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2 years ago

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What is physical change ?

1 answer:

ss7ja [257]2 years ago

3 0

Answer:

A physical change is a change to the physical—as opposed to chemical—properties of a substance. They are usually reversible. The physical properties of a substance include such characteristics as shape (volume and size), color, texture, flexibility, density, and mass.

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In an LC circuit containing a 40-mH ideal inductor and a 1.2-mF capacitor, the maximum charge on the capacitor is 45mC during os

GalinKa [24]

Answer:

E) 6.5 A

Explanation:

Given that

L = 40 m H

C= 1.2 m F

Maximum charge on capacitor ,Q= 45 m C

The maximum current I given as

I = Q.ω

ω =angular frequency

$\omega=\dfrac{1}{\sqrt{CL}}$

By putting the values

$\omega=\dfrac{1}{\sqrt{CL}}$

$\omega=\dfrac{1}{\sqrt{40\times 10^{-3}\times 1.2 \times 10^{-3}}}$

ω = 144.33 rad⁻¹

Maximum current

I = 45 x 10⁻³ x 144.33 A

I= 6.49 A

I = 6.5 A

E) 6.5 A

3 0

3 years ago

Physics question, please help?

Ludmilka [50]

0.4823 m/s

The initial velocity u1 of the ball=0

From the law of conservation of linear momentum.

m1u1+m2u2=m1v1+m2v2

(160×0)+(170×u1)=(160×0.3)+(170×0.2)

u1=0.4823m/s

6 0

3 years ago

A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge

Nesterboy [21]

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

$\tau=RC$

where

$R=50,000 \Omega$ is the resistance

$C=2.0\mu F=2.0\cdot 10^{-6}F$ is the capacitance

Substituting,

$\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s$

The charge on a charging capacitor is given by

$Q(t)=Q_0 (1-e^{-t/\tau} )$ (1)

where

$Q_0$ is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

$Q(t)=0.90 Q_0$

Substituting this into eq.(1) we find

$0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s$

4 0

3 years ago

A system consists of two particles. Which of the following scenarios would the force on each particle increase the most? If mult

MrMuchimi

Answer:

c

Explanation:

7 0

3 years ago

A particle moves along the x axis. It is initially at the position 0.180 m, moving with velocity 0.060 m/s and acceleration -0.3

shusha [124]

Answer:

a

$x_2 = -2.3356$

b

$v = -1.384 \ m/s$

Explanation:

From the question we are told that

The initial position of the particle is $x_1 = 0.180 \ m$

The initial velocity of the particle is $u = 0.060 \ m/s$

The acceleration is $a = -0.380 \ m/s^2$

The time duration is $t = 3.80 \ s$

Generally from kinematic equation

$v = u + at$

=> $v = 0.060 + (-0.380 * 3.80)$

=> $v = -1.384 \ m/s$

Generally from kinematic equation

$v^2 = u^2 + 2as$

Here s is the distance covered by the particle, so

$(-1.384)^2 = (0.060)^2 + 2(-0.380)* s$

=> $s = -2.5156 \ m$

Generally the final position of the particle is

$x_2 = x_1 + s$

=> $x_2 = 0.180 + (-2.5156)$

=> $x_2 = -2.3356$

6 0

3 years ago