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2 years ago

Question in picture.

Physics

2 answers:

Ber [7]2 years ago

4 0

The answers are to small to see.

Agata [3.3K]2 years ago

4 0

<h2>Answer:</h2>

<u>The correct choice is </u><u>C) 17.78 m/s</u>

<h2>Explanation:</h2>

According to the addition of vectors we add the magnitudes of vectors and the direction is seen in accordance to the resultanat vector. If the direction is same fro both vectors then simply the magnitude is added and direction remains the same. If the direction is opposite then the magnitude is subtracted. In the given two vectors the direction o f both the vectors is same so we simply add them as 12.57 + 5.21 = 17.78

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2 years ago

What is the value of ΔVBA=VB−VA if the charge on the plates is 1.00 x 10-9 C, the area of the plates is 2.00 m2 and the distanc

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3 years ago

A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction

rusak2 [61]

I'll bite:

-- Since the sled's mass is 'm', its weight is 'mg'.

-- Since the coefficient of kinetic friction is μk, the force acting opposite to the direction it's sliding is    (μk) times (mg) .

-- If the pulling force is constant 'F', then the horizontal forces on the sled
are 'F' forward and (μk · mg) backwards.

-- The net force on the sled is (F - μk·mg).
(I regret the visual appearance that's beginning to emerge,
but let's forge onward.)

-- The sled's horizontal acceleration is (net force) / (mass) = (F - μk·mg) / m.
This could be simplified, but let's not just yet.

-- Starting from rest, the sled moves a distance 's' during time 't'.
We know that s = 1/2 a t² , and we know what 'a' is. So we can write

   s = (1/2 t²) (F - μk·mg) / m .

Now we have the distance, and the constant force.
The total work is (Force x distance), and the power is (Work / time).
Let's put it together and see how ugly it becomes. Maybe THEN
it can be simplified.

Work = (Force x distance) = F x (1/2 t²) (F - μk·mg) / m

Power = (Work / time) = <em>F (t/2) (F - μk·mg) / m </em>

Unless I can come up with something a lot simpler, that's the answer.

To simplify and beautify, make the partial fractions out of the
2nd parentheses:
   <em> F (t/2) (F/m - μk·m)</em>

I think that's about as far as you can go. I tried some other presentations,
and didn't find anything that's much simpler.

Five points,ehhh ?

4 0

2 years ago

Read 2 more answers

One of the harmonics on a string 1.30m long has a frequency of 15.60 Hz. The next higher harmonic has a frequency of 23.40 Hz. F

Alja [10]

Answer:

$\large \boxed{\text{(a) 7.800 Hz; (b) 20.3 m/s; 40.6 m/s; 60.8 m/s}}$

Explanation:

a) Fundamental frequency

A harmonic is an integral multiple of the fundamental frequency.

$\dfrac{\text{23.40 Hz}}{\text{15.60 Hz}} = \dfrac{1.500}{1} \approx \dfrac{3}{2}$

$f = \dfrac{\text{24.30 Hz}}{3} = \textbf{7.800 Hz}$

b) Wave speed

(i) Calculate the wavelength

In a fundamental vibration, the length of the string is half the wavelength.

$\begin{array}{rcl}L & = & \dfrac{\lambda}{2}\\\\\text{1.30 m} & = & \dfrac{\lambda}{2}\\\\\lambda & = & \text{2.60 m}\\\end{array}$

(b) Calculate the speed s

$\begin{array}{rcl}v_{1}& = & f_{1}\lambda\\& = & \text{7.800 s}^{-1} \times \text{2.60 m}\\& = & \textbf{20.3 m/s}\\\end{array}$

$\begin{array}{rcl}v_{2}& = & f_{2}\lambda\\& = & \text{15.60 s}^{-1} \times \text{2.60 m}\\& = & \textbf{40.6 m/s}\\\end{array}$

$\begin{array}{rcl}v_{3}& = & f_{3}\lambda\\& = & \text{23.40 s}^{-1} \times \text{2.60 m}\\& = & \textbf{60.8 m/s}\\\end{array}$

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3 years ago

Meg goes swimming on a hot afternoon. When she comes out of the pool, her foot senses that the prevement is unbearably hot. Supp

mart [117]

1.Record her observation with the time it was hot.
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2 years ago