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3 years ago

What is the path that an electric current follows called

Physics

2 answers:

Gwar [14]3 years ago

5 0

I’m sure it’s called a circuit:)

lyudmila [28]3 years ago

3 0

I think its a electric circuit

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An object of mass 0.40 kg, hanging from a spring with a spring constant of 8.0 N/m, is set into an up-and-down simple harmonic m

Sergeeva-Olga [200]

Answer:

a = 2 m/s2

Explanation:

we know from newtons 2nd law

F = ma.

we also know that from hookes law we have

F = kx

equate both value of force to get value of acceleration

kx = ma,

where,

k is spring constant = 8.0 N/m

x is maximum displacement 0.10 m

m is mass of object 0.40 kg

a = \frac{kx}{m}

= \frac{8 *0 .10}{0.40}

a = 2 m/s2

5 0

3 years ago

The moon has a diameter of 3.48 x 106 m and is a distance of 3.85 x 108 m from the earth. The sun has a diameter of 1.39 x 109 m

Mrrafil [7]

Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

Explanation:

s = Diameter of the object

r = Distance between the Earth and the object

Angle subtended is given by

$\theta=\frac{s}{r}$

For the Moon

$\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad$

The angle subtended by the Moon is 0.00903 rad

For the Sun

$\theta_s=\dfrac{1.39\times 10^9}{1.5\times 10^{11}}\\\Rightarrow \theta_s=0.00926\ rad$

The angle subtended by the Sun is 0.00926 rad

Area ratio is given by

$\frac{A_m}{A_s}=\dfrac{\pi r_m^2}{\pi r_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{d_m^2}{d_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{(3.48\times 10^{6})^2}{(1.39\times 10^9)^2}\\\Rightarrow \frac{A_m}{A_s}=6.268\times 10^{-6}$

The area ratio is $6.268\times 10^{-6}$

3 0

3 years ago

An alpha particle (α), which is the same as a helium-4 nucleus, is momentarily at rest in a region of space occupied by an elect

beks73 [17]

Answer:

Speed of the alpha particle is $v=1.8180\times 10^3m/sec$

Explanation:

We have given charge on alpha particle $q=3.2\times 10^{-19}C$

Mass of the alpha particle $m=6.68\times 10^{-27}kg$

Potential difference $V=-3.45\times 10^{-3}volt$

We have to find the speed of the alpha particle

From energy conservation we know that

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

$\frac{1}{2}\times 6.68\times 10^{-27}\times v^2=3.2\times 10^{-19}\times 3.45\times 10^{-3}$

$v=1.8180\times 10^3m/sec$

4 0

3 years ago

A student is building a simple circuit with a battery, light bulb, and copper wires. When she connects the wires to the battery

denis23 [38]

It’s will be B because the circuit had a open or close so if that doesn’t work than it’s because it’s open

5 0

3 years ago

Read 2 more answers

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago