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murzikaleks [220]
3 years ago
15

Consider the image above. Vi = the initial velocity and Vf = the final velocity. Is there acceleration? Explain your answer.

Physics
2 answers:
patriot [66]3 years ago
7 0

I just took the quiz, answer is d.

lutik1710 [3]3 years ago
3 0
<span><span>Velocity is a vector, and the initial and final ones are in opposite directions.
There must have been acceleration in order to change the direction of motion.</span>

A) No. The initial and final velocities are the same.
This is all wrong, and not the correct choice.
It's "Yes", and the initial and final velocities are NOT the same.

B) Yes. The ball had to slow down in order to change direction.
This is poor, and not the correct choice.
The "Yes" is correct, but the explanation is bad.
Acceleration does NOT require any change in speed.

C) No. Acceleration is the change in velocity. The ball's velocity is constant.
This is all wrong, and not the correct choice.
It's "Yes", there IS acceleration, and the ball's velocity is NOT constant.

D) Yes. Even though the initial and final velocities are the same, there is a change in direction for the ball.
This choice is misleading too.
The "Yes" is correct ... there IS acceleration.
The change in direction is the reason.
The initial and final velocities are NOT the same.  Only the speeds are.
</span>
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A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an
maksim [4K]

Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

   The second amplitude is  E_{max}2 =  81.87 \  (V/m)

   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

=>   \alpha  =  \frac{0.18}{90}

=>    \alpha =  0.002 Np/m

       

4 0
3 years ago
What is the acceleration for an object moving with constant speed? Explain your answer
Oduvanchick [21]

Answer:

What is the acceleration of an object moving at a constant speed?

The Meaning of Constant Acceleration

The data table above show an object changing its velocity by 10 m/s in each consecutive second. This is referred to as a constant acceleration since the velocity is changing by a constant amount each second.

3 0
3 years ago
Read 2 more answers
Which vector is the sum of vectors a and b
jeka57 [31]

Answer:

C

Explanation:

Vector A points up

Vector B points right

The combination must be both up and right which is C

5 0
3 years ago
Bond formed when one or more electrons are transferred from one atom to another. True or False
Finger [1]

Answer: true

Explanation:

8 0
3 years ago
Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of (a) 100 V, (b) 1.0 kV
forsale [732]

Answer:

(a) \lambda=1.227\ A

(b) \lambda=0.388\ A

(c) \lambda=0.038\ A

Explanation:

Given that,

(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :

\lambda=\dfrac{h}{\sqrt{2meV} }

Where

m and e are the mass of and charge on an electron

On solving,

\lambda=\dfrac{12.27}{\sqrt{V} }\ A

V = 100 V

\lambda=\dfrac{12.27}{\sqrt{100} }\ A

\lambda=1.227\ A

(b) V = 1 kV = 1000 V

\lambda=\dfrac{12.27}{\sqrt{V} }\ A

\lambda=\dfrac{12.27}{\sqrt{1000} }\ A

\lambda=0.388\ A

(c) If V=100\ kV=10^5\ V

\lambda=\dfrac{12.27}{\sqrt{10^5} }\ A

\lambda=0.038\ A

Hence, this is the required solution.

7 0
3 years ago
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