Ask question

Ask question

All categories

2 years ago

9

Describe the difference between ‘conventional current’ and ‘electron flow’

1 answer:

Tamiku [17]2 years ago

6 0

Answer:

Screenshot attached

Explanation:

You might be interested in

Which expression shows the amount of work a 6-kilogram mass is capable of performing if it is dropped from a height of 2 meters?

Ipatiy [6.2K]

The acceleration of gravity is 9.8 m/s^2

3 0

2 years ago

Read 2 more answers

Solid state has ____________ intermolecular force of attraction.

castortr0y [4]

solid state has <u>the </u><u>most</u> intermolecular force of attraction.

6 0

3 years ago

94 points if you hurry <br><br><br><br> The phase of matter with the least kinetic energy:

OlgaM077 [116]

Answer:

solid

Explanation:

6 0

2 years ago

In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e

Zanzabum

Answer:

The velocity is $v_t = 0.02175 \ m/s$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 0.024 \ kg$

The initial speed of the bullet is $u_b = 1200 \ m/s$

The mass of the target is $m_t = 320 \ kg$

The initial velocity of target is $u_t = 0 \ m/s$

The final velocity of the bullet is is $v_b = 910 \ m/s$

Generally according to the law of momentum conservation we have that

$m_b * u_b + m_t * u_t = m_b * v_b + m_t * v_t$

=> $0.024 * 1200 + 320 * 0 = 0.024 * 910 + 320 * v_t$

=> $v_t = 0.02175 \ m/s$

3 0

3 years ago

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien

n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

$\phi=NBA$

Put the value into the formula

$\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2$

$\phi=7.85\times10^{-7}\ Tm^2$

We need to calculate the induced emf

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{7.85\times10^{-7}}{dt}$

Put the value of emf from ohm's law

$\epsilon =IR$

$IR=\dfrac{7.85\times10^{-7}}{dt}$

$Idt=\dfrac{7.85\times10^{-7}}{R}$

$Idt=\dfrac{7.85\times10^{-7}}{0.50}$

$Idt=0.00000157=1.57\times10^{-6}\ C$

We know that,

$Idt=dq$

$dq=1.57\times10^{-6}\ C$

We need to calculate the voltage across the capacitor

Using formula of charge

$dq=C dV$

$dV=\dfrac{dq}{C}$

Put the value into the formula

$dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}$

$dV=1.57\ V$

Hence, The voltage across the capacitor is 1.57 V.

5 0

3 years ago