We apply the gravity calculation expressed in the formula: g=GM/r2
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1) Radius = √6.674e-11*5.972e24/8 = 7058 kms Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2 Radius=√6.674e-11*5.972e24/4.9 = 9019 kms Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.
Answer:
Both objects travel the same distance.
(c) is correct option
Explanation:
Given that,
Mass of first object = 4.0 kg
Speed of first object = 2.0 m/s
Mass of second object = 1.0 kg
Speed of second object = 4.0 m/s
We need to calculate the stopping distance
For first particle
Using equation of motion

Where, v = final velocity
u = initial velocity
s = distance
Put the value in the equation

....(I)
Using newton law

Now, put the value of a in equation (I)

Now, For second object
Using equation of motion

Put the value in the equation

....(I)
Using newton law


Now, put the value of a in equation (I)

Hence, Both objects travel the same distance.
The correct answer is (B)
Which is (kQ1Q2) / d^2
Answer:
Explanation:
Ball rises and falls in equal periods. Because it goes by same speeds and same route. Therefore t_{rise} = t_{fall} =2.5 s
In free fall and vertical throw movement: x= (1/2).a.t^{2}
Here, we can find the height of the ball
x= (1/2).5.(2.5^{2}) = 15.625 m
From the attaches V-t graphics, x=V.t/2
15.624=V.(2.5)/2
V=12.5 m/s