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2 years ago

1. How are each of Newton’s Laws of motion demonstrated

Physics

1 answer:

stepan [7]2 years ago

5 0

Answer:

Newton's law of inertia - His first law states that every object remains at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force. ... This is the first part cited in Newton's first law; "there is no net force on the airplane and it travels at a constant velocity in a straight line."

Newton's law of acceleration - "a net external force changes the velocity of the object. The drag of the aircraft depends on the square of the velocity. So the drag increases with increased velocity."

Newton's law of Action/Reaction - "As a plane flies, the force of the air hitting the plane is always equal and opposite to the force of the plane pushing against the air. The force generated by the engine pushes against air while the air pushes back with an equal and opposite force."

Hope this helps! god bless :)

Please give me brainliest!

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As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over th

Fofino [41]

Answer:

$5.04\cdot 10^8 A$

Explanation:

The work function of the metal corresponds to the minimum energy needed to extract a photoelectron from the metal. In this case, it is:

$\phi = 3.950\cdot 10^{-19}J$

So, the energy of the incoming photon hitting on the metal must be at least equal to this value.

The energy of a photon is given by

$E=\frac{hc}{\lambda}$

where

h is the Planck's constant

c is the speed of light

$\lambda$ is the wavelength of the photon

Using $E=\phi$ and solving for $\lambda$ , we find the maximum wavelength of the radiation that will eject electrons from the metal:

$\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.950\cdot 10^{-19} J}=5.04\cdot 10^{-7}m$

And since

1 angstrom = $10^{-15}m$

The wavelength in angstroms is

$\lambda=\frac{5.04\cdot 10^{-7} m}{10^{-15} m/A}=5.04\cdot 10^8 A$

3 0

3 years ago

Which planet is represented by the letter "h"?

Mandarinka [93]

The planet is represented as Saturn ♄

7 0

2 years ago

Kinetic energy of an object is equal to

Gnom [1K]

See
K.E=1/2(mass*velocity²)
so option B is the correct answer.
Brainliest pls :-)

8 0

2 years ago

A rocket travels in the x-direction at speed 0.70c with respect to the earth. An experimenter on the rocket observes a collision

marishachu [46]

Answer:

A) The space time coordinate x of the collision in Earth's reference frame is

$x \approx 103,46x10^{9}m$ .

B) The space time coordinate t of the collision in Earth's reference frame is

$t=377,29s$

Explanation:

We are told a rocket travels in the x-direction at speed v=0,70 c (c=299792458 m/s is the exact value of the speed of light) with respect to the Earth. A collision between two comets is observed from the rocket and it is determined that the space time coordinates of the collision are (x',t') = (3.4 x 10¹⁰ m, 190 s).

An event indicates something that occurs at a given location in space and time, in this case the event is the collision between the two comets. We know the space time coordinates of the collision seen from the reference frame of the rocket and we want to find out the space time coordinates in Earth's reference frame.

<em>Lorentz transformation</em>

The Lorentz transformation relates things between two reference frames when one of them is moving with constant velocity with respect to the other. In this case the two reference frames are the Earth and the rocket that is moving with speed v=0,70 c in the x axis.

The Lorentz transformation is

$x'=\frac{x-vt}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y'=y$

$z'=z$

$t'=\frac{t-\frac{v}{c^{2}}x}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

prime coordinates are the ones from the rocket reference frame and unprimed variables are from the Earth's reference frame. Since we want position x and time t in the Earth's frame we need the inverse Lorentz transformation. This can be obtained by replacing v by -v and swapping primed an unprimed variables in the first set of equations

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$y=y'$

$z=z'$

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

First we calculate the expression in the denominator

$\frac{v^{2}}{c^{2}}=\frac{(0,70)^{2}c^{2}}{c^{2}} =(0,70)^{2}$

$\sqrt{1-\frac{v^{2}}{c^{2}}} =0,714$

then we calculate t

$t=\frac{t'+\frac{v}{c^{2}}x'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$t=\frac{190s+\frac{0,70c}{c^{2}}.3,4x10^{10}m}{0,714}$

$t=\frac{190s+\frac{0,70c .3,4x10^{10}m}{299792458\frac{m}{s}}}{0,714}$

$t=\frac{190s+79,388s}{0,714}$

finally we get that

$t=377,29s$

then we calculate x

$x=\frac{x'+vt'}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$x=\frac{3,4x10^{10}m+0,70c.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+0,70.299792458\frac{m}{s}.190s}{0,714}}$

$x=\frac{3,4x10^{10}m+39872396914m}{0,714}}$

$x=\frac{73872396914m}{0,714}}$

$x=103462740775,91m$

finally we get that

$x \approx 103,46x10^{9} m$

5 0

3 years ago

The force of attraction between a -130.0 C and +180.0 C charge is 8.00 N. What is the separation between these two charges in me

kondaur [170]

Answer with Explanation:

The force of attraction between 2 charges of magnitude $q_1,q_2$ separated by a distance 'r' is given by

$F=\frac{1}{4\pi \epsilon _o}\frac{q_1\times q_2}{r^2}=k\frac{q_1\times q_2}{r^2}$

where

$\epsilon _o$ is a constant known as permitivity of free space

$k=9\times 10^{9}Nm^2/C^2$

Applying the given values in the above relation we get

$8=9\times 10^{9}\times \frac{130\times 180}{r^{2}}\\\\r^{2}=\frac{9\times 10^{9}\times 130\times 180}{8}\\\\r^{2}=26325\times 10^{9}\\\\\therefore r=\sqrt{26325\times 10^{9}}=5.131\times 10^{6}meters$

5 0

2 years ago