Answer:
Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.
Explanation:
Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.
When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.
The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.
Rutherford proposed new model known as the Planetary model of atom, which described atom as containing a nucleus which is revolved by electron, just like planets revolve round the sun. And this nucleus contains opposite charge to electron which is proton, to balance the motion.
Answer:
The force of static friction acting on the luggage is, Fₓ = 180.32 N
Explanation:
Given data,
The mass of the luggage, m = 23 kg
You pulled the luggage with a force of, F = 77 N
The coefficient of static friction of luggage and floor, μₓ = 0.8
The formula for static frictional force is,
Fₓ = μₓ · η
Where,
η - normal force acting on the luggage 'mg'
Substituting the values in the above equation,
Fₓ = 0.8 x 23 x 9.8
= 180.32 N
Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N
Answer:
10.16 degrees
Explanation:
Apply Snells Law for both wavelenghts
\(n_{1}sin\theta_{1} = n_{2}sin\theta_{2}\)
For red
(1.620)(sin 25.5) = (1)(sin r)
For red, the angle is 35.45degrees
For violet
(1.660)(sin 25.5) = (1)(sin v)
For violet, the angle is 45.6 degrees
The difference is 45.6- 35.45 = 10.16 degrees
Answer:
1/3 the distance from the fulcrum
Explanation:
On a balanced seesaw, the torques around the fulcrum calculated on one side and on another side must be equal. This means that:

where
W1 is the weight of the boy
d1 is its distance from the fulcrum
W2 is the weight of his partner
d2 is the distance of the partner from the fulcrum
In this problem, we know that the boy is three times as heavy as his partner, so

If we substitute this into the equation, we find:

and by simplifying:

which means that the boy sits at 1/3 the distance from the fulcrum.