On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C
Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)
on the DT graph im going to assume that movement B is a positive velocity with A and C being negative.
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity.
Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question.
i'll skip my working out as its the same as before:
C, B, A and then D
the same idea as on Q30 applies to Q31 part b,
D,C,B then A
In the modern atomic model the number of electrons in a neutral atom is the same as the number of protons.
An atom contain, three fundamental particles, electron, proton, and neutron, neutron is charge less, there the charge of the atom is determined by the number of proton, and number of electron, Atoms are electrically neutral, it is only possible when number of electrons in an atom is equal to the number of proton.
1. Friction (Example: when you rub your hands together and you start to feel heat.)
2. Attract each other. (Example: Every object has a gravitational pull of some sort. The larger the object, the stronger the gravitational pull. This is why the sun is able to keep all eight planets in orbit around it.)
3. Hope this helped :)
When breathing out, it pushes your diaphragm, the muscle below your lungs, up, which then causes the air to leave your lungs.
Answer:
It corresponds to 1mm-10 mm range.
Explanation:
- Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
- As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:
- Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:
