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3 years ago

From Smallest to Largest Moon asteroid meteorite meteoroid

Physics

2 answers:

Yanka [14]3 years ago

7 0

Meteorite meteoroid asteroid moon give brainlest please

Korolek [52]3 years ago

7 0

In my view
Meteorite
Meteoroid
Asteroid
Moon

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A particle with charge -5.60 nC is moving in a uniform magnetic field →B=−(1.25T) k The magnetic force on the particle is measur

chubhunter [2.5K]

Answer:

Explanation:

Force = q ( v x B)

- 5.6 x 10⁻⁹ (v x - 1.25 k )

- 3.4x 10⁻⁷i + 7.4 x 10⁻⁷j

Let v = ai+bj +ck

Force = - 5.6 x 10⁻⁹ [(ai+bj +ck) x - 1.25 k )]

= - 5.6 x 10⁻⁹ ( 1.25aj - 1.25bi )

= - 7 a j + 7 b i

( 7bi - 7aj ) x 10⁻⁹

Comparing with given force

7b x 10⁻⁹ b = - 3.4 x 10⁻⁷

b = - 48.57

- 7 a x 10⁻⁹ = 7.4 x 10⁻⁷

a = - 105.7

velocity

= -105.7 i - 48.57 j + ck

b ) Component along k can not be obtained .

c ) v . F = ( -105.7 i - 48.57 j + ck ) . −(3.40×10−7N) ˆı +(7.40×10−7N) ˆȷ

= 105.7 x 3.4 x 10⁻⁷ - 48.57 x 7.4 x 10⁻⁷

= 359.38 x 10⁻⁷ - 359.38 x 10⁻⁷

angle between v and F = 90 degree

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3 years ago

Now assume that Alice and Bob are twins, and Alice left Earth and Bob stayed behind fixing his spaceship. If Alice spent some ti

alexgriva [62]

我們的確認為我可以幫你們解決問題，我們要不要買不起房屋貸款土地

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3 years ago

Based on the information in the chart, which of these acids would be the best conductor of electricity?

Novay_Z [31]

Answer:

sulphuric acid

Explanation:

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4 years ago

The maximum torque on a set of 20 square loops with edge lengths of 6 cm in a 4.0 T magnetic field is:A) 0.288 N-m.B) 4.8 N-m.C)

Vesnalui [34]

Answer:

option (D

Explanation:

Torque is given by

torque = N x i x A x B x sinФ

where, N is number of turns, A is area, b is the magnetic field and Ф be the angle between the area vector and the magnetic field vector, i be the current.

So, torque depends on the current.

option (D)

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3 years ago

The closely packed cones in the fovea of the eye have a diameter of about 2 Î¼m. For the eye to discern two images on the fovea

Marina CMI [18]

Answer:

The distance between the object is $l=0.0056\ cm$

Explanation:

The free body diagram of this setup is on the first uploaded image

From the question

The diameter of closely packed cones in the fovea of the eye is = $2 \mu m$

The distance of separation by one cone(not excited ) is $d = 4\mu m = 4*10^{-4}cm$

The distance between the two point-like object is l

The diameter of the eye is D = 2 cm

The distance of the two point-like object from the near point of the eye is A = 28 cm

From the diagram we see that the light from the two point-like object form a triangle of similar base l and d and height D and A

So for a triangle with similar base we have that

$\frac{l}{A} =\frac{d}{D}$

$\frac{l}{28} = \frac{4*10^{-4}}{2}$

making l the subject we have

$l = \frac{28 *4*10^{-4}}{2}$

$l=0.0056\ cm$

5 0

3 years ago

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