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3 years ago

In a transverse wave ,if you increase of the wave you will get what kind of wavelength

1 answer:

5 0

Using the formula v = f λ,
where v is the velocity of the wave, f is the frequency of the wave and λ is the wavelength of the wave. 

assuming that the velocity of the wave remains constant, by increasing the wavelength the frequency must decrease.

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An alpha particle (Î±), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elec

vaieri [72.5K]

Answer:

Speed = 575 m/s

Mechanical energy is conserved in electrostatic, magnetic and gravitational forces.

Explanation:

Given :

Potential difference, U = $-3.45 \times 10^{-3} \ V$

Mass of the alpha particle, $m_{\alpha} = 6.68 \times 10^{-27} \ kg$

Charge of the alpha particle is, $q_{\alpha} = 3.20 \times 10^{-19} \ C$

So the potential difference for the alpha particle when it is accelerated through the potential difference is

$U=\Delta Vq_{\alpha}$

And the kinetic energy gained by the alpha particle is

$K.E. =\frac{1}{2}m_{\alpha}v_{\alpha}^2$

From the law of conservation of energy, we get

$K.E. = U$

$\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$

$v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$

$v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$

$v_{\alpha} \approx 575 \ m/s$

The mechanical energy is conserved in the presence of the following conservative forces :

-- electrostatic forces

-- magnetic forces

-- gravitational forces

5 0

3 years ago

The purpose of the little Albert experiment?

HACTEHA [7]

Answer:

The aim of Watson and Rayner was to condition a phobia in an emotionally stable child.

Explanation:

Does this help?

3 0

3 years ago

Why is the physics of hssc2 so difficult?

Ivan

Could be easy for some people and hard for some people.

8 0

3 years ago

Read 2 more answers

An avant-garde composer wants to use the Doppler effect in his new opera. As the soprano sings, he wants a large bat to fly towa

Allisa [31]

Answer:

v’= 9.74 m / s

Explanation:

The Doppler effect is due to the relative movement of the sound source and the receiver of the sound, in this case we must perform the exercise in two steps, the first to find the frequency that the bat hears and then the frequency that the audience hears that also It is sitting.

Frequency shift heard by the murciela, in case the source is still and the observer (bat) moves closer

f₁ ’= f₀ (v + v₀)/v

Frequency shift emitted by the speaker in the bat, in this case the source is moving away from the observer (public sitting) that is at rest

f₂’= f₁’ v/(v - vs)

Note that in this case the bat is observant in one case and emitter in the other, called its velocity v’

v’= vo = vs

Let's replace

f₂’= f₀ (v + v’)/v v/(v -v ’)

f₂’= f₀ (v + v’) / (v -v ’)

(v –v’ ) f₂’ / f₀ = v + v ’

v’ (1+ f₂’ /f₀) = v (f₂’/fo - 1)

v’ (1 + 1.059) = 340 (1.059 - 1)

v’= 20.06 / 2.059

v’= 9.74 m / s

6 0

3 years ago

What is the change in potential energy of a 2.00 nC test charge, Uelectric, b - Uelectric, a, as it is moved from point a at x

lyudmila [28]

The question is incomplete. Here is the complete question.

A uniform electric field of 2kN/C points in the +x-direction.

(a) What is the change in potential energy of a +2.00nC test charge, $U_{electric,b} - U_{electric,a}$ as it is moved from point a at x = -30.0 cm to point b at x = +50.0 cm?

(b) The same test charge is released from rest at point a. What is the kinetic energy when it passes through point b?

(c) If a negative charge instead of a positive charge were used in this problem, qualitatively, how would your answers change?

Answer: (a) ΔU = 3.2× $10^{-6}$ J

(b) KE = 2× $10^{-6}$ J

Explanation: Potential Energy (U) is the amount of work done due to its position or condition and its unit is Joule (J). Kinetic Energy (KE) is the ability to do work by virtue of velocity and the unit is also (J). Mechanical Energy is the sum of Potential and Kinetic Energies of a system.

(a) Related to electricity, Potential Energy can be calculated as:

ΔU = Eqd

where E is the electric field (in N/C);

q is the charge (in C);

d is the distance between plaques (in m);

For a at x = - 30cm and b at x = 50 cm:

E = 2× $10^{3}$ N/C

q = 2× $10^{-9}$ C

d = 50 - (-30) = 80× $10^{-2}$ = 8× $10^{-1}$ m

ΔU = $U_{electric,b} - U_{electric,a}$ = Eqd

$U_{electric,b} - U_{electric,a}$ = 2× $10^{3}$ . 2× $10^{-9}$ . 8× $10^{-1}$

ΔU = 3.2× $10^{-6}$ J

(b) Mechanical Energy is constant, so:

$KE_{i} + U_{i} = KE_{f} + U_{f}$

Since the initial position is zero and there is no initial kinetic energy:

$KE_{f} = - U{f}$

$KE_{f} =$ - (2× $10^{3}$ . 2× $10^{-9}$ . 5× $10^{-1}$ )

$KE_{f} = - 2.10^{-6}$ J

(c) If the charge is negative, electric field does positive work, which diminishes the potential energy. The charge flows from the negative side towards the positive side and stays, not doing anything.

8 0

3 years ago