A ___ is the unit of measurement for force.

What is the highest degrees above the horizon the moon ever gets during the year in the Yakima Valley ?

Ivahew [28]

The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.

Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima. The latitude in the middle
of that intersection is 46.585° North. That's the number we need.

Here's how I would do it:

-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.

-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°.

-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.

This sets the limit of the highest in the sky that the moon can ever appear.

90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .

That doesn't happen regularly. It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).

Depending on the time of year, that can be any time of the day or night.

The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.

In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky. Then it's going to be somewhere near
67° above the horizon at midnight.

5 0

2 years ago

What is the longest wavelength of light that will emit electrons from a metal whose work function is 3.70 eV

SSSSS [86.1K]

Answer:

h f = Wf + K

where the total energy available is h f, Wf is the work function or the work needed to remove the electron and K is the kinetic energy of the removed electron

If K = zero then hf = Wf

Wf = h f = h c / λ or

λ = h c / Wf = 6.63E-34 * 3.0E8 / (3.7 * 1.6E-19)

λ = 6.63 * 3 / (3.7 * 1.6) E-7 = 3.36E-7

This would be 3360 angstroms or 336 millimicrons

Visible light = 400-700 millimicrons

6 0

1 year ago

This force will cause the path of the particle to curve. Therefore, at a later time, the direction of the force will ___________

melisa1 [442]

Answer:

have a component along the direction of motion that remains perpendicular to the direction of motion

Explanation:

In this exercise you are asked to enter which sentence is correct, let's start by writing Newton's second law.

circular movement

F = m a

a = v² / r

F = m v²/R

where the force is perpendicular to the velocity, all the force is used to change the direction of the velocity

in linear motion

F = m a

where the force is parallel to the acceleration of the body, the total force is used to change the modulus of the velocity

the correct answer is: have a component along the direction of motion that remains perpendicular to the direction of motion

8 0

3 years ago

A toy car has 2A of current if the car runs one four 1.5V batteries, what is the resistance?

lesya [120]

Answer:

3 ohm

Explanation:

Resistance = Voltage/Current

= 6.0/2

= 3 ohm

5 0

2 years ago

When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of t

choli [55]

Answer:

Part a)

$\tau = 23.1 Nm$

Part b)

$\tau = 17.05 Foot pound force$

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

$\tau = \vec r \times \vec F$

now we have

$\tau = (0.140)(165)$

$\tau = 23.1 Nm$

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

$\tau = 23.1 Nm$

$\tau = 23.1 (0.225 Lb)(3.28 foot)$

$\tau = 17.05 Foot pound force$

3 0

2 years ago