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3 years ago

6

The teachers of Scott, Christopher, Dianne, and Kailee have last names Koeninger, Dannemiller, Briscoe, and Carter. Match the ki

ds to their teacher.
1. Neither Ms. Dannemiller nor Ms. Briscoe is Kailee's teacher.
2. Christopher's teacher, who is not Carter, said that Christopher is one of his favorite students.
3. Dianne's teacher is not Dannemiller.

1 answer:

Pani-rosa [81]3 years ago

4 0

Scott-Dannemiller, Koeninger, Briscoe, and Carter

Christopher-Briscoe, Dannemiller, and Koeninger

Dianne-Koeninger, Briscoe, and Carter

Kailee-Koeninger and Carter

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How many new facilities does an average facility manager build during his or her career?

wlad13 [49]

An average facility manager can build one new facility during his or her career.

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2 years ago

Two wheel gears are connected by a chain. The larger gear has a radius of 8 centimeters and the smaller gear has a radius of 3 c

dezoksy [38]

Answer:

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

Explanation:

As we know that the small gear completes 24 revolutions in 20 seconds

so the angular speed of the smaller gear is given as

$\omega = 2\pi\frac{24}{20}$

$\omega = 2.4\pi rad/s$

Now we know that the tangential speed of the chain is given as

$v = r \omega$

so we have

$v = (3 cm)(2.4\pi)$

$v = 7.2 \pi cm/s$

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Since both gears are connected by same chain so both have same linear speed and hence correct answer will be

B.The linear velocity of the gears is the same. The linear velocity is 432π centimeters per minute.

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3 years ago

Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T

djverab [1.8K]

Answer:

$v_{2}$ will be less than $v_{1}$ and $P_{2}$ will be greater than $P_{1}$ .

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass ( $m_{1}$ ) is entering in a system is equal to the rate at which the same amount of fluid mass ( $m_{2}$ ) is leaving the system.

Rate of mass flow can be written as,

$m = \rho A v$

where $\rho$ is the density of the fluid, $A$ is the area through which the fluid is flowing and $v$ is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

$&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}$

where $A_{1}$ and $A_{2}$ are the cross-sectional areas through which the fluid is passing and $v_{1}$ and $v_{2}$ are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, $A_{2} > A_{1}$ , so from the above formula $v_{2} < v_{1}$ .

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case $v_{2} < v_{1}$ the pressure in the large cross-sectional area $P_{2}$ will be greater than the pressure $P_{1}$ in the small cross sectional area, i.e.,

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3 years ago

How would a parachute fall on the surface of the moon ? Describe. The mass of the Jupiter is

GREYUIT [131]

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2 years ago

Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher

siniylev [52]

Answer:

$v = 4.44 \times 10^5 m/s$

Explanation:

By Einstein's Equation of photoelectric effect we know that

$h\nu = W + \frac{1}{2}mv^2$

here we know that

$h\nu$ = energy of the photons incident on the metal

$W$ = minimum energy required to remove photons from metal

$\frac{1}{2}mv^2$ = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

$W = \frac{hc}{351 nm}$

here we know that

$hc = 1242 eV-nm$

now we have

$W = \frac{1242}{351} = 3.54 eV$

now by energy equation above when photon of 303 nm incident on the surface

$\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2$

$4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2$

$(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2$

$8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2$

$v = 4.44 \times 10^5 m/s$

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3 years ago