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3 years ago

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If you put a total of 8.05×106×106 electrons on an intially electrically neutral wire of length 1.03 m, what is the magnitude of

the electric field a perpendicular distance of 0.201 m away from the center of the wire?

1 answer:

olga_2 [115]3 years ago

6 0

Answer:

The magnitude of the electric field is 0.1108 N/C

Explanation:

Given;

number of electrons, e = 8.05 x 10⁶

length of the wire, L = 1.03 m

distance of the field from the center of the wire, r = 0.201 m

Charge of the electron;

Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)

Q = 1.2896 x 10⁻¹² C

Linear charge density;

λ = Q / L

λ = (1.2896 x 10⁻¹² C) / (1.03 m)

λ = 1.252 x 10⁻¹² C/m

The magnitude of electric field at r = 0.201 m;

$E = (\frac{1}{4 \pi \epsilon_o} )\frac{ 2 \lambda}{r} \\\\E = k \frac{ 2 \lambda}{r}\\\\E = (8.89*10^9)*\frac{2*1.252*10^{-12}}{0.201} \\\\E = 0.1108 \ N/C$

Therefore, the magnitude of the electric field is 0.1108 N/C

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A 150 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. Calculate the magn

xenn [34]

Answer:

T = 2010 N

Explanation:

m = mass of the uniform beam = 150 kg

Force of gravity acting on the beam at its center is given as

W = mg

W = 150 x 9.8

W = 1470 N

T = Tension force in the wire

θ = angle made by the wire with the horizontal = 47° deg

L = length of the beam

From the figure,

AC = L

BC = L/2

From the figure, using equilibrium of torque about point C

T (AC) Sin47 = W (BC)

T L Sin47 = W (L/2)

T Sin47 = W/2

T Sin47 = 1470

T = 2010 N

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3 years ago

You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1

drek231 [11]

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

The given parameters;

<em>Mass of the pendulum, = M </em>
<em>Length of the pendulum, = L</em>
<em>Initial angular speed, </em> $\omega _i$ <em> = 1 rad/s</em>

The moment of inertia of the rod about the end is given as;

$I_i = \frac{1}{3} ML^2$

The moment of inertia of the rod between the middle and the end is calculated as;

$I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}$

Apply the principle of conservation of angular momentum as shown below;

$I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f$

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129

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3 years ago

En un m.A.S. La amplitud tiene un valor de 10 centimetros y el periodo es de 2 segundos calcular el valor de la velocidad de 0.8

Ivanshal [37]

Answer:

v1=18.46m/s

v2=29.8cm/s

Explanation:

We know that

$A=10cm\\T=2s$

the equation of the motion is

$x=Acos(\omega t)\\$

we can calculate w by using

$\omega=\frac{2\pi}{T}=\frac{2\pi}{2}=\pi$

Hence, we have that

$x=10cm*cos(\pi t)\\$

the speed will be

$v=-\omega*Asin(\omega t)\\|v(0.8)|=|\pi*10cm*sin(\pi *0.8)|=18.46\frac{cm}{s}\\|v(1.4)|=|\pi*10cm*sin(\pi *1.4)|=29.8\frac{cm}{s}$

hope this helps!

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3 years ago

A 6 kg weight is lifted off the ground to a height that gives it 70.56 j of gravitational potential energy. what is its height?

Marina CMI [18]

GPE= 70.56 J -------------------> GPE= mgh-------------> X= height
70.56 = 6(kg) * 9.8(m/s/s) * X
70.56 = 58.8X
70.56/58.8= 58.8X/58.8
X= 1.2
The height is 1.2 feet or meters (whatever unit you are using in this problem)

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3 years ago

Read 2 more answers

A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff

strojnjashka [21]

(a) $6.43\cdot 10^5 J$

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

$E=K+U$

The initial kinetic energy is:

$K=\frac{1}{2}mv^2$

where m = 58.0 kg is the mass of the projectile and $v=140 m/s$ is the initial speed. Substituting,

$K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J$

The initial potential energy is given by

$U=mgh$

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

$U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J$

So, the initial mechanical energy is

$E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J$

(b) $-1.67 \cdot 10^5 J$

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

$K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J$

while the potential energy is

$U=(58.0 kg)(9.8 m/s^2)(336 m)=1.91\cdot 10^5 J$

So, the mechanical energy is

$E=K+U=2.85\cdot 10^5 J+1.91 \cdot 10^5 J=4.76\cdot 10^5 J$

And the work done by friction is equal to the difference between the initial mechanical energy of the projectile, and the new mechanical energy:

$W=E_f-E_i=4.76\cdot 10^5 J-6.43\cdot 10^5 J=-1.67 \cdot 10^5 J$

And the work is negative because air friction is opposite to the direction of motion of the projectile.

(c) 88.1 m/s

The work done by air friction when the projectile goes down is one and a half times (which means 1.5 times) the work done when it is going up, so:

$W=(1.5)(-1.67\cdot 10^5 J)=-2.51\cdot 10^5 J$

When the projectile hits the ground, its potential energy is zero, because the heigth is zero: h=0, U=0. So, the projectile has only kinetic energy:

E = K

The final mechanical energy of the projectile will be the mechanical energy at the point of maximum height plus the work done by friction:

$E_f = E_h + W=4.76\cdot 10^5 J +(-2.51\cdot 10^5 J)=2.25\cdot 10^5 J$

And this is only kinetic energy:

$E=K=\frac{1}{2}mv^2$

So, we can solve to find the final speed:

$v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(2.25\cdot 10^5 J)}{58 kg}}=88.1 m/s$

4 0

3 years ago