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3 years ago

10

Name three different avenues by which Thomas Edison received an education

1 answer:

klio [65]3 years ago

5 0

Edison's education is most unique and relevant.

1. The first teacher he had was his mother
2. He found vital lessons and was influenced greatly by the book of R.G. Parker called School of Natural Philosophy
3. Another educating piece he had was a book entitled The Cooper Union for the Advancement of Science and Art

His style of learning was though reading books on a variety of subjects, a self-educating environment that fosters independent learning which can be useful through his life.

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son4ous [18]

Answer: 292.95 J

Explanation:

change in internal energy= Heat transfer - work done

ΔU =Q -PΔV

Here, Q = 0 as there is no heat transfer.

P =2.00 atm = 2.00 × 101235 Pa = 202470 Pa

ΔV = final volume - initial volume = 0.8 V -V = -0.2 V

where V is the initial volume.

Volume of a spherical ball, $V = \frac{4}{3}\pi r^3$

r = d/2 = 23.9 cm / 2 = 0.12 m

$V = \frac{4}{3}\times 3.14 \times (0.12m)^3= 7.23\times10^{-3}m^3$

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Hence, internal energy would change by 292.95 J.

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3 years ago

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What has a positive charge a comb or rubber band?

Gelneren [198K]

Id say the rubber band

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3 years ago

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A car moves with an initial velocity of 19 m/s due north. (Part A) Find the velocity of the car after 5.6 s if its acceleration

galben [10]

Answer

Assuming

east is the positive x direction

north is the positive y direction

initial velocity , u = 19 j m/s

a) acceleration , a = 1.6 j m/s^2

Using first equation of motion

v = u + a × t

v = 19 + 5.6× 1.6

v = 28 j m/s

the velocity of the car after 5.6 s is 28 m/s north

b)

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Using first equation of motion

v = u + a × t

v = 19 - 5.6 ×1.5

v = 10.6 j m/s

the velocity of the car after 5.6 s is 10.6 m/s north

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3 years ago

A bullet is fired straight up from a gun with a

Katena32 [7]

<span>v(4 seconds)= 300 m/s - 9.8 (m/s^2)(4s) = 260.8 m/s </span>, hope this helps:)

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A spring with constant k = 78 N/m is at the base of a frictionless, 30.0°-inclined plane. A 0.50-kg block is pressed against the

olasank [31]

Explanation:

The given data is as follows.

Spring constant (k) = 78 N/m, $\theta = 30^{o}$

Mass of block (m) = 0.50 kg

According to the formula of energy conservation,

mgh sin $\theta = \frac{1}{2}kx^{2}$

h = $\frac{1}{2} \times \frac{kx^{2}}{mg Sin \theta}$

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