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3 years ago

Where on the Earth would you have to be to see the North Celestial Pole exactly halfway between your horizon and zenith?

Physics

1 answer:

mario62 [17]3 years ago

4 0

Answer:

The point straight overhead on the celestial sphere for any observer is called the zenith and is always 90 degrees from the horizon. The arc that goes through the north point on the horizon, zenith, and south point on the horizon is called the meridian.

From any location on Earth you see only half of the celestial sphere, the half above the horizon.

If you stood at the North Pole of Earth, for example, you would see the north celestial pole overhead, at your zenith. The celestial equator, 90° from the celestial poles, would lie along your horizon.

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A hawk leaves its nest in Fort Worth and travels 3,000 meters east. After snacking on a tasty rodent the hawk flies another 1,50

Paha777 [63]

I think the logical question here is to either find the distance or the displacement. They differ in such a way that distance is a scalar quantity that does not focus on the direction. Displacement is a vector quantity that covers the distance from the starting point to end point. Because it travels only in one direction (to the east), in this condition, distance is equal to displacement.

Distance = Displacement = 3,000 m + 1,500 m = 4,500 m

6 0

3 years ago

The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum

Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.

5 0

3 years ago

An electron is moving horizontally east in an electric field that points vertically downward. The electric force on the electron

saveliy_v [14]

Answer:

d. )directed upward.

Explanation:

As the electron has a negative charge, when under the influence of an electric field, is subject to an electric force, which direction is the opposite to the direction of the electric field.

This is because the electric field has the same direction that the force on a positive test charge at the same point.

As the electric field points vertically downward, the electric force on the electron (a negative charge) points vertically upward.

So, the statement d. is the one that results to be true.

7 0

3 years ago

Please left home at 8 AM to spend the day at in amusement park. He arrived at the park, which was 150 KM from his house, at 10 A

antiseptic1488 [7]

Answer:

$v=\frac{150km}{2h}=75\frac{km}{h}$

Explanation:

We can use the equation for the speed

$v=\frac{x}{t}$

where x is the distance and t the time. In this case we know that the time spent was 2 hours and the distance was 150km. By replacing we have

$v=\frac{150km}{2h}=75\frac{km}{h}$

I hope this useful for you

regards

3 0

3 years ago

Young's interference experiment demonstrated the particle nature of light T/F?

MAVERICK [17]

I think the answer is false

6 0

3 years ago