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nata0808 [166]
2 years ago
15

Betty is six years old and often asks her parents many questions. Her parents respond to her questions however trivial they may

be. How does this behavior affect Betty? This behavior helps Betty in development.
Physics
2 answers:
Brums [2.3K]2 years ago
5 0

This behavior helps Betty in <u>intellectual  </u>development.

KiRa [710]2 years ago
4 0

Answer:

Intellectual

Explanation:

a person possessing a highly developed intellect

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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
2 years ago
If you wanted to
VLD [36.1K]

Answer:

Option C

Explanation:

According to the formula

  • \\ \boxed{\sf R=\rho\dfrac{\ell}{A}}

So

If we use wide wire we increase the area of cross section so resistance decreases

5 0
2 years ago
A T-Bar is similar to a_______
dezoksy [38]

Answer:

Donkey

Explanation:

3 0
2 years ago
A particular 12 V car battery can send a total charge of 110 A·h (ampere-hours) through a circuit, from one terminal to the othe
DiKsa [7]
<h2>Answer:</h2>

(a) 3.96 x 10⁵C

(b) 4.752 x 10⁶ J

<h2>Explanation:</h2>

(a) The given charge (Q) is 110 A·h (ampere hour)

Converting this to A·s (ampere second) gives the number of coulombs the charge represents. This is done as follows;

=> Q = 110A·h

=> Q = 110 x 1A x 1h          [1 hour = 3600 seconds]

=> Q = 110 x A x 3600s

=> Q = 396000A·s

=> Q = 3.96 x 10⁵A·s = 3.96 x 10⁵C

Therefore, the number of coulombs of charge is 3.96 x 10⁵C

(b) The energy (E) involved in the process is given by;

E = Q x V           -----------------(i)

Where;

Q = magnitude of the charge = 3.96 x 10⁵C

V = electric potential = 12V

Substitute these values into equation (i) as follows;

E = 3.96 x 10⁵ x 12

E = 47.52 x 10⁵ J

E = 4.752 x 10⁶ J

Therefore, the amount of energy involved is 4.752 x 10⁶ J

8 0
3 years ago
An engine absorbs 1749 J from a hot reservoir and expels 539 J to a cold reservoir in each cycle. a. What is the engine’s effici
Masja [62]

Answer:

a)η = 69.18 %

b)W= 1210 J

c)P=3967.21 W

Explanation:

Given that

Q₁ = 1749 J

Q₂ = 539  J

From first law of thermodynamics

Q₁   = Q₂ +W

W=Work out put

Q₂=Heat rejected to the cold reservoir

Q₁ =heat absorb by hot  reservoir

W= Q₁- Q₂

W= 1210 J

The efficiency given as

\eta=\dfrac{W}{Q_1}

\eta=\dfrac{1210}{1749}

\eta=0.6918

η = 69.18 %

We know that rate of work done is known as power

P=\dfrac{W}{t}

P=\dfrac{1210}{0.305}\ W

P=3967.21 W

8 0
3 years ago
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