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Anarel [89]
3 years ago
14

Was leeuwenhoek the first person to build a microscope?

Physics
1 answer:
Elena L [17]3 years ago
4 0

Answer:

No. He was not the first person who built a microscope. But he made simple (one lens), but the microscopes that he did built were the best ones for that time.

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PLEASE HELP ME WITH THIS ONE QUESTION
Vinil7 [7]

Answer:

b

Explanation:

7 0
3 years ago
What is your understanding of repetition and replication in science?
strojnjashka [21]
From my understanding, the two are opposite. Let me give you each word and what it scientifically means to me.

repetition
occurs when an activity is repeated by the same person

replication
occurs when an activity is repeated by a different person


I hope this completely answers your question! If you need further explanation, please let me know!
~Brooke❤️
5 0
3 years ago
A spring (k = 20 N/m) sits on a frictionless surface. The spring is compressed 20 cm and a 2 kg mass is placed before the spring
Olenka [21]

Answer:

E.)none of the above

Explanation:

Elastic energy = kinetic energy

½ kx² = ½ mv²

v = x √(k / m)

v = 0.2 m √(20 N/m / 2 kg)

v = 0.632 m/s

Time to travel 2.0 m:

t = (2.0 m) / (0.632 m/s)

t = 3.16 s

7 0
4 years ago
If a man walks 600 meters in 300 seconds, what was his speed in meters per second?​
kifflom [539]

Answer:

His speed in meters per second was 2. Or, 2 meters per second.

Explanation:

6 0
3 years ago
Problem 7.17 A horizontal spring with stiffness 0.8 N/m has a relaxed length of 12 cm (0.12 m). A mass of 23 grams (0.023 kg) is
Jlenok [28]

The speed of the mass v = 0.884 m/s.

<u>Explanation</u>:

Let

K1 represents the kinetic energy of the mass when it is released,

U1 represents the potential energy of the spring when the mass is released,

K2 represents the kinetic energy of the mass when the spring returns to relaxed length,

U2 represents the potential energy of the spring when the spring returns to relaxed length

The spring is stretched by 0.27 - 0.12 = 0.15 m

                                K1 = 0

                                 U1 = (1/2) \times 0.8 \times (0.15)^2

                                     = 0.009 J

                                U2 = 0

By conservation of energy,

                        K2 + U2 = K1 + U1

                           K2 + 0 = 0 + 0.009 J

                                 K2 = 0.009 J

Let v = speed of the mass

                                K2 = 1/2 \times m \times v^2

m = 23 g = 0.023 kg

                           0.009 = 1/2 \times 0.023 \times v^2

                           0.009 = 0.0115 \times v^2

                                   v = √(0.009 / 0.0115)

                                  v = 0.884 m/s.

5 0
4 years ago
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