<span>Work is required to pull a nucleon out of an atomic nucleus. It has more mass outside the nucleus.</span>
Answer:
1.56 × 10^-3 cm.
Explanation:
So, we are given the following parameters from the question above;
Length = 3.67 cm, breadth = 2.93 cm, and the number of embedded transistors = 3.5 million.
Step one: find the area of the computer chip.
Therefore, Area = Length × breadth.
Area = 3.67 cm × 2.93 cm.
Area of the computer chip = 10.7531 cm^2. = 10.75 cm^2.
Step two: find the area of one transistor
The area of one transistor is; (area of the computer chip) ÷ (number of embedded transistors).
Hence;
The area of one transistor= 10.7531/4.4 × 10^6.
The area of one transistor= 2.44 × 10^-6 cm^2.
=> Note that We have our transistors as square, therefore;
The maximum dimension = √ (2.44 × 10^-6) cm^2.
The maximum dimension= 1.56 × 10^-3 cm.
ANSWER
0s to 3s
EXPLANATION
To find the time interval where the object was traveling more quickly, we have to find the velocity of the object during each interval.
From a position-time graph, we can obtain the velocity of the object from the slope of the graph in each interval. To find those slopes, we just have to divide the vertical difference by the time interval.
The interval from 0s to 3s:
The interval from 3s to 5s and the interval from 7s to 8s have a horizontal line, so the slope is zero and therefore the velocity is zero - meaning that the object was not moving during these periods.
The interval from 5s to 7s,
And the interval from 8s to 12s,
Two of these three velocities are negative. Negative velocity indicates that the object is moving backward.
From these velocities, the greatest one, in absolute value, is the one between 0s to 3s. During this interval, the object is moving backward but at the greatest velocity.
Answer:
217.28 m/s
Explanation:
u = 0, t 32.3 s, v = 47.1 m/s, s = 607 m
Let a be the acceleration.
Use third equation of motion.
v^2 = u^2 + 2 a s
47.1 x 47.1 = 0 + 2 a x 607
a = 1.83 m/s^2
For small plane
a = 1.83 m/s^2 , v = 28.2 m/s, u = 0, Let teh distance be s.
Use third equation of motion
28.2^2 = 0 + 2 x 1.83 x s
s = 217.28 m/s
