All categories

2 years ago

If you are standing at Earth’s North Pole, which of the following will be directly overhead?

Physics

1 answer:

Sidana [21]2 years ago

7 0

The north celestial pole (and also the south) are described as two imaginary points that project on the axis of rotation of the earth, indefinitely towards space and that intersects with the celestial sphere (Remember that the earth has a slight inclination with respect to to its axis of rotation). The north and south celestial poles are directly above the head of someone located at the north or south pole of the earth.

Correct Answer is A.

You might be interested in

Compare green and orange light from the visible spectrum. You are currently in a labeling module. Turn off browse mode or quick

77julia77 [94]

Green: nm 495–570. Yellow: nm 570–590. 590–620 nm for orange. Red: 620-750 nm (400–484 THz frequency)

Solids' molecules are strongly attracted to one another. As a result, the molecules are barely moving and tightly packed. Because of this, shape and volume are fixed.

The forces of attraction and repulsion in liquids are comparable. Compared to the solid state, they move a little bit more. They then assume the shape of the container while still having a fixed capacity.

The attraction forces between the molecules in gases are quite weak. They move quite freely and grow in an effort to fill as much space as they can. Consequently, their volume and shape vary (adopt the shape of the container).

You can learn more about states of the matter here:

brainly.com/question/18538345

#SPJ4

6 0

1 year ago

A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnifica

grin007 [14]

Answer:

$f_e = 1.51 cm$

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

$m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}$

$400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}$

$9.6 = \dfrac{16-f_e}{f_e}$

$9.6f_e = 16-f_e$

$10.6 f_e = 16$

$f_e = 1.51 cm$

3 0

3 years ago

Compared with a force, a torque involves

ycow [4]

A rotational movement.

6 0

2 years ago

A test charge is placed at a distance of 2.5 × 10-2 meters from a charge of 6.4 × 10-5 coulombs. What is the electric field at t

Novosadov [1.4K]

Using the formula: E = kQ / d² where E is the electric field, Q is the test charge in coulomb, and d is the distance.

E = kQ / d²

k = 9 x 10^9 N-m²/C²
Q = 6.4 x 10^-5 C
d = 2.5 x 10^-2 m

Substituting the given values to the equation, we have:
E = (9 x 10^9)(6.4 x 10^-5) / (2.5 x 10^-2) ²

Electric field at the test charge is 921600000 N/C

8 0

2 years ago

Read 2 more answers

How to find out the heat capacity of a material?

DochEvi [55]

$\huge\underline{\underline{\boxed{\mathbb {EXPLANATION}}}}$

The heat capacity is given by the expression:

$\longrightarrow \sf{\triangle Q= m \triangle C \triangle T}$

$\longrightarrow \sf{Q= \: Heat}$

$\longrightarrow \sf{M= \: Mass}$

$\longrightarrow \sf{C= \: Specific \: Heat}$

$\longrightarrow \sf{T= \: Temperature}$

$\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}$

$\leadsto$ When the $\bm{heat}$ is measured in the calorimeter, we obtain a value, and since we know the mass of the material and we control the change in $\bm{temperature}$ , we can then determine the specific heat "C" by simply remplazing in the expression.

5 0

1 year ago