Answer:
Explanation:
Gravitational force between two objects having mass m₁ and m₂ at a distance R
F = G m₁ m₂ / R²
Force between baby and father F₁ = 6.67x10⁻¹¹ x 4.1 x 120 / .18²
= 1.01 x 10⁻⁶ N
b )
Force between baby and Jupiter
F₂ = 6.67x10⁻¹¹ x 1.9x 10²⁷ x 4.1 / ( 6.29 x 10¹¹ )²
= 1.31 x 10⁻⁶ N
c )
Ratio = 1.01 / 1.31
= .77
First we have to calculate the time taken to travel the distance 30 m, is
.
Now from equation of motion,

Given,
.
As object starts from rest, so u = 0.
Substituting these values in above equation, we get
.
Thus, the acceleration is 
Answer:
Explanation:
a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force
R = mg where m is mass of the block
Force of friction F = μ x mg
= .173 x 12.2 x 9.8
= 20.68 N
b ) Only force of friction is acting on the body so
deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²
acceleration = - 1.7 m /s²
c )
v² = u² - 2 a s
v = 0 , u = 3.9 m /s
a = 1.7 m /s
0 = 3.9² - 2 x 1.7 x s
s = 4.47 m
since child is moving along with the wagon and we need to find the acceleration of child inside that wagon then in this case the system of interest must be child + wagon
System of interest will be the system that is used to find the force or acceleration using Newton's law
Here we have to assume that system on which if we will calculate the forces then the net value of force on that system will help to calculate the unknown quantities
So here our system will be boy + wagon
Answer:
The thrown rock will strike the ground
earlier than the dropped rock.
Explanation:
<u>Known Data</u>


, it is negative as is directed downward
<u>Time of the dropped Rock</u>
We can use
, to find the total time of fall, so
, then clearing for
.
![t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s](https://tex.z-dn.net/?f=t_%7BD%7D%3D%5Csqrt%5B2%5D%7B%5Cfrac%7B300m%7D%7B4.9m%2Fs%5E%7B2%7D%7D%7D%20%3D%5Csqrt%5B2%5D%7B61.22s%5E%7B2%7D%7D%20%3D7.82s)
<u>Time of the Thrown Rock</u>
We can use
, to find the total time of fall, so
, then,
, as it is a second-grade polynomial, we find that its positive root is
Finally, we can find how much earlier does the thrown rock strike the ground, so 