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2 years ago

Can u pls help me ........

Physics

1 answer:

OLEGan [10]2 years ago

5 0

Answer:

solidification

Explanation:

a liquid becoming a solid, is called solidification. For any pure substance, the temperature at which melting occurs—known as the melting point—is a characteristic of that substance.

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An electron moving in a direction perpendicular to a uniform magnetic field at a speed of 1.6 107 m/s undergoes an acceleration

umka2103 [35]

Answer:

B = 0.024T positive z-direction

Explanation:

In this case you consider that the direction of the motion of the electron, and the direction of the magnetic field are perpendicular.

The magnitude of the magnetic force exerted on the electron is given by the following formula:

$F=qvB$ (1)

q: charge of the electron = 1.6*10^-19 C

v: speed of the electron = 1.6*10^7 m/s

B: magnitude of the magnetic field = ?

By the Newton second law you also have that the magnetic force is equal to:

$F=qvB=ma$ (2)

m: mass of the electron = 9.1*10^-31 kg

a: acceleration of the electron = 7.0*10^16 m/s^2

You solve for B from the equation (2):

$B=\frac{ma}{qv}\\\\B=\frac{(9.1*10^{-31}kg)(7.0*10^{16}m/s^2)}{(1.6*10^{-19}C)(1.6*10^7m/s)}\\\\B=0.024T$

The direction of the magnetic field is found by using the right hand rule.

The electron moves upward (+^j). To obtain a magnetic forces points to the positive x-direction (+^i), the direction of the magnetic field has to be to the positive z-direction (^k). In fact, you have:

-^j X ^i = ^k

Where the minus sign of the ^j is because of the negative charge of the electron.

Then, the magnitude of the magnetic field is 0.024T and its direction is in the positive z-direction

8 0

2 years ago

What is the new acceleration (in m/s/s)?

Delicious77 [7]

The new acceleration is $108 m/s^2$

Explanation:

We can answer this problem by applying Newton's second law, which states that:

$F=ma$

where

F is the net force on an object

m is the mass of the object

a is its acceleration

The equation can be rewritten as

$a=\frac{F}{m}$

In this problem, the initial acceleration is

$a=18.0 m/s^2$

Later:

- The net force is tripled: $F'=3F$

- The mass is halved: $m'=\frac{m}{2}$

Therefore, the new acceleration is:

$a'=\frac{F'}{m'}=\frac{3F}{m/2}=6\frac{F}{m}=6a$

which means that the new acceleration is 6 times the original acceleration, therefore

$a'=6(18)=108 m/s^2$

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0

2 years ago

When an electric stove element is hot enough, it gives off a dull red glow. When it cools to the point that it no longer glows,

DochEvi [55]

Answer:

It will have a longer wavelength

Explanation:

When an electric stove is hot and gives dull red glow. a part of the energy dissipated is emitted as visible light and part as infrared radiation in the form of heat. When the stove cools down, and no longer glows all the energy is now in the form of infrared radiation.In the electromagnetic spectrum infrared rays have a higher wavelength than visible light. Hence for the reason the radiation will have a higher wavelength since visible light is cut off.

4 0

2 years ago

Consider the four quantum numbers of an electron in an atom, n, l, ml, and ms. The energy of an electron in an isolated atom dep

Effectus [21]

Answer:

The energy of an electron in an isolated atom depends on b. n only.

Explanation:

The quantum number n, known as the principal quantum number represents the relative overall energy of each orbital.

The sets of orbitals with the same n value are often referred to as an electron shell, in an isolated atom all electrons in a subshell have exactly the same level of energy.

The principal quantum number comes from the solution of the Schrödinger wave equation, which describes energy in eigenstates $E_n$ , and for the case of an hydrogen atom we have:

$E_n=-\cfrac{13.6}{n^2}\, eV$

Thus for each value of n we can describe the orbital and the energy corresponding to each electron on such orbital.

6 0

2 years ago

Block B is attached to a massless string of length L = 1 m and is free to rotate as a pendulum. The speed of block A after the c

Amanda [17]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The minimum velocity of A is $v_A= 4m/s$

Explanation:

From the question we are told that

The length of the string is $L = 1m$

The initial speed of block A is $u_A$

The final speed of block A is $v_A = \frac{1}{2}u_A$

The initial speed of block B is $u_B = 0$

The mass of block A is $m_A = 7kg$ gh

The mass of block B is $m_B = 2 kg$

According to the principle of conservation of momentum

$m_A u_A + m_B u_B = m_Bv_B + m_A \frac{u_A}{2}$

Since block B at initial is at rest

$m_A u_A = m_Bv_B + m_A \frac{u_A}{2}$

$m_A u_A - m_A \frac{u_A}{2} = m_Bv_B$

$m_A \frac{u_A}{2} = m_Bv_B$

making $v_B$ the subject of the formula

$v_B =m_A \frac{u_A}{2 m_B}$

Substituting values

$v_B =\frac{7 u_A}{4}$

This $v__B$ is the velocity at bottom of the vertical circle just at the collision with mass A

Assuming that block B is swing through the vertical circle(shown on the second uploaded image ) with an angular velocity of $v__B'$ at the top of the vertical circle

The angular centripetal acceleration would be mathematically represented

$a= \frac{v^2_{B}'}{L}$

Note that this acceleration would be toward the center of the circle

Now the forces acting at the top of the circle can be represented mathematically as

$T + mg = m \frac{v^2_{B}'}{L}$

Where T is the tension on the string

According to the law of energy conservation

The energy at bottom of the vertical circle = The energy at the top of

the vertical circle

This can be mathematically represented as

$\frac{1}{2} m(v_B)^2 = \frac{1}{2} mv^2_B' + mg 2L$

From above

$(T + mg) L = m v^2_{B}'$

Substitute this into above equation

$\frac{1}{2} m(\frac{7 v_A}{4} )^2 = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = \frac{1}{2} (T + mg) L + mg 2L$

$\frac{49 mv_A^2}{16} = T + 5mgL$

The value of velocity of block A needed to cause B be to swing through a complete vertical circle is would be minimum when tension on the string due to the weight of B is zero

This is mathematically represented as

$\frac{49 mv_A^2}{16} = 5mgL$

making $v_A$ the subject

$v_A = \sqrt{\frac{80mgL}{49m} }$

substituting values

$v_A = \sqrt{\frac{80* 9.8 *1}{49} }$

$v_A= 4m/s$

6 0

2 years ago