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marysya [2.9K]
2 years ago
10

Name three different avenues by which Thomas Edison received an education

Physics
1 answer:
klio [65]2 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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If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
Gnom [1K]

The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2


and the average current density is

J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2

8 0
2 years ago
Read 2 more answers
Solve the equation x=3logy2 for y.
melisa1 [442]

X = 3 · log(Y²)

X = 3 · 2·log(Y)

X/6 = log(Y)

10^(X/6) = 10^log(Y)

Y = 10^(X/6)

6 0
2 years ago
Read 2 more answers
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
2 years ago
Determine the change in thermal energy of 100 g of copper (M = 63,5, Debye 348K) if it is cooled from
Setler [38]

Answer:

given,

mass of copper = 100 g

latent heat of liquid (He) = 2700 J/l

a) change in energy

Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (300 - 4)

Q = 11153.63 J

He required

Q = m L

11153.63 = m × 2700

m = 4.13 kg

b) Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (78 - 4)

Q = 2788.41 J

He required

Q = m L

2788.41 = m × 2700

m = 1.033 kg

c) Q = m Cp (T₂ - T₁)

Q = 0.1 × 376.812 × (20 - 4)

Q = 602.90 J

He required

Q = m L

602.9 = m × 2700

m =0.23 kg

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2 years ago
Apply the impulse-momentum relation and the work-energy theorem to calculate the maximum value of t if the cake is not to end up
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Puto chupame el semen ok? right?
8 0
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