The amount of gravity between 1 kg of lead and Earth is the same as the amount of gravity between 1 kg of marshmallows and Earth.
Answer: Option C
<u>Explanation:</u>
According to universal law of gravity, the gravitational force is directly proportionate to the product of masses of the interacting objects and inversely proportionate to its distance squared between them.
Since the mass of Earth is very high compared to the mass of the other objects, be it lead or marshmallows, and the distance will be mostly same. Thus, there will be no change in the amount of gravity acting between them. Thus, the amount of gravity between 1 kg of lead and Earth is the same as the amount of gravity between 1 kg of marshmallows and Earth.
A) T=2*pi(L/g)^1/2
T=2*3.14(2.5/9.81)^1/2
T=3.171s
b) T=1/f
f=1/3.171
f=0.315Hz
Answer:
7. 1.56
8. A curved line that slopes upward
Explanation:
7. The curve for quartz at the left side of the graph (400 nm) is just above the middle of the distance between 1.5 and 1.6, so is best approximated by 1.56.
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8. The graph shows n vs. wavelength. The question asks about n vs. frequency. Frequency is inversely proportional to wavelengh, which means you need to describe the curve from right to left, rather than left to right. From right to left, the curve rises, so can be called "a curved line that slopes upward."
The answer is vaporization, then condensation.
Here is a set of flash card that can help with this subject.
To develop this problem we will apply the concepts related to the potential energy per unit volume for which we will obtain an energy density relationship that can be related to the electric field. From this formula it will be possible to find the electric field required in the problem. Our values are given as
The potential energy, 
The volume, 
The potential energy per unit volume is defined as the energy density.
The energy density related with electric field is given by
Here, the permitivity of the free space is
Therefore, rerranging to find the electric field strength we have,
Therefore the electric field is 2.21V/m
