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Olenka [21]
3 years ago
13

A ball is falling from the sky down to the left. What two forces are being used?

Physics
2 answers:
ioda3 years ago
8 0

Answer: Gravity for sure and...air resistance.

Explanation: I hope this is correct.

suter [353]3 years ago
4 0

Answer:

push and pull

Explanation:

i beleve good luck❤️❤️

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In which material does the light wave have the larger wavelength?
pochemuha

Answer: in the smaller/ thinner pieces.

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What alleles are always both expressed
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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
If the mass of both weights is 225 gm, the first mass is located 20∘ north of east, the second mass is located 20∘ south of east
Montano1993 [528]

Answer:

The voltage is 2.114 V.

Explanation:

Given that,

Mass of both weights = 225 gm

Transducer sensitivity = 0.5 V/N

The first mass is located 20∘ north of east, the second mass is located 20∘ south of east,

We need to calculate the net equivalent force

Using formula of force

F_{3}=m_{1}g\cos\theta+m_{2}g\cos\theta

F_{3}=2mg\cos\theta

Put the value into the formula

F_{3}=2\times0.225\times10\cos20^{\circ}

F_{3}=4.228\ N

We need to calculate the voltage

Using formula of voltage

Voltage =sensitivity\times F_{3}

Put the value into the formula

Voltage=0.5\times4.228

Voltage =2.114\ V

Hence, The voltage is 2.114 V.

7 0
3 years ago
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