector u has magnitude 16.6
u makes an angle of 50.5° from x axis 
So 
Vertical component 
So vector u will be u = 10.558 i+12.808 j
Unit vector 
Now in second case let vector v has a magnitude of 24.5
Making an angle with -15.3° from x axis 
So horizontal component 
Vertical component 
So vector v will be 23.6316 i - 6.464 j
Unit vector of v 
 
        
             
        
        
        
a closed system does not allow matter or energy to pass through
 
        
             
        
        
        
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
 N-m²/kg²
g = 6.67 ×  ( 4.8×10^22 ) / ( 1550 × 10³ )
 ( 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