<em><u>Answer:</u></em>
<em><u>Answer:the number of electrons in their outermost shell </u></em>
<u>Question:</u>
Two particles are separated by a distance d. At this distance, the strong interaction between them is much greater than the electrical interaction. What are the possible values of d?
a. extremely large values, such as the distances between the planets and sun
b. extremely small values, such as the distances between particles in an atomic nucleus
c. values between 1 mm and 1 m
d. values between 1 m and 1 km
<u>Answer:
</u>
The possible values of d are extremely small values, such as the distances between particles in an atomic nucleus.
<u>Explanation:
</u>
Any kind of interaction requires force between the interacting materials and that force tends to directly proportional to the product of mass of the materials and inversely proportionate to distance of separation of the interacting materials.
So, mostly based on the distance of separations, an interaction can be termed as strong or weak interactions. If the interacting particles are placed at higher distance, then the force acting between them tend to decrease leading to weak interaction.
An interaction stronger than electrical interactions can occur within subatomic particles. This leads to binding of the subatomic particles as an atom. So the distance d should be extremely small in subatomic particles like distance between the particles in an atomic nucleus.
Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)
The formula for speed is speed = distance ÷ time. To work out what the units are for speed, you need to know the units for distance and time. In this example, distance is in metres (m) and time is in seconds (s), so the units will be in metres per second (m/s).
Answer:
Power output = 96.506 watts
Explanation:
Drag coefficient (Cd) = 0.9
V = 7.3 m/s
Air density (ρ) = 1.225 kg/m^(3)
Area (A) = 0.45 m^2
Let's find the drag force ;
Fd=(1/2)(Cd)(ρ)(A)(v^(2))
So Fd = (1/2)(0.9)(1.225)(0.45)(7.3^(2)) = 13.22N
Drag power = Drag Force x Drag velocity.
Thus drag power, = 13.22 x 7.3 = 96.506 watts