Tor F: When you exert a force on a baseball, the equal and opposite force on the

Name three different avenues by which Thomas Edison received an education

klio [65]

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.

5 0

3 years ago

Describe Darwin’s<br> observations on the Galápagos islands<br> during his voyage on the HMS Beagle.

Helen [10]

Answer:

During the voyage Charles Darwin explored the Galapagos islands and noticed the same species have different adaptations in places. ... Charles noticed that each species has the same ancestor but they evolve to adapt over time so they can live longer.

Explanation:

4 0

3 years ago

When the fission of uranium-235 is carried out, about 0.1 percent of the mass of the reactants is lost during the reaction.

Dmitry [639]

The lost mass is converted into energy.

7 0

3 years ago

A horizontal 745 N merry-go-round of radius

Arturiano [62]

Answer:

The kinetic energy of the merry-goround after 3.62 s is 544J

Explanation:

Given :

Weight w = 745 N

Radius r = 1.45 m

Force = 56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62 = ?

Solution:

Step 1: Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I = $0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I

=> $0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

Step 4: Finding the Angular acceleration

Angular acceleration , $\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

Step 4: Finding the Final angular velocity

Final angular velocity , $\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE = $0.5 \times 79.91 \times 3.69^2$

KE = 544J

6 0

4 years ago

If a rock is thrown upward on the planet Mars with a velocity of 10 m/s, its height in meters t seconds later is given by y= 10t

spin [16.1K]

Answer:

a)

i) v = 4.42 m/s

ii) v = 5.36 m/s

iii) v = 6.1 m/s

iv) v = 6.26 m/s

v) v = 6.28 m/s

b) The instantaneous velocity at t = 1 is 6.28 m/s

Explanation:

a) The average velocity is the variation of the position over time. It is expressed as follows:

v = Δy/Δt

Where

v = average velocity

Δy = displacement = final position - initial position

Δt = variation of time = final time - initial time

i) Let´s find the position at both times and then apply the equation for the average velocity:

y(t) = 10 · t - 1.86 · t²

y(1 s) = 10 m/s · 1 s - 1.86 m/s² · (1 s)²

y = 8.14 m

y (2 s) = 10 m/s · 2 s - 1.86 m/s² · (2 s)²

y = 12.56 m

Then, the average velocity will be:

v = final position - initial position / final time - initial time

v = 12.56 m - 8.14 m / 2 s - 1 s = 4.42 m/s

ii) We proceed in the same way as in i)

y(1.5 s) = 10 m/s · 1.5 s - 1.86 m/s² · (1.5 s)²

y = 10.82 m

v = 10.82 m - 8.14 m / 1.5 s - 1 s = 5.36 m/s

iii)

y(1.1 s) = 10 m/s · 1.1 s - 1.86 m/s² · (1.1 s)²

y = 8.75 m

v = 8.75 m - 8.14 m / 1.1 s - 1 s = 6.1 m/s

iv)

y(1.01 s) = 10 m/s · 1.01 s - 1.86 m/s² · (1.01 s)²

y = 8.20 m

v = 8.20 m - 8.14 m / 1.01 s - 1 s = 6 m/s ( 6.26 m/s without rounding the y-final value)

v)

y(1.001 s) = 10 m/s · 1.001 s - 1.86 m/s² · (1.001 s)²

y = 8.146

v = 8.146 m - 8.14 m / 1.001 s - 1 s = 6 m/s (6.28 m/s without rounding the value of y-final)

b) The instantaneous velocity is given by the derivative of the position function:

y = 10 · t - 1.86 · t²

dy/dt = 10 - 2 · 1.86 · t = 10 - 3.72 · t

At t = 1

v = 10 m/s - 3.72 m/s² · 1 s = 6.28 m/s

4 0

3 years ago