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m_a_m_a [10]
3 years ago
11

Increasing the mass attached to a spring will increase its vibrational period.

Physics
1 answer:
iren2701 [21]3 years ago
8 0
Increasing the mass attached to a spring will increase it's vibrational period - this is true. Imagine placing an elephant at the end of a bungee cord vs an apple, the apple will recoil faster than the elephant.
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What is the significance of a standard system of measurement?
dimulka [17.4K]

Because scientists all over the world are working together, looking for answers to the same questions, just as much as if they all worked in the same physical laboratory in the same building.  They need a way to share data and experimental results in a form that everyone can understand. ( D )

Let's say I perform an experiment and get very exciting results. I'm a good scientist, so the next thing I want to do is to publish a complete description of how I did my experiment, and include all of my results.  That way, scientists around the world can read about what I did, they can find any mistakes that I made, and they can even repeat my experiment for themselves and see if they get the same results.

Now let's say that my results looked like this:

Result #1). 

The reaction stabilized when it reached the rate of 1.26 briligs per tove.

Result #2).

After running at that constant rate for 35 toves, a pile of product was produced whose mass was exactly 61.284 wibbles.

Result #3).

When the pile of product was allowed to settle for another 20 toves, it had spread out, and covered an area of 14.907 square filks.

Do YOU understand my results ?

All those other scientists would have a tough time trying to decide whether my results made sense.  And if they repeated my experiment, they would have no way to tell whether their results matched mine or not.

Without a standard system of measurement, and units that mean the same thing to everybody, us scientists literally could not communicate.


3 0
2 years ago
Read 2 more answers
A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

N_2 = 40

A = 7.5*10^{-4}m^2

l = 0.5m

Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0
3 years ago
You have a grindstone (a disk) that is 95.2 kg, has a 0.399 m radius, and is turning at 93 rpm, and you press a steel axe agains
olya-2409 [2.1K]

Answer:

angular acceleration is -0.2063  rad/s²

Explanation:

Given data

mass m = 95.2 kg

radius r = 0.399 m

turning ω = 93 rpm

radial force N  = 19.6 N

kinetic coefficient of friction  μ = 0.2

to find out

angular acceleration

solution

we know frictional force that is = radial force × kinetic coefficient of friction

frictional force = 19.6 × 0.2

frictional force = 3.92 N

and

we know moment of inertia  that is

γ =  I ×α = frictional force × r

so

γ  = 1/2 mr²α

α  = -2f /mr

α  = -2(3.92) /95.2 (0.399)

α  = - 7.84 / 37.9848 = -0.2063

so angular acceleration is -0.2063  rad/s²

3 0
3 years ago
Every force has one and only one 3rd law pair force. every force has one and only one 3rd law pair force.
Likurg_2 [28]
I would have to say B failed because I think I read something about it being only 2law not 3
7 0
3 years ago
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A 6kg sled sliding on an icy surface experiences a 6-N frictional force exerted by the ice and the air resistive force of 0.6 N.
White raven [17]

The net force on the sled is 6.6 N pointing backwards, opposite to the direction it's sliding. That's why it's slowing down, and will eventually stop.

5 0
3 years ago
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