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xxTIMURxx [149]

3 years ago

10

What are two types of scientific investigations?

1 answer:

kvv77 [185]3 years ago

4 0

Experiments and fieldwork

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The Higher The Value Of Coefficient Of Friction The _____ The Resistance To Sliding.

Vadim26 [7]

Lower the resistance to sliding.

7 0

3 years ago

What is 16.558 m/s rounded to three significant figures?

In-s [12.5K]

Answer:

Option C. 16.6 m/s

Explanation:

To round this 16.558 m/s to 3sf, we need to count the number beginning from 1. When we get to the 3rd number( ie 5), we'll examine the fourth number(i.e 5)to see if it less than five or greater. If it less than five, then we'll discard it. But if it five or greater, we'll approximate it and add it to the 3rd number.

So.

16.558 m/s = 16.6m/s to 3sf

3 0

3 years ago

A spring attached to a mass is at rest in the initial

svet-max [94.6K]

Answer:

E=(1/2)kx^2

Explanation:

6 0

3 years ago

Read 2 more answers

A 350-g mass is attached to a spring whose spring constant is 64 N/m. Its maximum acceleration is 5.3 m/s2. What is the frequenc

max2010maxim [7]

The frequency of oscillation is 2.153 Hz

What is the frequency of spring?

Spring Frequency is the natural frequency of spring with a weight at the lower end. Spring is fixed from the upper end and the lower end is free.

For the mass-spring system in this problem,

The Frequency of spring is calculated with the equation:

$f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }$

Where,

f = frequency of spring

k = spring constant = 64 N/m

m = mass attached to spring = 350g = 0.350 kg

a = maximum acceleration = 5.3 m/s^2

Substituting the values in the equation,

$f = \frac{1}{2\pi } \sqrt{\frac{64}{0.350} }$

$f = \frac{1}{2\pi } ( 13.522)$

$f = 2.1535 Hz$

Hence,

The frequency of oscillation is 2.153 Hz

Learn more about frequency here:

<u>brainly.com/question/13978015</u>

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6 0

2 years ago

Ammonia gas occupies a volume of 0.450 L at a pressure of 96 kPa. What

Sauron [17]

Answer:

0.426 L

Explanation:

Boyles law is expressed as p1v1=p2v2 where

P1 is first pressure, v1 is first volume

P2 is second pressure, v2 is second volume.

Given information

P1=96 kPa, v1=0.45 l

P2=101.3 kpa

Unknown is v2

Making v2 the subject from Boyle's law

$v2=\frac {p1v1}{p2}$

Substituting the given values then

$v2=\frac {96*0.45}{101.3}=0.4264560710760l\approx 0.426 l$

Therefore, the volume is approximately 0.426 L

6 0

3 years ago