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Tasya [4]
3 years ago
13

What is the word we use for when we repeat the same thing more than once in an experiment?

Physics
2 answers:
AlladinOne [14]3 years ago
7 0

Answer:

The answer is "Repetition"

Explanation: when you say something more than 3 times, is called repetiton

Norma-Jean [14]3 years ago
3 0

Answer:

repetitive or adjective

Explanation:

Repetitive involves doing something over and over again.

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You have a spring-loaded air rifle. When it is loaded, the spring is compressed 0.3 m and has a spring constant of 150 N/m. In j
Feliz [49]

The potential energy of the spring is 6.75 J

The elastic potential energy stored in the spring is given by the equation:

E= \frac{1}{2} kx^2

where;

k is the spring constant

x is the compression/stretching of the string

In this problem, we have the spring as follows:

k = 150 N/m is the spring constant

x = 0.3 m is the compression

Substituting in the equation, we get

E=\frac{1}{2} (150) (0.3)^2

E=6.75J

Therefore. the elastic potential energy stored in the spring is 6.75J .

Learn more about potential energy here:

brainly.com/question/10770261

#SPJ4

6 0
1 year ago
What element has 5 valence electrons, but more than 7 protons and less than 33 protons?
MAVERICK [17]

Answer:

Phosphorus (P)

Explanation:

7 0
2 years ago
Two small nonconducting spheres have a total charge of 93.0 μC . Part A
Travka [436]

Answer:

charge on each

Q1 = 2.06 ×10^{-5}  C

Q2 = 7.23 × 10^{-5} C

when force were attractive

Q1 = 1.07 × 10^{-4} C

Q2 = -1.39 × 10^{-5} C

Explanation:

given data

total charge = 93.0 μC

apart distance r  = 1.14 m

force exerted  F = 10.3 N

to find out

What is the charge on each and What if the force were attractive

solution

we know that force is repulsive mean both sphere have same charge

so total charge on two non conducting sphere is

Q1 + Q2 = 93.0 μC  = 93 ×10^{-6} C

and

According to Coulomb's law force between two sphere is

Force F = \frac{K Q1 Q2 }{r^2}      .........1

Q1Q2 = \frac{F*r^2}{k}

here F is force and r is apart distance and k is 9 × 10^{9} N-m²/C² put all value we get

Q1Q2 = \frac{ 10.3*1.14^2}{9*10^9}

Q1Q2 = 1.49 ×  10^{-9} C²

and

we have  Q2 = 93 ×10^{-6} C - Q1

put here value

Q1²  - 93 ×10^{-6}  Q1 + 1.49 ×  10^{-9} = 0

solve we get

Q1 = 2.06 ×10^{-5}  C

and

Q1Q2 = 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = 1.49 ×  10^{-9}

Q2 = 7.23 × 10^{-5} C

and

if force is attractive we get here

Q1Q2 = - 1.49 ×  10^{-9} C²

then

Q1²  - 93 ×10^{-6}  Q1 - 1.49 ×  10^{-9} = 0

we get here

Q1 = 1.07 × 10^{-4} C

and

Q1Q2 = - 1.49 ×  10^{-9}

2.06 ×10^{-5}  Q2 = - 1.49 ×  10^{-9}

Q2 = -1.39 × 10^{-5} C

5 0
2 years ago
Questions 1
Aleonysh [2.5K]

Answer:

Radiant to Chemical

Explanation:

I think

8 0
2 years ago
Read 2 more answers
A skateboarder, with an initial speed of 2.0 m/s, rolls virtually friction free down a straight incline of length 18 m in 3.3 s.
lesya692 [45]

Initial speed of the skateboarder (u) = 2 m/s

Distance covered (s) = 18 m

Time taken = 3.3 seconds

Let the acceleration be a.

Using seconds equation of motion:

s = ut + \frac{1}{2} a t^2

18 = 2 \times 3.3 + \frac{1}{2}a \times 3.3^2

a = \frac{11.4 \times 2}{10.89}

a = 2.09 m/s^2

Now, Acceleration down the incline = g Sin Θ

g Sin Θ = a

9.8 × Sin Θ = 2.09

Sin Θ = \frac{2.09}{9.8}

Θ = 12.31°

Hence, the angle of the inclined plane is: Θ = 12.31°

4 0
3 years ago
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