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prohojiy [21]
3 years ago
12

Find an expression for the electric field e⃗ at the center of the semicircle. hint: a small piece of arc length δs spans a small

angle δθ=δs /r, where r is the radius.
Physics
1 answer:
ahrayia [7]3 years ago
8 0
Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by 

<span>C = 2L = 2*pi*R ---> R = L/pi </span>

<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>

<span>we can define a small charge dq as </span>

<span>dq = l*ds = l*R*da </span>

<span>So the electric field can be written as: </span>

<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>

<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>

<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>

<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>
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The earth has a vertical electric field at the surface, pointing down, that averages 100 N/C. This field is maintained by variou
zlopas [31]

Answer:

q = 3.6 10⁵  C

Explanation:

To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth

          r = 6 , 37 106 m

          E = k q / r²

          q = E r² / k

          q = \frac{100 \ (6.37 \ 10^6)^2}{9 \ 10^9}

          q = 4.5 10⁵ C

Now let's calculate the charge on the planet with E = 222 N / c and radius

           r = 0.6 r_ Earth

           r = 0.6 6.37 10⁶ = 3.822 10⁶ m

           E = k q / r²

            q = E r² / k

            q = \frac{222 (3.822 \ 10^6)^2}{ 9 \ 10^9}

            q = 3.6 10⁵  C

4 0
3 years ago
Greg is boiling water on the stove for some noodles. what object can he safely hold while draining the water from the noodles af
zmey [24]
In order for Greg to safely drain the water out of the noodles, he should use potholders or any thing that is does not conduct heat or transfer heat. Some pots are also equipped with handles that are made of plastics for safely transferring of its content to another container. 
5 0
3 years ago
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0
3 years ago
What is the Sl unit for speed
Akimi4 [234]

Answer: meter per second

Explanation: meter per second

Speed has the dimensions of distance divided by time. The SI unit of speed is the meter per second, but the most common unit of speed in everyday usage is the kilometer per hour or, in the US and the UK, miles per hour. For air and marine travel the knot is commonly used.

3 0
3 years ago
how much force is needed to cause a 15 kilogram bike to accelerate at a rate of 10 meters per second?
egoroff_w [7]
F = m*a, mass times acceleration.

F = 15*10 = 150 N
8 0
3 years ago
Read 2 more answers
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