Answer:
The magnitude of the average force on the wall during the collision is 6 N.
Explanation:
Given;
mass of snowball, m = 120 g = 0.12 kg
velocity of the snowball, v = 7.5 m/s
duration of the collision between the snowball and the wall, t = 0.15 s
Magnitude of the average force can be calculated by applying Newton's second law of motion;
F = ma
where;
a is acceleration = v / t
a = 7.5 / 0.15
a = 50 m/s²
F = ma
F = 0.12 x 50
F = 6 N
Therefore, the magnitude of the average force on the wall during the collision is 6 N.
Answer:
Strong nuclear force is 1-2 order of magnitude larger than the electrostatic force
Explanation:
There are mainly two forces acting between protons and neutrons in the nucleus:
- The electrostatic force, which is the force exerted between charged particles (therefore, it is exerted between protons only, since neutrons are not charged). The magnitude of the force is given by
where k is the Coulomb's constant, q1 and q2 are the charges of the two particles, r is the separation between the particles.
The force is attractive for two opposite charges and repulsive for two same charges: therefore, the electrostatic force between two protons is repulsive.
- The strong nuclear force, which is the force exerted between nucleons. At short distance (such as in the nucleus), it is attractive, therefore neutrons and protons attract each other and this contributes in keeping the whole nucleus together.
At the scale involved in the nucleus, the strong nuclear force (attractive) is 1-2 order of magnitude larger than the electrostatic force (repulsive), therefore the nucleus stays together and does not break apart.
Answer:
Explanation:
<u>Accelerated Motion
</u>
When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by
where a is the acceleration, and vo is the initial speed
.
The train has two different types of motion. It first starts from rest and has a constant acceleration of for 182 seconds. Then it brakes with a constant acceleration of until it comes to a stop. We need to find the total distance traveled.
The equation for the distance is
Our data is
Let's compute the first distance X1
Now, we find the speed at the end of the first period of time
That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0
Computing the second distance
The total distance is
The answer is either c or d but c is the best answer