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Klio2033 [76]
3 years ago
9

An electron is released from rest in a uniform electric field and accelerates to the north at a rate of 115 m/s squared. What ar

e the magnitude and direction of the electric field?
Physics
1 answer:
Iteru [2.4K]3 years ago
6 0
The direction of the electric field would be south. 

qE/m = 115 
<span>       E = 115*m/q </span>
<span>           = 115 * 9.1 * 10^(-31) / 1.67*10^(-19) </span>
<span>           = 762.87 * 10^(-12) </span>
<span>           = 6.27 x 10^-10 N/C
</span>
Hope this answers the question. Have a nice day. Feel free to ask more questions.
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Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball
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Answer:

k = 4422.35  KN/m

Explanation:

Given that

Frequency ,f= 29 Hz

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Natural frequency ω

ω = 2 π f

We also know that for spring mass system

ω ² m =k        

k=Spring constant

So we can say that

( 2 π f)² =  m k

By putting the values

(2 x π x 29)² = 7.5 x 10⁻³  k

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k = 4422.35  KN/m

Therefore spring constant will be 4422.35  KN/m

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A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the a
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Solution :

Let $m_1=m_2=4$ kg

$u_1 = 5$ m/s

Let $v_1$ and $v_2$ are the speeds of the disk $m_1$ and $m_2$  after the collision.

So applying conservation of momentum in the y-direction,

$0=m_1 .v_1_y -m_2 .v_2_y $

$v_1_y = v_2_y$

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$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$

$v_2=1.732 \times v_1$

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

$m_1.u_1=m_1.v_1_x+m_2.v_2_x$

$u_1=v_1_x+v_2_x$

$5=v_1. \cos 60 + v_2 . \cos 30$

$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$

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