A conversion factor is a ratio with a value of

An AC voltage source and a resistor are connected in series to make up a simple AC circuit. If the source voltage is given by ΔV

ra1l [238]

Answer:

t = 5.48 × 10⁻³ s

Explanation:

Given:

ΔV = ΔVmax × sin(2πft)

frequency, f = 16.9Hz

thus,

ΔV = ΔVmax × sin(2π×16.9×ft)

Now,

Let 'R' be the resistance

Also according to the ohms law

i = V/R

where,

i = current

V = voltage

hence,

$i=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}$

also, given at time 't' the current in the circuit is 55.0% of the peak current

thus

$i=\frac{55}{100}\times \frac{\Delta V_{max}}{R}=0.55\times \frac{\Delta V_{max}}{R}$

thus,

$0.55\times \frac{\Delta V_{max}}{R}=\frac{\Delta V_{max}sin(2\pi \times 16.9\times t)}{R}$

or

$0.55=sin(2\pi \times 16.9\times t)}$

or

$0.5823=(2\pi \times 16.9\times t)}$

or

t = 5.48 × 10⁻³ s (Answer)

8 0

3 years ago

A.

Aliun [14]

Answer:

the correct answer is c, they will accelerate away from each other at different speeds. the 80kg will go faster due to less mass

6 0

3 years ago

A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th

solniwko [45]

Answer:

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

Explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

1) tangential acceleration which is given as

$a_t = R\alpha$

2) Centripetal acceleration

$a_c = \omega^2 R$

here we know that

$\omega = \alpha t$

$a_c = (\alpha t)^2 R$

now we know that net linear acceleration is given as

$a = \sqrt{a_c^2 + a_t^2}$

so we have

$a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}$

$a = R\alpha\sqrt{1 + \alpha^2t^4}$

4 0

3 years ago

Facts about Alessandro Volta?

Mila [183]

Invented the battery
Italian
Has 3 kids

8 0

3 years ago

Read 2 more answers

4.5 billion km, the average separation between the sun and Neptune (report answer in hours). How long does it take light to trav

Liula [17]

Answer:

t = 4.17 hours

Explanation:

given,

The distance between Sun and Neptune, d = 4.5 billion Km

= 4.5 x 10⁹ Km

= 4.5 x 10¹¹ m

The velocity of light, c = 3 x 10⁸ m/s

The velocity is always equal to displacement by the time.

∴ t = d / V

= 4.5 x 10¹¹ m / 3 x 10⁸ m/s

= 15,000 s

= 4.17 h

Hence, the time taken by the light rays to reach the Neptune is, t = 4.17 h

4 0

3 years ago