Which item is a common bamer to physical activity?

According to the Academic Honesty Policy, plagiarism is ______.

hram777 [196]

UMUC. this is for the blank answer.

4 0

4 years ago

The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric fi

e-lub [12.9K]

Answer:

please the answer below

Explanation:

A general electromagnetic plane wave, traveling in the x direction, can be expressed in the form:

$\vec{E}=E_0e^{-i(k\cdot x-\omega t)}\hat{j}\\\\\vec{B}=B_0e^{-i(k\cdot x-\omega t)}\hat{k}\\\\$ (1)

1.

a. amplitudes

from (1) we can observe that E_0 and B_0 are the amplitudes.

2. frequency

3.

By replacing (1) we obtain:

$\vec{E}=E_0e^{-i(k(0)-\omega t)}\hat{j}=E_0[cos\omega t+sin\omega t]\hat{j}$

4.

the wave respond to the followinf physical variables: amplitude, frequency, time and position, as we can see in (1).

hope this helps!!

7 0

3 years ago

A rock is dropped at the same instant that a ball at the same height is thrown horizontally. which will will have the greater sp

Svetach [21]

They will both hit the ground at the same time due to gravity.
that means they took the same time to travel the distance to the ground

however, the ball traveled farther
speed=distance (or displacement) divided by time
so the greater the distance, the greater the speed

the ball traveled faster in the same time so it traveled faster

answer is ball

5 0

3 years ago

Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

Oksana_A [137]

Answer:

av=0.333m/s, U=3.3466J

b.

$v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s$

Explanation:

a. let $m_A$ be the mass of block A, and $m_B=10.0kg$ be the mass of block B. The initial velocity of A, $\rightarrow v_A_1=2.0m/s$

-The initial momentum =Final momentum since there's no external net forces.

$pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}$

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

$v_A_1-v_B_1=v_{B2}-v_{A2}$

-Applying the conservation of momentum. The blocks have the same velocity after collision:

$v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s$

#Total Mechanical energy before and after the elastic collision is equal:

$K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J$

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, $m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0$

We plug these values in the equation:

$m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}$

$2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s$

7 0

4 years ago

I need help with #40 need answer and show me what you multiplied please

ale4655 [162]

(0.5)×(0squared)×(3)=(1.5j)

3 0

3 years ago