Answer:
Add Ff from Fa
Explanation:
Fnet = sum of all force
horizontal net force = Ff + Fa
Answer: 0.137 m
Explanation:
Given
Mass of brick, m = 3 kg
Angle of inclination, Φ = 34°
Force constant of the spring, k = 120 N/m
The force of the brick, F can be gotten using the relation
F = mg
F = 3 * 9.8
F = 29.4 N
Now, the force parallel to the incline, F(p) can be gotten using the formula,
F(p) = F sinΦ, so that
F(p) = 29.4 * sin 34
F(p) = 29.4 * 0.559
F(p) = 16.4 N
The stretch distance then is,
d = F(p) / k * 1 m
d = 16.4 / 120
d = 0.137 m
Thus, the spring stretched by a distance of 0.137 m
They hit positive charges in the gold atoms.
As alpha particles are positively charged particles, they will experience repulsion force as soon as they approach positive charges in the center of the atom (in the nucleus).
Also remember that back at Rutherford time, they did not have the right visualization of the atom, they had a model of it like if it was a budding of positive charges and the negative charges are put into them. So this experiment gave a really better and more reliable imagination of the atom structure.
Hope this helps.
The velocity of the two balls after the collision is 0.73 m/s.
The velocity of the two balls after the collision can be calculated using the formula below.
<h3>Formula:</h3>
- mu+m'u' = V(m+m')............... Equation 1
<h3>Where:</h3>
- m = mass of the first ball
- m' = mass of the second ball
- u = initial velocity of the first ball
- u' = initial velocity of the second ball
- V = velocity of the two balls after the collision.
make V the subject of the equation
- V = (mu+m'u')/(m+m')................ Equation 2
From the question,
<h3>Given:</h3>
- m = 0.25 kg
- m' = 0.3 kg
- u = 1 m/s
- u' = 0.5 m/s
Substitute these values into equation 2
- V = [(0.25×1)+(0.3×0.5)](0.25+0.3)
- V = 0.4/0.55
- V = 0.73 m/s.
Hence, the velocity of the two balls after the collision is 0.73 m/s
Learn more about collision here: brainly.com/question/7694106
Answer: i guess d is the correct answer