Two thermometers are calibrated, one in de

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

The function h(t) = 162 +64t models the height, in feet, of a ball t seconds after it was kicked into the air. Which statement i

melamori03 [73]

Answer:

the ball didn't not reach the Maximum height because of the time interval

3 0

3 years ago

(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

$D\sin\theta=n\lambda$

$D=\dfrac{n\lambda}{\sin\theta}$

Where, D = width of slit

$\lambda$ = wavelength

Put the value into the formula

$D=\dfrac{n\lambda}{\sin\theta}$

Here, $\sin\theta$ should be maximum.

So. maximum value of $\sin\theta$ is 1

Put the value into the formula

$D=\dfrac{1\times\lambda}{1}$

$D=\lambda$

(b). If the minimum number is 50

Then, the width is

$D=\dfrac{50\times\lambda}{1}$

$D=50\lambda$

(c). If the minimum number is 1000

Then, the width is

$D=\dfrac{1000\times\lambda}{1}$

$D=1000\lambda$

Hence, The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

4 0

3 years ago

When you let the air go out of a balloon the balloon moves or shoots like a rocket. Would this happen if there was no surroundin

Pie

Answer:

The balloon would still move like a rocket

Explanation:

The principle of work of this system is the Newton's third law of motion, which states that:

"When an object A exerts a force on an object B (action), object B exerts an equal and opposite force (reaction) on object A"

In this problem, we can identify the balloon as object A and the air inside the balloon as object B. As the air goes out from the balloon, the balloon exerts a force (backward) on the air, and as a result of Newton's 3rd law, the air exerts an equal and opposite force (forward) on the balloon, making it moving forward.

This mechanism is not affected by the presence or absence of surrounding air: in fact, this mechanism also works in free space, where there is no air (and in fact, rockets also moves in space using this system, despite the absence of air).

3 0

3 years ago

SOMEONE PLEASE HELP!!

Arisa [49]

Hello!

$\large\boxed{\text{C. 7,350,000 J}}$

Use the equation:

$PE = mgh$

Where:

m = mass of the object (kg)

g = acceleration due to gravity (≈9.8 m/s)

h = height above ground (m)

Plug the given values into the equation:

PE = 7500 · 9.8 · 100

PE = 7,350,000 Joules.

7 0

3 years ago