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Sergeeva-Olga [200]
3 years ago
13

Respostas 100 J 200 J 300 J 400 J 500 J

Physics
2 answers:
Aneli [31]3 years ago
7 0

Answer:

Explanation:

400 J

:> :)

Fiesta28 [93]3 years ago
5 0

400J

hope it's help

now is my exam and your

pls welcome to me

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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
An object is placed at 1.5 m from a convex lens with a focal length of 1.0 m.
AleksAgata [21]
A. Use the thin lens equation to determine the image distance from the lens. Is the correct answer because I have no idea why
8 0
3 years ago
A 0.5 kg ball moves in a circle that is 0.4 m in radius at a speed of 4.0 m/s. Calculate the force exerted on the ball.
blsea [12.9K]

Answer:

Explanation:

Given a ball of mass m= 0.5kg

The ball moves in as circle of radius r= 0.4m

Speed of the ball is v = 4m/s

Centripetal force is exerted on ball and it is given as

Fc = m•ac

ac is centripetal acceleration and it is given as

ac = v²/r

Then,

Fc = mv²/r

Fc = 0.5 × 4²/0.4

Fc = 20N.

The force exerted on the ball is 20N

5 0
3 years ago
A singly charged ion of 7Li (an isotope of lithium which lost only one electron) has a mass of 1.16 ×10^-26 kg. It is accelerate
MaRussiya [10]

Explanation:

It is given that,

Mass of lithium, m=1.16\times 10^{-26}\ kg

It is accelerated through a potential difference, V = 224 V

Uniform magnetic field, B = 0.724 T

Applying the conservation of energy as :

\dfrac{1}{2}mv^2=qV

v=\sqrt{\dfrac{2qV}{m}}

q is the charge on an electron

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 224\ V}{1.16\times 10^{-26}\ kg}}

v = 78608.58 m/s

v=7.86\times 10^4\ m/s

To find the radius of the ion's path in the magnetic field. The centripetal force is balanced by the magnetic force as :

qvB=\dfrac{mv^2}{r}

r=\dfrac{mv}{qB}

r=\dfrac{1.16\times 10^{-26}\ kg\times 7.86\times 10^4\ m/s}{1.6\times 10^{-19}\ C\times 0.724\ T}

r = 0.0078 meters

So, the radius of the path of the ion is 0.0078 meters. Hence, this is the required solution.

4 0
3 years ago
10 N pushes a 10 kg crate to the right. Determine the acceleration of the crate.
oksano4ka [1.4K]
The rate of acceleration of the crate would be 1 m/s^2 because the equation for force is F=ma and when you plug in your numbers you get 10=10a so a=1
8 0
3 years ago
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