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Dima020 [189]
3 years ago
10

An evacuated long tube contains a coin and a feather. If both objects fall together starting from the top of the tube, it is exp

ected that: If this experiment is repeated at a place 2000 kilometers above the sea-level, the acceleration due to gravity gexp is expected to :
Physics
1 answer:
Natasha_Volkova [10]3 years ago
5 0

Answer:

gexp = 3.65 m/s²

Explanation:

The value of acceleration due to gravity changes with the altitude. The following formula gives the value of acceleration due to gravity at some altitude from the sea level:

gexp = g(1 - 2h/Re)

where,

gexp = expected value of g at altitude = ?

g = acceleration due to gravity at sea level = 9.8 m/s²

h = altitude = 2000 km = 2 x 10⁶ m

Re = Radius of Earth = 6.37 x 10⁶ m

Therefore,

gexp = (9.8 m/s²)(1 - 2*2 x 10⁶ m/6.37 x 10⁶ m)

<u>gexp = 3.65 m/s²</u>

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3: the electricity is generated in the permanent magnet

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Two identical plastic cups contain the same amount of water at two different temperatures, as shown to the left. Both cups are p
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the molecules will begin to move slowly and will turn to ice

Explanation:

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A man ran the 200m race within 19.19 seconds. Calculate the speed of the man during the race
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Answer:

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what is the energy of an electromagnetic wave that has a frequency of 8.0 x 10^15 Hz? Use the equation...
Katarina [22]

(C)

Explanation:

E = hf = (6.626×10^{-34}\:\text{J•s})(8.0×10^{15}\:\text{Hz})

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4 0
3 years ago
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
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