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4 years ago

Humberto builds two circuits using identical components.

2 answers:

6 0

Humberto should expect to see that all bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

B) All bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

Bumek [7]4 years ago

5 0

Answer:

B) All bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

Explanation:

When we connect more bulbs in circuit 1 where all are connected in series then the net resistance of the circuit will increase and voltage across each bulb will decrease.

So as per the formula

$P = \frac{V^2}{R}$

so here since voltage decreases across each bulb so we can say that power across each bulb will decrease and bulb gets more dimmy then initially connected bulbs

When we connect bulbs in parallel then voltage across each bulb will remain same and then the brightness of each bulb will remain constant.

So correct answer will be

B) All bulbs in circuit 1 will shine more dimly than the original bulbs, while all bulbs in circuit 2 will have the same brightness as the original bulbs.

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the threshold of hearing is defined as the minimum discernible intensity of the sound. it is approximately 10−12w/m2 . find the

vivado [14]

The distance is 97720.5 m

From the question, we have

P = 0.06 W × 2 = 0.12 W

d = ?

Sound intensity, I = P/4πd²

I = 10⁻¹² W/m²

10⁻¹² = 0.12/4πd²

d = 97720.5 m

The distance is 97720.5 m

Sound intensity :

The power carried by sound waves per unit area in the direction perpendicular to that region is known as sound intensity or acoustic intensity. The watt per square meter (W/m2) is the SI unit of intensity, which also covers sound intensity. Sound intensity is a measure of how quickly energy moves across a given space. The unit area in the SI measurement system is 1 m2. So Watts per square meter are used to measure sound intensity. As there will be energy flow in certain directions but not in others, sound intensity also provides a measure of direction.

To learn more about Sound intensity visit: brainly.com/question/12899113

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1 year ago

Which type of energy is released when a bond between atoms is broken

scoray [572]

Answer:

Explanation:

7 0

3 years ago

Which technology is most efficient in mapping large areas of the ocean floor?

qaws [65]

''Sonar'' is the answer.

7 0

3 years ago

A ball is thrown with an initial speed vi at an angle i with the horizontal. The horizontal range of the ball is R, and the ball

adell [148]

Answer:

Part a)

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

$v_y = \sqrt{Rg/3}$

Part d)

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

$\theta_i = 33.7 degree$

Part f)

$H = \frac{13R}{8}$

Part g)

$X = \frac{13R}{4}$

Explanation:

Initial speed of the launch is given as

initial speed = $v_i$

angle = $\theta_i$ degree

Now the two components of the velocity

$v_x = v_i cos\theta_i$

similarly we have

$v_y = v_i sin\theta_i$

Part a)

Now we know that horizontal range is given as

$R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}$

maximum height is given as

$H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}$

so we have

$v_i sin\theta = \sqrt{Rg/3}$

time of flight is given as

$T = \frac{2v_isin\theta_i}{g}$

$T = \frac{2\sqrt{Rg/3}}{g}$

$T = 2\sqrt{\frac{R}{3g}}$

Part b)

Now the speed of the ball in x direction is always constant

so at the peak of its path the speed of the ball is given as

$R = v_x T$

$R = v_x 2\sqrt{\frac{R}{3g}}$

$v_x = \frac{\sqrt{3Rg}}{2}$

Part c)

Initial vertical velocity is given as

$v_y = v_i sin\theta_i$

$v_i sin\theta = \sqrt{Rg/3}$

Part d)

Initial speed is given as

$v = \sqrt{v_x^2 + v_y^2}$

so we will have

$v = \sqrt{Rg/3 + 3Rg/4}$

$v = \frac{1}{2}\sqrt{13Rg}$

Part e)

Angle of projection is given as

$tan\theta_i = \frac{v_y}{v_x}$

$tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}$

$\theta_i = 33.7 degree$

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

$H = \frac{v^2}{2g}$

$H = \frac{13R}{8}$

Part g)

For maximum range the angle should be 45 degree

so maximum range is

$X = \frac{v^2}{g}$

$X = \frac{13R}{4}$

3 0

3 years ago

Suppose you first walk 12.0 m in a direction 200 west of north and then 20.0 m in a direction 40.00 south of west. How far are y

Tpy6a [65]

Complete Question

The complete question is shown on the first uploaded image

Answer:

the compass direction of the resultant displacement is $\theta =4.7^o$ south of west

Explanation:

Generally using cosine we can obtain the resultant R as follows

$R^2 = A^2 + B^2 -2ABcos(70)$

=> $R = \sqrt{12^2 + 20^2 - 2(12 ) * (20) cos 70}$

=> $R = 19.48 \ m$

We can obtain the direction of the resultant by first using sine rule to obtain angle C as follows

$\frac{A}{sin C} = \frac{R}{sin70 }$

=> $C= sin ^{-1} [\frac{A * (sin 70)}{R} ]$

=> $C = sin ^{-1} [\frac{20 * (sin 70)}{19.48} ]$

=> $C = 74.7 ^o$

Then the direction is obtained as

$\theta = C - 70$

=> $\theta = 74.7 - 70$

=> $\theta =4.7^o$

Hence the compass direction of the resultant displacement is $\theta =4.7^o$ south of west

7 0

3 years ago