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2 years ago

2.5: Một người nặng 72kg ngồi trên sàn treo nặng 12kg như hình vẽ. Hỏi người đó

Physics

1 answer:

Ivahew [28]2 years ago

5 0

Answer:

english

Explanation:

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A 2.3 m long wire weighing 0.075 N/m is suspended directly above an infinitely straight wire. The top wire carries a current of

ZanzabumX [31]

Answer:

Explanation:

Magnetic field near current carrying wire

= $\frac{\mu_0}{4\pi} \frac{2i}{r}$

i is current , r is distance from wire

B = 10⁻⁷ x $\frac{2\times49}{r}$

force on second wire per unit length

B I L , I is current in second wire , L is length of wire

= 10⁻⁷ x $\frac{2\times49}{r}$ x 33 x 1

= 3234 x $\frac{10^{-7}}{r}$

This should balance weight of second wire per unit length

3234 x $\frac{10^{-7}}{r}$ = .075

r = $\frac{3234}{.075}$ x 10⁻⁷

= .0043 m

= .43 cm .

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3 years ago

What is an electric fuse? What is the working principle of electric fuse?

user100 [1]

<h2>What is an electric fuse?</h2><h3>Answer: Electric fuse is a safety device used to limit the current in an electric circuit which melts and breaks the circuit whenever there is an excess flow of current through the circuit.</h3><h2>What is the working principle of electric fuse?</h2><h3>An electric fuse is based on the principle of heating effect of electric current. It is made up of thin metallic wire of non-combustible material. A fuse is always connected between the ends of the terminal in a series connection with the circuit.</h3>

4 0

3 years ago

The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A

kotykmax [81]

Answer:

$2.1\times 10^{-12} c$

Explanation:

We are given that

Surface area of membrane= $5.3\times 10^{-9} m^2$

Thickness of membrane= $1.1\times 10^{-8} m$

Assume that membrane behave like a parallel plate capacitor.

Dielectric constant=5.9

Potential difference between surfaces=85.9 mV

We have to find the charge resides on the outer surface of membrane.

Capacitance between parallel plate capacitor is given by

$C=\frac{k\epsilon_0 A}{d}$

Substitute the values then we get

Capacitance between parallel plate capacitor= $\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}$

$C=0.25\times 10^{-12}F$

V= $85.9 mV=85.9\times 10^{-3}$

$Q=CV$

$Q=0.25\times 10^{-12}\times 85.9\times 10^{3}=2.1\times 10^{-12} c$

Hence, the charge resides on the outer surface= $2.1\times 10^{-12} c$

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3 years ago

Read 2 more answers

a typical cmall flashlight contains two batteries each having na emf of 2.0 v connected in series with a bulb havin ga resistanc

Helen [10]

Answer:

P = 0.25 W

Explanation:

Given that,

The emf of the battry, E = 2 V

The resistance of a bulb, R = 16 ohms

We need to find the power delivered to the bulb. We know that, the formula for the power delivered is given by :

$P=\dfrac{V^2}{R}\\\\P=\dfrac{2^2}{16}\\\\=0.25\ W$

So, 0.25 W power is delivered to the bulb.

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2 years ago

(a) If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.5

Tju [1.3M]

Answer:

The fraction of its energy that it radiates every second is $3.02\times10^{-11}$ .

Explanation:

Suppose Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by

$\dfrac{dE}{dt}=\dfrac{q^2a^2}{6\pi\epsilon_{0}c^3}$