Spring tides will occur during a full moon and an new moon. The sun and moon are on a certain angle causing these tides to be neap tides, they aren't high tide nor are they low tides so you would verbalize they are in between and that is why it is called neap tides as well.
1,000 milligrams = 1 gram
2,000 milligrams = 2 grams
3,000 milligrams = 3 grams
4,000 milligrams = 4 grams
Answer:
The current is halved
Explanation:
The relationship between the current and the resistance is given by Ohm's Law, as follows:

where,
V = Voltage
I = Current
R = Resistance
Therefore, if we double the resistance:

Hence the correct option is:
<u>The current is halved</u>
This equation is one of the most useful in classical physics. It is a concise statement of Isaac Newton's<span> Second Law of Motion, holding both the proportions and vectors of the Second Law. It translates as: The net force on an object is </span>equal<span> to the </span>mass<span>of the object multiplied by the </span>acceleration<span> of the object.</span>
Answer:
The fraction of its energy that it radiates every second is
.
Explanation:
Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

Given that,
Kinetic energy = 6.2 MeV
Radius = 0.500 m
We need to calculate the acceleration
Using formula of acceleration

Put the value into the formula

Put the value into the formula


We need to calculate the rate at which it emits energy because of its acceleration is

Put the value into the formula


The energy in ev/s


We need to calculate the fraction of its energy that it radiates every second


Hence, The fraction of its energy that it radiates every second is
.