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Pepsi [2]
2 years ago
14

Is more of north america north or south of the tropic of cancer

Physics
1 answer:
TEA [102]2 years ago
6 0

Explanation:

This parallel is located in the northern hemisphere, located at a latitude of 23 ° 26′12.2 ″ (or 23,43671 °) north of the equator. Approximately in the geographic half of Mexico, crossing the states of Baja California Sur, Sinaloa, Durango, Zacatecas, San Luis, Potosí, Nuevo León, Tamaulipas. Therefore, there is more of north america, at north of the tropic of cancer.

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An 7.5 × binocular has 3.7-cm-focal-length eyepieces. What is the focal length of the objective lenses? Express your answer to t
elixir [45]

To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,

\mu = \frac{f_0}{f_e}

Here,

\mu = Magnification

f_e = Focal length eyepieces

f_0 = Focal length of the Objective

Rearranging to find the focal length of the objective

f_0 = \mu f_e

Replacing with our values

f_0 = 7.5* 3.7cm

f_0 = 27.75cm

Therefore the focal length of th eobjective lenses is 27.75cm

5 0
3 years ago
The mere exposure effect is
gulaghasi [49]
DescriptionThe mere-exposure effect is a psychological phenomenon by which people tend to develop a preference for things merely because they are familiar with them.
6 0
3 years ago
A contestants spin a wheel when it is their turn in a game show. One contestant gives the wheel an initial angular speed of 3.40
guajiro [1.7K]

Answer:

4.62 s

Explanation:

We are given that

Initial angular speed,\omega=3.4 rad/s

\theta=1\frac{1}{4} rev=\frac{5}{4}\times 2\pi=2.5\pi rad

\omega'=0

\omega'^2-\omega^2=2\alpha \theta

Substitute the values

0-(3.4)^2=2\times 2.5\pi \alpha

\alpha=\frac{-(3.4)^2}{2\times 2.5\pi}=-0.736 rad/s^2

\omega'=\omega+\alpha t

0=3.4-0.736 t

-0.736t=-3.4

t=\frac{-3.4}{-0.736}=4.62 s

Hence, the wheel takes 4.62 s to come to rest.

3 0
3 years ago
What is the greatest distance an image can be located behind a convex spherical mirror?
Afina-wow [57]

Answer:

Maximum distance of image from mirror is equal to focal length of the mirror

Explanation:

As we know by the equation of mirror we have

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

here we know for convex mirror

object position is always negative as it will be placed behind the mirror always

while the focal length of the convex mirror is always taken positive

So here we have

\frac{1}{d_i} + \frac{1}{-d_o} = \frac{1}{f}

\frac{1}{d_i} = \frac{1}{d_o} + \frac{1}{f}

so here maximum value of image distance is equal to focal length of the mirror

6 0
3 years ago
Which option describes nonmetals?
jonny [76]

Answer:

D: Conducts electrcity poorly

Explanation:

Most conductors are metals, and non- metals are insulators, therefore they dont conduct electricity

8 0
3 years ago
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