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3 years ago

A four-year college-level degree is called

2 answers:

8 0

A four-year college-level degree is called a bachelor degree.
If you decide to continue your education after receiving your bachelor degree, you will then pursue an associate degree.
If you decide to further pursue a higher degree after receiving an associate degree, you will then pursue a master degree.
Finally, once you have received your master degree, if you want to pursue an even higher education, you will then pursue a doctorate degree, or PhD.

FinnZ [79.3K]3 years ago

7 0

A four year degree is a bachelor's degree. It is the next step after an associate's degree, although most students who have a bachelor's will not have an associate's.

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The amplitude of a sound wave determines its

Savatey [412]

The Loudness i believe.

5 0

3 years ago

Read 2 more answers

Calculate the kinetic energy of Earth due to its spinning about its axis. Krot = J Compare your answer with the kinetic energy o

Studentka2010 [4]

Answer:

Explanation:

Answer: Let ke = 1/2 IW^2 = 1/2 kMr^2 W^2 be Earth's rotational KE. W = 2pi/24 radians per hour rotation speed and k = 2/5 for a solid sphere M is Earth mass, r = 6.4E6 m.

Then ke = 1/2 2/5 6E24 (6.4E6)^2 (2pi/(24*3600))^2 = ? Joules. You can do the math, note W is converted to radians per second for unit consistency.

Let KE = 1/2 KMR^2 w^2 be Earth's orbital KE. w = 2pi/(365*24) radians per hour K = 1 for a point mass. Note I used 365 days, a more precise number is 365.25 days per year, which is why we have Leap Years.

Find KE/ke = 1/2 KMR^2 w^2//1/2 kMr^2 W^2 = (K/k)(w/W)^2 (R/r)^2 = (5/2) (365)^2 (1.5E11/6.4E6)^2 = 7.81E9 ANS

8 0

4 years ago

Learning Goal: To review the concept of conservative forces and to understand that electrostatic forces are, in fact, conservati

CaHeK987 [17]

Explanation:

The electrostatic forces are conservative forces!

The mainly property of the conservative fields is $\vec{\nabla} \times \vec E=\vec 0$

In spherical coordinates the field's expression is:

$\vec E=\frac{Q}{4\pi \epsilon _0 r^2} .\^r$

and the curl expression is:

$\nabla\times \vec E=\frac{1}{r^2{\sin}\,\theta}\left|\begin{matrix}\hat{r} & r\,\hat{\theta} & r\,{\sin}\,\theta\,\hat{\varphi} \\& & \\\frac{\partial}{\partial r} & \frac{\partial}{\partial \theta} & \frac{\partial}{\partial \varphi}\\ & & \\E_r & rE_\theta & r{\sin}\,\theta\, E_\varphi\end{matrix}\right|=(0, 0, 0)$

to find the expression for the potential function associated:

$\vec E=\vec \nabla . V, \Delta V= V_b-V_a=-\int _c \vec E.d\vec l=-\int _c E\^r.dr\^r=-\int _c Edr=\int \limits^a_b \frac{Q}{4\pi \epsilon _0 r^2} dr= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{r}|^b_a)= \frac{Q}{4\pi \epsilon _0}.(\frac{1}{b}-\frac{1}{a})$

5 0

3 years ago

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30° angle. The block’s initial speed is 10 m/s. What vert

Romashka [77]

Answer:

Vertical Height = 0.784 meter, Speed back at starting point = 10 m/s

Explanation:

Given Data:

V is the overall velocity vector, $Vi$ and $Ui$ are its initial vertical and horizontal components

$R = 10 m/s\\ Projection Angle (theta) = 30 degrees\\Vi = 10*sin(30) = 5 m/s\\Ui = 10*cos(30) = 8.66 m/s$

To find:

Max Height $h$ achieved

Calculation:

1) Using the $3^{rd}$ equation of motion, we know

$2*a*s = Vf^{2} - Vi^{2}$

2) In terms of gravity $g$ height $h$ and the vertical component of Velocity $Vf , Vi$ .

3) As $Vf = 0$ as at maximum height the vertical component of velocity is zero maximum height achieved

$2*g*h = Vf^{2} -Vi^{2}$

putting values

4) $h = 0.784 m/s$

5) As for the speed when it reaches back its starting point, it will have a speed similar to its launching speed, the reason being the absence of air friction (Air drag)

3 0

4 years ago

A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed

AleksandrR [38]

(a) 3.5 Hz

The angular frequency in a spring-mass system is given by

$\omega=\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

Here in this problem we have

k = 160 N/m

m = 0.340 kg

So the angular frequency is

$\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s$

And the frequency of the motion instead is given by:

$f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz$

(b) 0.021 m

The block is oscillating up and down together with the upper end of the spring. The block will lose contact with the spring when the direction of motion of the spring changes: this occurs when the spring is at maximum displacement, so at

x = A

where A is the amplitude of the motion.

The maximum displacement is given by Hook's law:

$F=kA$