Answer:
230.4 s
Explanation:
The speed of car A is
and the distance travelled is
so the time taken for car A is
The speed of car B is
and the distance travelled is
so the time taken for car B is
So the difference in time is
Which corresponds to
so car B arrived 230.4 s before car A.
Answer:
electrons exist in specified energy levels
Explanation:
In its gold-foil scattering with alpha particles, Rutherford proved that the plum-pudding model of the atom theorised by Thomson was wrong.
From his experiment, Rutherford inferred that the atom actually consists of a very small nucleus, where all the positive charge is concentrated, and the rest of the atom is basically empty, with the electrons (negatively charged) orbiting around the nucleus at very large distance.
However, Rutherford did not specify anything about the orbits of the electrons. Later, Bohr predicted that the electrons actually orbit the nucleus in specific orbits, each orbit corresponding to a specific energy level. Bohr's model found confirmation in the observation of the emission spectrum lines: when an electron in one of the higher energy level jumps down into an orbit with lower energy, the atom emits a photon which has an energy exactly equal to the difference in energy between the two orbits (and this energy of the photon corresponds to a precise wavelength).
Answer:
a) ω = 9.86 rad/s
b) ac = 194. 4 m/s²
c) minimum coefficient of static friction, µs = 19.8
Explanation:
a) angular speed, ω = 2πf, where f is frequency of revolution
1 rps = 6.283 rad/s, π = 3.142
ω = 2 * 3.14 * 0.25 * 6.28
ω = 9.86 rad/s
b) centripetal acceleration, a = rω²
where r is radius in meters; r = 200 cm or 2 m
a = 2 * 9.86²
a = 194. 4 m/s²
c) µs = frictional force/ normal force
frictional force = centripetal force = ma; where a is centripetal acceleration
normal force = mg; where g = 9.8 m/s²
µs = ma/mg = a/g
µs = 194.4 ms⁻²/9.8 ms⁻²
c) minimum coefficient of static friction, µs = 19.8
Answer:
a) 145.6kgm^2
b) 158.4kg-m^2/s
c) 0.76rads/s
Explanation:
Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and
(c) the angular speed of the merry-go-round and child after the child has jumped on.
a) From I = MK^2
I = (160Kg)(0.91m)^2
I = 145.6kgm^2
b) The magnitude of the angular momentum is given by:
L= r × p The raduis and momentum are perpendicular.
L = r × mc
L = (1.20m)(44.0kg)(3.0m/s)
L = 158.4kg-m^2/s
c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:
L = Iw
158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]
w = 158.6/208.96
w = 0.76rad/s
<h2>
Answer: It is highly flammable.</h2>
Explanation:
Liquid oxygen is created from oxygen atoms that have been forced to assume the liquid state due to <u>compression (change of pressure) and temperature modification.
</u>
Specifically this is achieved by cooling the oxygen enough to change it to its liquid state. So,<u> as the temperature drops, the atoms move more slowly because they have less energy.
</u>
In this sense, in the liquid state it is easier to store and mobilize oxygen, taking into account that it is a highly flammable gas.
