Which is located furthest from our sun?

Newton’s Laws of Motion are absolute in classical physics. One example that uses all three laws simultaneously is the firing of

Debora [2.8K]

I think that by "Classical physics" is meant low speed things. By low speed, I think is meant speed far below very roughly half the speed of light, so that Relativistic, special or general, effects can be ignored. Or at least it is hoped that they can be ignored.
Fire extinguishers and rockets get propelled by forcing out large amounts of material (gases under very high pressure) through a nozzle, and the RECOIL from that propels something forward. So, if the action is the ejection of material, the reaction (recoil) is the ejector moving along the same line in the other direction. And that's an example of Newton's third law.
Given a propulsion system, the magnitude of the force recoiling on the ejector will change the momentum of the ejector, often written as the equation F=ma where F is the force, m is the mass being accelerated, and a being the acceleration.
Just as something will stay still until it is moved - inertia - so once set in uniform motion in a straight line, the thing will continue in that motion, theoretically for ever or until something alters its momentum. Newton's first law is to the effect of "every body continues in a state of rest or uniform motion in a straight line unless acted on by a resultant external force". Which, I think, is where the concept of inertia stems from.
I think that the above mostly tcuches on the 3 laws.Any more help needed, please ask.

6 0

2 years ago

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A 20-cm long solenoid consists of 100 turns of a coil of radius r = 3.0 cm. A current of Io in the coiled wire produces a magnet

Romashka-Z-Leto [24]

Answer:

vi) Double the current in the wire, and double the number of turns in the 20-cm long solenoid

Explanation:

The magnetic field inside the solenoid and the current flowing in the coil of solenoid are related to each other by the following equation

B₀=μ₀nI₀

Where,

B₀ is the magnetic field in the middle of solenoid

n is the number of turns in the coil of solenoid

I₀ is the current flowing in the coil of solenoid

In the above equation, as μ₀ is a constant so the magnetic field will be directly proportional to the number of turns multiplied by the current. So, changing the radius of the coil or length of the coil will have no effect on the magnetic field.

As we have to increase the magnetic field by 4 times, we need to double the current as well as the number of turns as mentioned in the option vi.

3 0

3 years ago

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An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.

Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }$

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }$

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }$

$B = \boxed{7.07 *10^{-10} T}$

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

$B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}$

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

$\boxed{B = 0 T}$

3 0

2 years ago

A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.

kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight= $\sqrt{\frac{2H}{g}}$

substituting value

= $\sqrt{\frac{2*25}{9.8}}$
=2.26seconds

<h3>the ball is in flight before striking the ground for 2.26seconds.</h3>

b. How far from the building does the ball strike the ground?

Horizontal range=?

we have

Horizontal range=u* $\sqrt{\frac{2H}{g}}$

=8.25*2.26
=18.63m

<h3>The ball strikes 18.63m far from building. </h3>

7 0

2 years ago

_____ = force × distance A. Work B. Velocity C. Pressure D. Momentum

gavmur [86]

Work is the correct answer

8 0

3 years ago