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Tems11 [23]
3 years ago
11

Determine the potential difference between two charged parallel plates that are 0.10 cm apart and have an electric field strengt

h of 12 V/cm. V
Physics
1 answer:
NikAS [45]3 years ago
5 0
Formula for Electric Field strength , E =  V/d

where V = Voltage in volts, and d = distance of separation in meters.

d = 0.1 cm ,    E = 12 V/cm

E = V/d =  15 / 0.025

V = E*d

V = 12 * 0.1 = 1.2

Potential difference, V = 1.2 V 
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Answer:

Explanation:

Given

Ship A velocity is 40 mph and is traveling 35 west of north

Therefore in 2 hours it will travel 40\times 2=80 miles

thus its position vector after two hours is

r_A=-80sin35\hat{i}+80cos35\hat{j}

similarly B travels with 20 mph and in 2 hours

=20\times 2=40 miles Its position vector[tex]r_B=40sin80\hat{i}+40cos80\hat{j}

Thus distance between A and B  is

r_{AB}=\left ( -40sin80-80sin35\right )\hat{i}+\left ( 80cos35-40cos80\right )\hat{j}

|r_{AB}|=\sqrt{\left ( -40sin80-80sin35\right )^2+\left ( 80cos35-40cos80\right )^2}

|r_{AB}|=103.45 miles

Velocity of A

v_A=-40sin35\hat{i}+40cos35\hat{j}

Velocity of B

v_B=20sin80\hat{i}+20cos80\hat{j}

Velocity of A w.r.t B

v_{AB}=v_A-v_B

v_{AB}=\left ( -20sin80-40sin35\right )\hat{i}+\left ( 40cos35-20cos80\right )\hat{j}

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What are two things that determine our traits? If someone could answer asap, that would be awesome!
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The greenhouse effect keeps earth warmer than it would normally be. True False
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Answer:

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Explanation:

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6 0
3 years ago
A 0.750 kg block is attached to a spring with spring constant 17.5 N/m. While the block is sitting at rest, a student hits it wi
Dmitriy789 [7]

Answer:

a

 A =  0.081 \  m

b

The value is  u =  0.2569 \  m/s

Explanation:

From the question we are told that

   The mass is  m  =  0.750 \ kg

   The spring constant is  k  =  17.5 \  N/m

    The instantaneous speed is  v  =  39.0 \  cm/s= 0.39 \  m/s

    The position consider is  x =  0.750A  meters from equilibrium point

   

Generally from the law of  energy conservation we have that

        The kinetic energy induced by the hammer  =  The energy stored in the spring

So

          \frac{1}{2} *  m * v^2  =  \frac{1}{2}  *  k  *  A^2

Here a is the amplitude of the subsequent oscillations

=>      A =  \sqrt{\frac{m *  v^ 2 }{ k} }

=>      A =  \sqrt{\frac{0.750 *  0.39 ^ 2 }{17.5} }

=>       A =  0.081 \  m

Generally from the law of  energy conservation we have that

The kinetic energy  by the hammer  =  The energy stored in the spring at the point considered   +   The kinetic energy at the considered point

             \frac{1}{2}  * m *  v^2 = \frac{1}{2}  * k x^2 + \frac{1}{2}  * m *  u^2

=>          \frac{1}{2}  * 0.750 *  0.39^2 = \frac{1}{2}  * 17.5* 0.750(0.081 )^2 + \frac{1}{2}  * 0.750 *  u^2

=>          u =  0.2569 \  m/s

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Answer:

608

Explanation:

Trust me

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