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vladimir1956 [14]
2 years ago
9

What is the magnitude of electric field between 2 charged plates that are separated by a distance of 2.4cm, if the voltage acros

s the plates is 200V? Answer in N/C.
How fast would an electron be moving when it hit the positive plate if it launched across the gap from the negative plate? answer in m/s
Physics
1 answer:
Radda [10]2 years ago
7 0

Answer:

V = E * d    

E = (200 J/C) / .024 m =   8330 J / C-m  

1/2 m v^2 = V q      potential energy of electron

v^2 = 2 * 200 J/C * 1.6E-19 C / 9.11 E-31 kg

v^2 = 400 * 1.6E-19 / 9.11 E-31   N-m  / kg

v = 8.4E6 m/s

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HELP ME Please!!!!!!<br>Select ALL that apply
k0ka [10]

Answer:

Increase the number of coils.

Explanation:

By increasing the number of coils we can increase the electromagnet.

8 0
3 years ago
An electromagnetic wave in a vacuum traveling in the +x direction generated by a variable source initially has a wavelength λ of
djverab [1.8K]

Answer:

B = E/c = 14.04T₁ = 11 pT

Explanation:

We know c = E/B where E = maximum electric field = 3.30 × 10⁻³ V/m, B = maximum magnetic field and c = speed of light

B = E/c also c = fλ = λ/T where λ = wavelength = 235 μm = 235 × 10⁻⁶ m and T = period

c = λ₁/T₁ = λ₂/T₂    T₂ = 2.8T₁ where λ₁,λ₂ are the initial and final wavelengths and T₁,T₂ are the initial and final periods.

T₁ =  λ₁/c = 235 × 10⁻⁶ m/3 × 10⁸ m/s = 7.833 × 10⁻¹³ s = 0.7833 ps

T₂ = 2.8T₁ = 2.8 × 7.833 × 10⁻¹³ s = 21.93 × 10⁻¹³ s = 2.193 ps

λ₁/T₁ = λ₂/2.8T₁

λ₂ = 2.8λ₁ = 2.8 × 235 μm = 658 μm

c = λ₂/T₂ = 2.8λ₁/2.8T₁ = λ₁/T₁ , since the speed of light c is constant.

B = E/c = E/λ₁/T₁ = ET₁/λ₁

B = ET₁/λ₁ = 3.30 × 10⁻³ V/m × T₁/235 × 10⁻⁶ m = 14.04T₁  Tesla

B = 14.04 × 7.833 × 10⁻¹³ s = 10.99 × 10⁻¹² T ≅ 11 pT

5 0
4 years ago
At night, large bodies of water release heat to the atmosphere quickly. True False
Nikolay [14]

Answer:

false I hope. have a good day!

3 0
2 years ago
Someone please help me
mars1129 [50]

The answer is parallel

8 0
3 years ago
Guys please help me ​
Likurg_2 [28]

Answer:

1)t=2.26\: s

2)S=33.9\: m

3)v=26.77\: m/s

4)\alpha=55.92

Explanation:

1)

We can use the following equation:

y_{f}=y_{0}+v_{iy}t-0.5*g*t^{2}

Here, the initial velocity in the y-direction is zero, the final y position is zero and the initial y position is 25 m.

0=25-0.5*9.81*t^{2}

t=2.26\: s

2)

The equation of the motion in the x-direction is:

v_{ix}=\frac{S}{t}

15=\frac{S}{2.26}

S=33.9\: m

3)

The velocity in the y-direction of the stone will be:

v_{fy}=v_{iy}-gt

v_{fy}=0-(9.81*2.26)

v_{fy}=-22.17\: m/s

Now, the velocity in the x-direction is 15 m/s then the velocity will be:

v=\sqrt{v_{x}^{2}+v_{fy}^{2}}=\sqrt{15^{2}+(-22.17)^{2}}

v=26.77\: m/s

4)

The angle of this velocity is:

tan(\alpha)=\frac{22.17}{15}

\alpha=tan^{-1}(\frac{22.17}{15})

\alpha=55.92

Then α=55.92° negative from the x-direction.

I hope it helps you!

6 0
3 years ago
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