Answer: 1.4 x 10^-8N
Explanation:
Given that,
Mass of Particle 1 (m1) = 12kg
Mass of Particle 2 (m2) = 25kg
distance between particles (r) = 1.2m. Gravitational force (F) =
Apply the formula for gravitational force:
F = Gm1m2/r²
where G is the gravitational constant with a value of 6.7 x 10^-11 Nm2/kg2
Then, F = (6.7 x 10^-11 Nm²/kg² x 12kg x 25kg) / (1.2m)²
F = (2.01 x 10^-8Nm²) / (1.44m²)
F = 1.396 x 10^-8N (Rounded to the nearest tenth as 1.4 x 10^-8N)
Thus, the magnitude of the gravitational force acting on the particles is 1.4 x 10^-8 Newton
Momentum of the wagon increases by (200 x 3)
= 600 newton-sec
= 600 kg-m/sec
When the velocity increases, then the acceleration will be positive and when the velocity decrease then the acceleration will be negative.
During the first hour, the velocity was 70 mph and during the seconds hour the velocity was 60 mph. Hence, the velocity decrease in the seconds hour. So, the acceleration will be negative during the second hour.
Now, during the third hour the velocity increases as it is 80 mph. Hence, the acceleration will be positive during the third hour.
Answer:
V = 49.05 [m/s]
Explanation:
We can easily find the result using kinematics equations, first, we will find the distance traveled during the 5 seconds.

where:
Yo = initial position = 0
y = final position [m]
Vo = initial velocity = 0
t = time = 5 [s]
g = gravity aceleration = 9.81 [m/s^2]
The initial speed is zero, as the body drops without imparting an initial speed. Therefore:
y = 0 + (0*5) + (0.5*9.81*5^2)
y = 122.625[m]
Now using the following equation we can find the speed it reaches during the 5 seconds.
![v_{f} ^{2}= v_{i} ^{2}+(2*g*y)\\v_{f}=\sqrt{2*9.81*122.625} \\v_{f}=49.05 [m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%3D%20v_%7Bi%7D%20%5E%7B2%7D%2B%282%2Ag%2Ay%29%5C%5Cv_%7Bf%7D%3D%5Csqrt%7B2%2A9.81%2A122.625%7D%20%5C%5Cv_%7Bf%7D%3D49.05%20%5Bm%2Fs%5D)