Newton's second law of motion pertains to the behavior of objects for which all existing forces are not balanced. The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. The acceleration of an object depends directly upon the net force acting upon the object, and inversely upon the mass of the object. As the force acting upon an object is increased, the acceleration of the object is increased. As the mass of an object is increased, the acceleration of the object is decreased.
And the correct answer is A) on the surface of the moon; because Newton's second law provides the explanation for the behavior of objects upon which the forces do not balance. The law states that unbalanced forces cause objects to accelerate with an acceleration that is directly proportional to the net force and inversely proportional to the mass.
So the correct answer is A) on the surface of the moon
Hope I helped.
Answer:222 meters
Explanation:
Speed=80kph=(80x5/18)=(80x5)/18=22.2m/s
Time=10 seconds
distance=speed x time
distance=22.2 x 10
distance=222
distance=222 meters
Answer:
The magnitude of the electric field at a point equidistant from the lines is 
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle
We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field
Put the value into the formula
Hence, The magnitude of the electric field at a point equidistant from the lines is
Answer:
c
Explanation:
force is how hard it is pulled or pushed
- Height (h) = 10 m
- Density (ρ) = 1000 Kg/m^3
- Acceleration due to gravity (g) = 10 m/s^2
- We know, pressure in a fluid = hρg
- Therefore, the pressure exerted by a column of fresh water
- = hρg
- = (10 × 1000 × 10) Pa
- = 100000 Pa
<u>Answer</u><u>:</u>
<u>1000</u><u>0</u><u>0</u><u> </u><u>Pa</u>
Hope you could understand.
If you have any query, feel free to ask.
