All categories

3 years ago

What does physics have to do with science and chemistery

1 answer:

5 0

Physics is known as the fundamentals of science, as all the branches of science, whether it is biology, chemistry, ecology, botany, environmental science, geology, geography, and more all follow the laws of physics, such as laws of thermodynamics, etc. Chemistry is existent because of physics due to the Conservation of mass. Once you get to a much deeper level of science, you will notice that all the fields of science start to blend together and all add up to one huge painting!

You might be interested in

What is the measurement of loudness of sounds

AURORKA [14]

The measurement of sound is in decibels.

4 0

4 years ago

The black woodpecker's beak is connected to its skull with ____ to prevent jolts.

Nina [5.8K]

Answer:

hyoid bone

Explanation:

Woodpeckers have a special bone that acts like a seat-belt for its skull. It's called the hyoid bone, and it wraps all the way around a woodpecker's skull. Every time the bird pecks, the hyoid acts like a seat-belt for the bird's skull and the delicate brain it protects.

5 0

3 years ago

Read 2 more answers

A box sits on a table. A long arrow labeled F subscript P = 22 N points right. A short arrow labeled F subscript f = ? N points

anygoal [31]

Answer:

b -4 N

may the force be with you

8 0

3 years ago

Read 2 more answers

A bare helium nucleus has two positive charges and a mass of 6.64×10−27kg(a) Calculate its kinetic energy in joules at 2.00% of

grandymaker [24]

To solve this problem we will apply the concepts related to kinetic energy and energy conservation. The kinetic energy will be expressed in terms of mass and speed, as well as load and voltage. From this last expression we will find the charges by electron and by Helium nucleus.

a ) Kinetic Energy is given as

$KE = \frac{1}{2} mv^2$

Replacing with our values we have that

$KE = \frac{1}{2} (6.64*10^{-27})(2.0\% (3.00*10^8))^2$

$KE = 1.1935*10^{-13}J$

Therefore the kinetic energy of the helium nucleus is $1.1935*10^{-13}J$

PART B) Now for calculate the electron volts we use the kinetic energy as a expression between the charge and the voltage, that is

$KE = qV$

Here,

q = Charge of an electron

V = Voltage

Rearranging to find the potential we have,

$V = \frac{KE}{q}$

$V = \frac{1.19*10^{-13}}{1.6*10^{-19}}$

$V = 743750eV$

Therefore the kinetic energy in electron vols is $743750eV$

PART C) Applying the same relationship but now using the Helium core load, we will have to

$KE = QV$

Here,

Q = Charge of a helium nucleus

V = Voltage

Rearranging to find the potential we have

$V = \frac{KE}{Q}$

But we need to note that the charge is equal to the number of charge for the unit charge, then

$Q = \text{No. Charge} \times \text{Unit Charge}$

$Q = (2)(1.6*10^{-19}C)$

$Q = 3.2*10^{-19}C$

Now replacing we have that

$V= \frac{1.19*10^{-13}}{32*10^{-19}}$

$V = 371875V$

Therefore the voltage applied is 371875V

5 0

3 years ago

An air-filled parallel-plate capacitor is connected to a battery and allowed to charge up. Now a slab of dielectric material is

natita [175]

Answer:

The charge on the capacitor had increased

Explanation:

The expression for the capacitance of an air-filled parallel-plate capacitor is as follows as;

$C=\frac{\epsilon _{0}A}{d}$

Here, C is the capacitance, $\epsilon _{0}$ is the permittivity of free space, A is the area and d is the distance between the parallel plate capacitor.

When a slab of dielectric material is placed between the plates of the capacitor then the expression for the capacitance is as follows;

$C=\frac{K\epsilon _{0}A}{d}$

Here, K is the dielectric constant.

In the given problem, a slab is inserted between the plates of the capacitor then the capacitance of the capacitor will increase in this case. Therefore, the option (a) is true.

The expression for the charge stored in the capacitor is as;

Q= CV

Here, Q is the charge and V is the potential.

The charge will also increase in this case as the charge stored in the capacitor is directly proportional to the capacitance. Therefore, the option (d) is not true.

The expression for the energy stored in the capacitor is as follows;

$E=\frac{CV^{2}}{2}$

The voltage is constant in the given problem but the capacitance increases then the energy stored in the capacitor will increase. Therefore, the option (b) is not true.

The voltage across the capacitor will remain same as the capacitor is still connected to the battery. Therefore, the option (c) is not true.

Therefore, only option (a) is true.

3 0

3 years ago