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rusak2 [61]
2 years ago
12

Rectangular plate has a voltage of +180V and plate to the first plate) has a 'voltage of -5V. Determie another rectangular plate

(parallel to between these two plates? The distance of the field strength at any point separation of the plates is finite, say 8.6mm.​
Physics
1 answer:
3241004551 [841]2 years ago
4 0

For a Rectangular plate has a voltage of +180V and a 'voltage of -5V. , the second plate has the Electric field mathematically given as

E=21.5*10^3v/m

<h3>What is the field strength?</h3>

Generally, the equation for the Electric field   is mathematically given as

E=v/d

Where

v={180-(-5)}v

v=185v

Therefore

E=185/8.6*10^{-3}

E=21.5*10^3v/m

In conclusion, Electric field

E=21.5*10^3v/m

Read more about electric field

brainly.com/question/9383604

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A ball is thrown with a speed of 20 m/s at an angle of 40o above the horizontal from the top of a 22-m tall building.
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Answer:

(a). The distance is  49.79 m.

(b). The speed of the ball is 24.39 m/s.

Explanation:

Given that,

Speed = 20 m/s

Angle = 40°

Height = 22 m

Time = 3.25 sec

(a). We need to calculate the distance

Using formula of distance

d=u\cos\theta\times t

Put the value into the formula

d=20\cos40\times3.25

d=49.79\ m

(b). We need to calculate the horizontal velocity

Using formula of velocity

v_{x}=u\cos\theta

Put the value into the formula

v_{x}=20\times\cos40

v_{x}=15.3\ m/s

We need to calculate the vertical velocity

Using equation of motion

v_{y}=u\sin\theta-gt

Put the value into the formula

v_{y}=20\sin40-9.8\times3.25

v_{y}=-19\ m/s

Negative sign shows the opposite direction.

We need to calculate the speed of ball

Using formula of speed

v=\sqrt{v_{x}^2+v_{y}^2}

v=\sqrt{(15.3)^2+(19)^2}

v=24.39\ m/s

Hence, (a). The distance is  49.79 m.

(b). The speed of the ball is 24.39 m/s.

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Anino acids are important to the body because they build
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Answer:

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Sam lifted his backpack with 5 Newtons of force a total of 400
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2000joules

Explanation:

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A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t
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Answer:

a) W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

b) W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

c) V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

d)  d_1 =0.183m or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J

Part b

For this case first we can convert the spring constant to N/m like this:

2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}

And the work donde by the spring on this case is given by:

W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i

And if we solve for the initial velocity we got:

V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

-1/2mV^2_i = W_g + W_{spring}

And replacing we got:

-1/2mV^2_i =mg d_1 -1/2 k d^2_1

And we can put the terms like this:

\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0

If we multiply all the equation by 2 we got:

k d^2_1 -2 mg d_1 -m V^2_i =0

Now we can replace the values and we got:

200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0

200 d^2_1 -2.058 d_1 -6.3525=0

And solving the quadratic equation we got that the solution for d_1 =0.183m or 18.3 cm because the negative solution not make sense.

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Answer:

They are conductors/conductive. Materials that can transfer thermal energy well are conductive.

Explanation:

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