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kotykmax [81]
3 years ago
7

The sharing of valence electrons between atoms indicates an ionic bond. True or false

Physics
1 answer:
morpeh [17]3 years ago
7 0

Answer:

False

Explanation:

Sharing valence electrons to make a bond creates a covalent bond, not an ionic bond.

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A space station worker found herself floating free 100 meters from the space station because her safety line became unhooked
yuradex [85]

Answer:

Explanation: is that suppose to be the question or are you just saying?

8 0
3 years ago
The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have w
anzhelika [568]

Answer:

594.8 Hz

Explanation:

Parameters given:

Speed of sound, v = 345 m/s

Wavelength = 58 cm = 0.58 m

Speed of a wave is given as:

Speed = wavelength * frequency

Therefore:

Frequency = Speed/Wavelength

Frequency = 345/0.58

Frequency = 594.8 Hz

8 0
3 years ago
Which situation has the most potential energy
balu736 [363]

Answer:

B

Explanation: you are staying longr to jump 3 meters than one meter

4 0
3 years ago
A record is spinning at the rate of 25rpm. If a ladybug is sitting 10cm from the center of the record.
marin [14]

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

A)

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

\omega = 25 rpm

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s

B)

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

\omega = 2 \pi f (1)

where

\omega is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

\omega=0.42 rev/s

Keeping in mind that 1 rev = 2\pi rad, the angular speed can be rewritten as

\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42

And re-arranginf eq.(1), we can find the frequency:

f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz

And the frequency is the number of complete revolutions made per second.

C)

For an object in circular motion, the tangential speed is related to the angular speed by the equation

v=\omega r

where

\omega is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

\omega = 2\pi \cdot 0.42 rad/s is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s

D)

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

d=vt

where

v is the tangential speed

t is the time elapsed

Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

d=(26.4)(20)=528 cm

Learn more about circular motion:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

7 0
4 years ago
Read 2 more answers
Someone answer? Please
dexar [7]
I don’t know ask your parents
6 0
2 years ago
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