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3 years ago

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A municipal water supply is provided by a tall water tower. Water from this tower flows to a building. How does the water flow o

ut of a faucet on the ground floor of a building compare with the water flow out of an identical faucet on the second floor of the building

1 answer:

stealth61 [152]3 years ago

4 0

Answer:

THE ANSWER IS: Water flows more rapidly out of the ground-floor faucet.

Explanation:

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A technician wishes to determine the wavelength of the light in a laser beam. To do so, she directs the beam toward a partition

Nana76 [90]

Answer:

λ = 5.656 x 10⁻⁷ m = 565.6 nm

Explanation:

Using the formula of fringe spacing from the Young's Double Slit experiment, which is given as follows:

$\Delta x = \frac{\lambda L}{d}\\\\\lambda = \frac{\Delta x\ d}{L}$

where,

λ = wavelength = ?

Δx = fringe spacing = 1.6 cm = 0.016 m

L = Distance between slits and screen = 4.95 m

d = slit separation = 0.175 mm = 0.000175 m

Therefore,

$\lambda = \frac{(0.016\ m)(0.000175\ m)}{4.95\ m}\\\\$

<u>λ = 5.656 x 10⁻⁷ m = 565.6 nm</u>

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3 years ago

If a 100 N net force act on a 50-kg car, what will the acceleration be

inna [77]

Acceleration = force / mass.

A = 100/50 = 2 m/s^2 .

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2 years ago

Blake is setting up his tent at a renaissance fair. If the tent is 8 feet tall and the tether can be staked no more than 2 feet

Mariana [72]

The stake, height and tether length of the tent form a right angle triangle where the tether length is the hypotenuse.
Applying Pythagoras theorem:
length² = height² + (stake distance)²
length = √(8² + 2²)
length = 8.5 feet

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2 years ago

Which of the following events may occur when IR radiation is absorbed?

kherson [118]

can you have all of them as possibilities ???

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3 years ago

In a historical movie, two knights on horseback start from rest 86 m apart and ride directly toward each other to do battle. Sir

Harlamova29_29 [7]

Answer:

Relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

Explanation:

Let the distance covered by Sir George be $S_{1}$

and the distance covered by Sir Alfred be $S_{2}$

Since the knights collide, hence they must have traveled for the same amount of time just before collision

From one of the equations of motion for linear motion

$S = ut + \frac{1}{2}at^{2}$

Where $S$ is the distance traveled

$u$ is the initial velocity

$a$ is the acceleration

and $t$ is the time

For Sir George,

$S = S_{1}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.21 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{1} = (0)t + \frac{1}{2}(0.21)t^{2}\\S_{1} = 0.105 t^{2}\\$

$t^{2} = \frac{S_{1}}{0.105}$

Now, for Sir Alfred

$S = S_{2}$

$u$ = 0 m/s (Since they start from rest)

$a =$ 0.26 m/s²

Hence,

$S = ut + \frac{1}{2}at^{2}$ becomes

$S_{2} = (0)t + \frac{1}{2}(0.26)t^{2}\\S_{2} = 0.13 t^{2}\\$

$t^{2} = \frac{S_{2}}{0.13}$

Since, they traveled for the same time, $t$ just before collision, we can write

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$

Since, the two nights are 86 m apart, that is, the sum of the distances covered by the knights just before collision is 86 m. Then we can write that

$S_{1} + S_{2} = 86$ m

Then, $S_{2} = 86 - S_{1}$

Then,

$\frac{S_{1}}{0.105}= \frac{S_{2}}{0.13}$ becomes

$\frac{S_{1}}{0.105}= \frac{86 -S_{1}}{0.13}$

$0.13{S_{1}}= 0.105({86 -S_{1}})\\0.13{S_{1}}= 9.03 - 0.105S_{1}}\\0.13{S_{1}} + 0.105S_{1}}= 9.03 \\0.235{S_{1}} = 9.03\\{S_{1}} =\frac{9.03}{0.235}$

$S_{1} = 38.43$ m

∴ Sir George covered a distance of 38.43 m just before collision.

Hence, relative to Sir George's starting point, the knights collide at a distance of 38.43 m from Sir George's starting point.

6 0

3 years ago