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denis-greek [22]

3 years ago

10

When the rate of motion is the same direction the speed will decrease. True False

1 answer:

slega [8]3 years ago

6 0

A)true is the answer

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1) how does reflection differ from diffraction?

Evgesh-ka [11]

Answer: (1) The correct answer is A.

(2) The correct answer is D.

Explanation:

(1)

Reflection is the sending back of light from the surface without absorbing it. In the reflection phenomenon, the wave does not continue moving forward.

Diffraction is the bending of the light around the obstacle. In the diffraction phenomenon, the wave travels forward after striking around the obstacle.

Therefore, the correct answer is A.

(2)

Amplitude is the maximum displacement in the medium from the rest position.

The amount of energy is related to the amplitude. Amplitude is related to the amount of energy carried by the wave. Low energy wave is characterized by a low amplitude. High energy wave is characterized by a high amplitude.

Therefore, the correct option is D.

5 0

3 years ago

Read 2 more answers

When doing yoga stretches, you are working on the cardiovascular endurance component of physical fitness.

Tomtit [17]

I believe it is true :D

6 0

2 years ago

Read 2 more answers

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago

Using only one management style with all people is the most effective leadership technique for any organization.

Softa [21]

the answer for this is false

4 0

3 years ago

A 0.900-V potential difference is maintained across a 1.5m length of 2

Daniel [21]

Answer:

I = 6.42 A

Explanation:

Given that,

Potential difference, V = 0.9 V

Length of the wire, l = 1.5 m

Area of cross section, $A=0.6\ mm^2=6\times 10^{-7}\ m^2$

We need to find the current in the wire. Let I is current. We can find it using Ohm's law as follows :

V = IR

Where R is the resistance of the wire

$I=\dfrac{V}{R}\\\\I=\dfrac{V}{\rho \dfrac{l}{A}}\\\\I=\dfrac{0.9}{5.6\times 10^{-8}\times \dfrac{1.5}{6\times 10^{-7}}}\\\\I=6.42\ A$

So, the current in the wire is 6.42 A.

3 0

3 years ago