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riadik2000 [5.3K]
3 years ago
5

Two identical pebbles are dropped. The first is dropped from a height of 256 feet and the second is dropped from a height of 400

feet. Use the graphs to tell how long it takes each pebble to reach the ground.
Physics
1 answer:
ehidna [41]3 years ago
8 0

Answer:

4.022 seconds and 4.99 seconds

Explanation:

Hello!

The free fall of the stone corresponds to a uniformly varied rectilinear movement

d=V_0*t+1/2*g*t^2

Being a free fall the initial speed is zero.

The distance is positive when considered in the same direction and direction as acceleration and speed.

256 feet stone

79.25 m=0 m⁄s*t+1/2*9,8 m⁄s^2 *t^2

t = 4.022 seconds

400 feet stone

121.92m=0 m⁄s*t+1/2*9,8 m⁄s^2 *t^2

t= 4,99 seconds

success with your homework!

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4 0
2 years ago
Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i
garik1379 [7]

Answer

given,

wavelength (λ) = 650 nm

angle = 5°

using bragg's law

sin \theta = \dfrac{n \lambda}{d}

d= \dfrac{n \lambda}{sin \theta}

d= \dfrac{1 \times 650 \times 10^{-9}\ m}{sin5^0}

d = 7.46 x 10⁻⁴ cm

number of slits per centimeter

  = \dfrac{1}{d}\\\Rightarrow \dfrac{1}{7.46\times 10^{-4}}\\\Rightarrow 1340 split per centimeter.

b) wavelength of two rays  650 nm and 420 nm

 d = \dfrac{1}{5000}

     d =  2 x 10⁻6 m

    we now,

sin \theta = \dfrac{n \lambda}{d}

for 650 nm

sin \theta = \dfrac{2\times 650\times 10^{-9}}{2\times 10^{-6}}

\theta =sin^{-1}(0.65)

θ = 40.54°

for 450 nm

sin \theta = \dfrac{2\times 450\times 10^{-9}}{2\times 10^{-6}}

\theta =sin^{-1}(0.45)

θ = 24.83°

now, difference

|θ_{650} -θ_{420}| =40.54°-24.83°

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8 0
4 years ago
You take a couple of capacitors and connect them in series, to which you observe a total capacitance
Zepler [3.9K]

Answer:

Approximately \rm 5.7\; \mu F and approximately 29\; \rm \mu F.

Explanation:

Let C_1 and C_2 denote the capacitance of these two capacitors.

When these two capacitors are connected in parallel, the combined capacitance will be the sum of C_1 and C_2. (Think about how connecting these two capacitors in parallel is like adding to the total area of the capacitor plates. That would allow a greater amount of charge to be stored.)

C(\text{parallel}) = C_1 + C_2.

On the other hand, when these two capacitors are connected in series, the combined capacitance should satisfy:

\displaystyle \frac{1}{C(\text{series})} = \frac{1}{C_1} + \frac{1}{C_2}.

(Consider how connecting these two capacitors in series is similar to increasing the distance between the capacitor plates. The strength of the electric field (V) between these plates will become smaller. That translates to a smaller capacitance if the amount of charge stored (Q) stays the same.)

The question states that:

  • C(\text{parallel}) = 35\; \rm \mu F, and
  • C(\text{series}) = 4.8\; \rm \mu F.

Let the capacitance of these two capacitors be x\; \rm \mu F and y\; \rm \mu F. The two equations will become:

\displaystyle \left\lbrace \begin{aligned}& x + y = 35 \\ & \frac{1}{x} + \frac{1}{y} = \frac{1}{4.8}\end{aligned}\right..

From the first equation:

y = 35 - x.

Hence, the y in the second equation here can be replaced with (35 - x). That equation would then become:

\displaystyle \frac{1}{x} + \frac{1}{35 - x} = \frac{1}{4.8}.

Solve for x:

\displaystyle \frac{x + (35 - x)}{x \, (35 - x)} = \frac{1}{4.8}.

x\, (35 - x) = 4.8.

x^2 - 35 \, x + 168 = 0.

Solve this quadratic equation for x:

x \approx 5.7 or x \approx 29.3.

Substitute back into the equation y = 35 - x for y:

  • x \approx 5.7 and y \approx 29.3, or
  • x \approx 29.3 and x \approx 5.7.

In other words, these two capacitors have only one possible set of capacitances (even though the previous quadratic equation gave two distinct real roots.) The capacitances of the two capacitors would be approximately 5.7\; \rm \mu F and approximately 29\; \rm \mu F (both values are rounded to two significant digits.)

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3 years ago
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Answer:

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Explanation:

We will either convert the distance to centimeters or speed to meters/second.

Converting distance to centimeters: 100 meters= 100*100centimeters =10000centimeters.

time=distance/speed

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5 0
3 years ago
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