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3 years ago

When a guitar string plays the note "a," the string vibrates at 440 hz ?

1 answer:

8 0

Yes, that's correct. The note "A" (which is used to tune the other strings of the guitar) corresponds to a frequency of 440 Hz.

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What is a constellation as astronomers define it today? What does it mean when an astronomer says, “I saw a comet in Orion last

MissTica

Explanation:

Constellation: The complete sky has been divided in 88 different areas, in a way we have divided Earth in countries, not necessarily having same shapes and size. These 88 areas are known as constellations. These contains a lot of stars. When we join the brightest stars together we can imagine a shape out of them which is called as Asterism. Most of the people are unaware of this difference. Some of the famous constellations are Orion, Taurus, Gemini, Hydra, Ursa Major etc.

When an astronomer says that there is a comet is in the Orion, he means that a comet is in the boundaries of Orion constellation.

4 0

3 years ago

What objects has large volume and small mass

viva [34]

Answer:

A hot air balloon

7 0

3 years ago

a train traveling at 30m/s comes to a complete stop in 20 sec.how far did the train go during the deceleration

elixir [45]

Answer:

600m

Explanation:

v=30m/s

t=20s

s=VT

20×30=600m

3 0

3 years ago

Fronts are termed by the temperature of the air mass that overtakes another air mass. A. True B. False

Vikki [24]

A. True
if cold air is replacing warm air it is a cold front and vice versa.

6 0

3 years ago

Read 2 more answers

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

6 0

3 years ago