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Phoenix [80]
3 years ago
9

What is a bond between a positive and a negative ion charge?

Physics
2 answers:
klasskru [66]3 years ago
5 0

Answer:

Ionic Bond

Explanation:

nasty-shy [4]3 years ago
5 0

Answer:

<u>An ionic bond</u>

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The diagram shows monochromatic light passing through two openings.
geniusboy [140]

Answer: constructive interference in which waves strengthen each other

Explanation:

4 0
3 years ago
What is the resultant of a pair of one pound forces at right angles to each other?
OleMash [197]

Answer:

F_R=\sqrt{2} \ pound.force

Explanation:

Given that there are two force of 1 pound each at right angles to each other.

The from the vector law of addition:

F_R=\sqrt{F_1^2+F_2^2}

where:

F_R= resultant force

F_1\ \&\ F_2 be the two of the forces to be added.

F_R=\sqrt{1^2+1^2}

F_R=\sqrt{2} \ pound.force

8 0
4 years ago
Write the adverbs use in sentences no.1-5 ty​
miv72 [106K]

Answer:

ok I have written in your face check it out fast don't have time see see see fast

5 0
3 years ago
A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa
andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

\omega=\frac{v}{r}

where

\omega is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

\omega=\frac{v}{r}

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s

(c) 5550 m

The maximum playing time of the CD is

t =74.0 min \cdot 60 s/min = 4,440 s

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) -6.4\cdot 10^{-3} rad/s^2

The angular acceleration of the CD is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 21.6 rad/s is the final angular speed (when the CD is scanned at the outermost part)

\omega_i = 50.0 rad/s is the initial angular speed (when the CD is scanned at the innermost part)

t=4440 s is the time elapsed

Substituting into the equation, we find

\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2

5 0
4 years ago
Read 2 more answers
Activity log - Standing Whole Body Record your time and activity below and in your unit fitness log. If a particular category do
sasho [114]

Answer:

Date: N/A

Warm-up: 5 minutes

Cool-down: 6 minutes

Type of Activity: Cardiovascular

Description of Activity and Time: Running in place for 10 minutes

Intensity Level: Moderate

Total Time: 21 minutes

Explanation: Mark me as brainiest hope this helps I just did it

6 0
4 years ago
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