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3 years ago

I will mark u brainlist

Physics

2 answers:

dedylja [7]3 years ago

7 0

Answer:

DragonFly or a dog

Explanation:

Damm [24]3 years ago

6 0

The dog does not go through metamorphosis.

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Why do women typically tend to have slightly greater stability than men?

stiks02 [169]

I think it would have to be B.

7 0

3 years ago

Read 2 more answers

An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m lon

telo118 [61]

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

$L = \frac{\lambda}{2} \\\\\lambda = 2L$

λ = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀ is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N + N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

6 0

3 years ago

Particles q1 = -53.0 uc, q2 = +105 uc, and

Nimfa-mama [501]

Answer:

$-180.38\ \text{N}$

Explanation:

$q_1=-53\ \mu\text{C}$

$q_2=105\ \mu\text{C}$

$q_3=-88\ \mu\text{C}$

r = Distance between the charges

$r_{12}=0.5\ \text{m}$

$r_{23}=0.95\ \text{m}$

$r_{13}=1.45\ \text{m}$

k = Coulomb constant = $9\times 10^9\ \text{Nm}^2/\text{C}^2$

Net force is given by

$F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}$