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lutik1710 [3]
3 years ago
7

What’s up world or people

Physics
1 answer:
Korolek [52]3 years ago
8 0

Hi how are you?

Are you a Libra?

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An atom with seven valence electrons would most likely lose an electron to become stable true or fasle
mel-nik [20]

Answer:

false

Explanation:

since it has seven valence electrons it means it's a non metal and non metals always gain electrons.

5 0
2 years ago
Explain the importance of measurement<br>Physics.<br><br>​
KiRa [710]

Answer:

Measurements are an important part of comparing things, as they provide the basis on comparing objects to other objects. Measurements allow us to recognize three hours and see how it's shorter than five hours, without having to observe the hours passing by themselves.

7 0
3 years ago
If you run at an average speed of 10mi/h how long will it take for you to run 2.5miles
givi [52]

Answer:

If the avg speed is 10mi/h and you want to know how long it will take to run 2.5mi/h you put that as a ratio 2.5/10 which is 1/4 of an hour so it will take 15 minutes to run 2.5 miles

Explanation:

3 0
3 years ago
7. CALCULATE: How much work is done in each of
kifflom [539]
<h3>Answer</h3>

a. 8J

b. 90J

<h3>Notes formula</h3>

W = F × s

W = Work done (J)

F = Force (N)

s = distance (m)

<h3>Known that </h3>

a. F = 4N

s = 2m

b. F = 30N

s = 3m

<h3>Question</h3>

a. W = ..?

b. W = ..?

<h3>Way to do it</h3>

a. W = F × s

= 4N × 2m = 8J

b. W = F × s

= 30N × 3m = 90J

<em>#Moderators please don't be mean, dont delete my answers just to get approval from your senior or just to get the biggest moderation daily rank.</em>

6 0
2 years ago
An object 5.Ocm in the length is placed at a distance of 20cm in front of convex mirror
andrew-mc [135]

Answer:

Position = \frac{60}{7}\ cm behind the mirror

Nature = Virtual and Erect

Size = \frac{15}{7}\ cm : Diminished

Explanation:

Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.

Object distance = u = -20 cm

Focal length = f = Radius of curvature/2 = 30/2 = 15 cm

We have to use mirror formula to find image distance.

\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\\ \frac{1}{-20}+\frac{1}{v}=\frac{1}{15}\\ \frac{1}{v}=\frac{7}{60}\\v=\frac{60}{7}\ cm

Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.

Magnification = =\frac{h_{image}}{h_{object}}=-\frac{v}{u}=\frac{60}{7\times20}=\frac{3}{7}

Height of the object = 5 cm

Height of the image = 5\times\frac{3}{7}=\frac{15}{7}\ cm

Since the height of the image is positive and less than the size of object,it is erect and diminished.

4 0
4 years ago
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