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3 years ago

55 Points ££which state of matter can undergo three types of thermal expansion

Physics

2 answers:

bija089 [108]3 years ago

7 0

Answer:

The three types of thermal expansions are Linear expansion , Superficial expansion and Cubical expansion

CaHeK987 [17]3 years ago

5 0

In my knowledge 

Your answer is Gases

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What causes an electrical current in a wire? Question 1 options: Electric fields cause atoms to move in a wire Electrons build u

Mademuasel [1]

The cause of electric current in a wire is that there is electron flow due to electrical attraction and repulsion.

What is Electric current?

The rate of flow of electric charge through a point is known as electric current. The electric current is nothing but the flow of electrons in general.

Consider a space, when electrons of charge (e) is flowing through the space in a given time interval t. Then,

I = e/t

here,

I is the electric current.

Moreover, there is a flow of electrons in a wire due to a potential difference existing at the circuit. Due to that, electrons always flow from high electric potential to low electric potential.

Thus, we can conclude that the cause of electric current in a wire is that there is electron flow due to electrical attraction and repulsion.

Learn more about the electric current here:

brainly.com/question/7643273

5 0

2 years ago

Which color in the rainbow has the shortest wavelength?

klio [65]

Answer:

violet

Explanation:

violet has shortest wavelength

8 0

2 years ago

What is the de Broglie wavelength for a proton with energy 50 keV? Due to the limitations of Canvas, please give the wavelength

ad-work [718]

Answer:

1.2826 x 10^-13 m

Explanation:

$\lambda = \frac{h}{\sqrt{2 m K}}$

Here, k be the kinetic energy and m be the mass

K = 50 KeV = 50 x 1.6 x 10^-16 J = 80 x 10^-16 J

m = 1.67 x 10^-27 kg

$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67\times 10^{-27}\times 80\times 10^{-16}}}$

λ = 1.2826 x 10^-13 m

6 0

3 years ago

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.

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

Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac

Serjik [45]

The strength of the gravitational field is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
$r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m$

And now we can use the previous equation to calculate the field strength at that altitude:
$g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2} = 8.95 m/s^2$

And we can see this value is a bit less than the gravitational strength at the surface, which is $g_s = 9.81 m/s^2$ .

4 0

3 years ago

55 Points ££which state of matter can undergo three types of thermal expansion​

55 Points ££which state of matter can undergo three types of thermal expansion