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2 years ago

Write newtone first law

Physics

2 answers:

nika2105 [10]2 years ago

6 0

an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.

gtnhenbr [62]2 years ago

4 0

Newtons first law is that an Object that is in motion will stay in motion until an unbalanced force acts upon it, and part 2 is that an object at rest will stay at rest until and unbalanced force acts upon it

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What is the difference between a bar magnet and an electromagnet? (check all that are

lukranit [14]

Answer:

D and A

Explanation:

Hint The basic difference between them is that a bar magnet is a permanent magnet whereas an electromagnet is a temporary magnet. An electromagnet is formed when an electric current is passed through wires wound around soft metalcore. An electromagnet loses its magnetism once the current flow is stopped.

5 0

2 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

Given:

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

2 years ago

In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 1.00 mm away. part a

mihalych1998 [28]

We can use Coulomb's law to find the force F acting on the proton that is released. F = k x Q1 x Q2 / r^2 k = 9 x 10^9 Q1 is the charge on one proton which is 1.6 x 10^{-19} C Q2 is the same charge on the other proton r is the distance between the protons F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2 F = 2.304 x 10^{-22} N We can use the force to find the acceleration. F = ma a = F / m a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg) a = 1.38 x 10^5 m/s^2 The initial acceleration of the proton is 1.38 x 10^5 m/s^2

8 0

3 years ago

Which medium will light travel through slowest? 1.Solid 2.Gas 3.Empty Space

Allushta [10]

It will travel slowest through gases.

6 0

2 years ago

What do radio waves and gamma rays have in common?

Reil [10]

Answer:

The only difference between radio waves, visible light and gamma rays is the energy of the photons.'

Bothrays have photons, but in radio waves they are weaker.