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2 years ago

Why should we convert units within the metric system?

Physics

1 answer:

dexar [7]2 years ago

6 0

We need it to calculate the how big it is, it’s mass, or even volume to get the measurements of lots of objects

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A record player has a velocity of 33.33 RPM. How fast is the record spinning in m/s at a distance of 0.085 m from the center?

anygoal [31]

Answer:

A record player has a velocity of 33.33 RPM. How fast is the record spinning in m/s at a distance of 0.085 m from the center? [0.297 m/s] 6. A merry-go-round a.k.a “the spinny thing” is rotating at 15 RPM, and has a radius of 1.75 m A.

5 0

3 years ago

Read 2 more answers

One has a frequency of 490 Hz and the other is 488 Hz. How many beats per second are heard?

tresset_1 [31]

When sounds at two different frequencies are combined,
two new sounds are created ... at the sum and difference
of the original frequencies.

Combining two sounds at 490 Hz and 488 Hz creates
beats at 978 Hz and 2 Hz.

The fluttering "wah wah" effect of the 2 Hz beat is much more
noticeable than the new sound at 978 Hz.

7 0

3 years ago

Alpha particles, each having a charge of +2e and a mass of 6.64 ×10-27 kg, are accelerated in a uniform 0.50 T magnetic field to

sergij07 [2.7K]

Answer:

$KE=1.2036\times 10^{-12}\ J$

Explanation:

Given:

charge on the alpha particle, $q=2e=3.2\times 10^{-19}\ C$

mass of the alpha particle, $m=6.64\times 10^{-27}\ kg$

strength of a uniform magnetic field, $B=0.5\ T$

radius of the final orbit, $r=0.5\ m$

<u>During the motion of a charge the magnetic force and the centripetal forces are balanced:</u>

$q.v.B=m.\frac{v^2}{r}$

$m.v=q.B.r$

where:

v = velocity of the alpha particle

$v=\frac{q.B.r}{m}$

$v=\frac{3.2\times 10^{-19}\times 0.5\times 0.5}{6.64\times 10^{-27}}$

$v=1.2048\times 10^{7}\ m.s^{-1}$

Here we observe that the velocity of the aprticle is close to the velocity of light. So the kinetic energy will be relativistic.

<u>We firstly find the relativistic mass as:</u>

$m'=\frac{1}{\sqrt{1-\frac{v^2}{c^2} } } \times m$

$m'=\frac{6.64\times 10^{-27}}{\sqrt{1-\frac{(1.2048\times 10^7)^2}{(3\times 10^8)^2} } }$

$m'=6.6533\times10^{-27}\ kg$

now kinetic energy:

$KE=m'.c-m.c$

$KE=6.6533\times 10^{-27}\times (3\times 10^8)^2-6.64\times 10^{-27}\times (3\times 10^8)^2$

$KE=1.2036\times 10^{-12}\ J$

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3 years ago

A 500kg car is driven forward with a thrust force of 1500N. Air resistance and friction acts against the motion of the car with

Wittaler [7]

2m/s^2, this is because F=ma, meaning a is also equal to F/m. The car applies 1500N in one direction and outside sources apply a total of -500N, meaning the 500kg car is moving forward with a total of 1000N of force. Taking the total 1000N and dividing it by 500kg gives you and acceleration of 2m/s^2. Hope this helps!

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2 years ago

A 0.48 kg circus monkey is about to be shot from a cannon as part of his thrilling circus act. Draw a free body diagram labeling

Alenkinab [10]

What is it that you need help on?

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2 years ago