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grigory [225]
3 years ago
14

Neutrons have a charge. Answer here

Physics
1 answer:
natta225 [31]3 years ago
3 0

Answer:

yes

Explanation:

i learn it

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Object 1 has a mass of 99.2 kg. Object 2 has a mass of 42.3 kg. If the 2 object pust against each other, what will be the speed
Vlad1618 [11]
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7 0
3 years ago
Two 10-cm-diameter metal plates 1.0 cm apart are charged to {12.5 nC. They are suddenly connected together by a 0.224-mm- diamet
Alekssandra [29.7K]

Answer:

(a).The maximum current in the wire is 4.217\times10^{5}\ A.

(b). The electric field in the wire is 11.2\times10^{5}\ N/C.

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

Explanation:

Given that,

Diameter of metal plates = 10 cm

Distance between the plates = 1.0 cm

Charged = 12.5 nC

Diameter of copper wire = 0.224 mm

We need to calculate the cross section area of the plates

Using formula of area

A=\pi r^2

Put the value into the formula

A=\pi\times(5\times10^{-2})^2

A=7.85\times10^{-3}\ m^2

We need to calculate the capacitor

Using formula of capacitor

C=\dfrac{\epsilon_{0}A}{d}

Put the value into the formula

C=\dfrac{8.85\times10^{-12}\times7.85\times10^{-3}}{1.0\times10^{-2}}

C=6.94\times10^{-12}\ F

We need to calculate the resistance of the wire

Using formula of resistivity

R=\dfrac{\rho l}{A}

Put the value into the formula

R=\dfrac{1.7\times10^{-8}\times1.0\times10^{-2}}{\pi\times(0.1125\times10^{-3})^2}

R=4.27\times10^{-3}\ \Omega

We need to calculate the voltage

Using formula of charge

q=CV

V=\dfrac{q}{C}

Put the value into the formula

V=\dfrac{12.5\times10^{-9}}{6.94\times10^{-12}}

V=1.801\times10^{3}\ V

(a). We need to calculate the current

Using formula of current

I=\dfrac{V}{R}

I=\dfrac{1.801\times10^{3}}{4.27\times10^{-3}}

I=421779.85\ A

I=4.217\times10^{5}\ A

(b). We need to calculate the electric field

Using formula of electric field

E=\dfrac{kq}{r^2}

Put the value into the formula

E=\dfrac{9\times10^{9}\times12.5\times10^{-9}}{(1.0\times10^{-2})^2}

E=11.2\times10^{5}\ N/C

The electric field in the wire is 11.2\times10^{5}\ N/C.

(c). In this case, the voltage between the capacitor plates decreases as the charge decreases with time.

The current is directly proportional to the voltage between the plates .

Hence, The current also decrease with time.

(d). We need to calculate the total amount of energy dissipated in the wire

Using formula of energy

E=\dfrac{1}{2}CV^2

Put the value into the formula

E=\dfrac{1}{2}\times6.94\times10^{-12}\times(1.801\times10^{3})^2

E=1.126\times10^{-5}\ J

The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

Hence, (a).The maximum current in the wire is 4.217\times10^{5}\ A.

(b). The electric field in the wire is 11.2\times10^{5}\ N/C.

(c).The current also decrease with time.

(d). The total amount of energy dissipated in the wire is 1.126\times10^{-5}\ J

8 0
3 years ago
A(n) ________ system is one in which energy moves freely in and out, but no matter enters or leaves the system.
3241004551 [841]
A CLOSED SYSTEM. In a closed system in which energy moves freely in and out, but no matter enters or leaves.
4 0
3 years ago
Read 2 more answers
Problem 9: Suppose you wanted to charge an initially uncharged 85 pF capacitor through a 75 MΩ resistor to 90.0% of its final vo
Bingel [31]

Answer:

t=14.678\times 10^{-3}s

Explanation:

Given:

Capacitance, C = 85 pF = 85 × 10⁻¹² F

Resistance, R = 75 MΩ = 75×10⁶Ω

Charge in capacitor at any time 't' is given as:

Q=Q_o(1-e^{-\frac{t}{RC}})

where,

Q₀ = Maximum charge = CE

E = Initial voltage

t = time

also, Q = CV

V= Final voltage = 90% of E = 0.9E

thus, we have

C\times 0.9E=CE(1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})

or

0.9=1-e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}

or

e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}}=1-0.9=0.1

taking log both sides, we get

ln(e^{-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}})=ln(0.1)=ln(\frac{1}{10})

or

-\frac{t}{75\times 10^6\ \times\ 85\times 10^{-12}}=-ln(10)

or

t=75\times 10^6\ \times\ 85\times 10^{-12}\times ln{10}

or

t=14.678\times 10^{-3}s

3 0
3 years ago
What is brownion motion
Alex777 [14]

Answer: Brownion motion is the erratic random movement of microscopic particles in a fluid, as a result of continuous bombardment from molecules of the surrounding medium.

Explanation:

Brownian motion is the random movement of particles in a fluid due to their collisions with other atoms or molecules.

4 0
3 years ago
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