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ioda
3 years ago
11

Two parallel plates are 1 cm apart and are connected to a 500 V source. What force will be exerted on a single electron half way

between the plates?
Physics
1 answer:
VladimirAG [237]3 years ago
3 0

Answer: 8*10^-15 N

Explanation: In order to calculate the force applied on an electron in the middle of the two planes at 500 V we know that,  F=q*E

The electric field between  the plates is given by:

E = ΔV/d = 500 V/0.01 m=5*10^3 N/C

the force applied to the electron is: F=e*E=8*10^-15 N

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A certain spring exerts a nonlinear force given by F(x) = −60x − 18x2 , where x is in meters and F is in newtons. A 0.90-kg bloc
tia_tia [17]

Answer:

a)  W = 6.75 J and b) v = 3.87 m / s

Explanation:

a) In the problem the force is nonlinear and they ask us for work, so we must use it's definition

      W = ∫ F. dx

Bold indicates vectors.  In a spring the force is applied in the direction of movement, whereby the scalar product is reduced to the ordinary product

     W = ∫ F dx

We replace and integrate

    W = ∫ (-60 x - 18 x²) dx

    W = -60 x²/2 -18 x³/3

Let's evaluate between the integration limits, lower W = 0 for x = -0.50 m, to the upper limit W = W for x = 0 m

    W = -30 [0- (-0.50) 2] -6 [0 - (- 0.50) 3]

    W = 7.5 - 0.75

    W = 6.75 J

b)  Work is equal to the variation of kinetic energy

    W = ΔK

    W = ΔK = ½ m v² -0

    v =√ 2W/m

    v = √(2 6.75/ 0.90)

    v = 3.87 m / s

8 0
3 years ago
A stone is thrown vertically upward with a speed of 18.0 . (a)How fast is it moving when it reaches a height of 11.0 ? (b)How lo
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For the first part, we are looking for Vf when dy=11.0
Upward is positive, downward is negative. 
So <span>Vf = square root [2(-9.8)(11.0) + (18.0)^2] </span>
<span>Vf = 10.4 m/s your answer is correct. 

For the part b, t is equals to the time took to reach and dy is equals to 11.0
you did, </span>11= 18t m/s-(1/2) 9.8t^2 then <span>-11 + 18t- 9.8t^2. By quadratic formula, for the way down the answer is 2.9 s while on it's way up, the answer is 0.77 s</span><span>
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3 years ago
Read 2 more answers
A train 471 m long is moving on a straight track with a speed of 75.1 km/h. The engineer applies the brakes at a crossing, and l
hram777 [196]

Answer:

t = 37.6 s

Explanation:

As we know that train is initially moving with the speed

v_i = 75.1 km/h

now we know that

v_i = 20.86 m/s

now the final speed of the train when it crossed the crossing

v_f = 15 km/h

v_f = 4.17 m/s

now we can use kinematics here

v_f^2 - v_i^2 = 2 a d

4.17^2 - 20.86^2 = 2 a(471)

a = -0.44 m/s^2

Now the time to cross that junction is given as

v_f - v_i = at

4.17 - 20.86 = a(-0.44)

t = 37.6 s

4 0
3 years ago
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