Answer:
1 greater distances fallen in successive seconds
Explanation:
When a body falls freely it is subjected to the action of the force of gravity, which gives an acceleration of 9.8 m / s2, consequently, we are in an accelerated movement
If we use the kinematic formula we can find the position of the body
Y = Vo t + ½ to t2
Where the initial velocity is zero or constant and the acceleration is the acceleration of gravity
Y = - ½ g t2 = - ½ 9.8 t2 = -4.9 t2
Let's look for the position for successive times
t (s) Y (m)
1 -4.9
2 -19.6
3 -43.2
The sign indicates that the positive sense is up
It can be clearly seen that the distance is greatly increased every second that passes
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Answer:
(a)
M = 1.898 x 10^27 kg
(b)
v = 13.74 km/s
(c) E = 0.28 N/kg
Explanation:
Time period, T = 3.55 days = 3.55 x 24 x 3600 second = 306720 s
Radius, r = 6.71 x 10^8 m
G = 6.67 x 10^-11 Nm^2/kg^2
(a) 


M = 1.898 x 10^27 kg
(b) Let v be the orbital velocity


v = 13739.5 m/s
v = 13.74 km/s
(b) The gravitational field E is given by


E = 0.28 N/kg
The coefficient of static friction is 0.234.
Answer:
Explanation:
Frictional force is equal to the product of coefficient of friction and normal force acting on any object.
So here the mass of the object is given as 2 kg, so the normal force will be acting under the influence of acceleration due to gravity.
Normal force = mass * acceleration due to gravity
Normal force = 2 * 9.8 = 19.6 N.
And the frictional force is given as 4.6 N, then

Coefficient of static friction = 4.6 N / 19.6 N = 0.234
So the coefficient of static friction is 0.234.