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Travka [436]
3 years ago
14

During a storm, a student sees a rock fall from a cliff and smash into another rock below. which of the following is an applicat

ion of Newton's Second law of Motion? A - The gravitational attraction of Earth initially pulled the rock toward the ground. B - the mass of the rock is related to the force with which it hit the ground. C - The rock's inertia is increased on collision.
Physics
1 answer:
Igoryamba3 years ago
6 0
Since Newton's second law states that Force = mass x acceleration (F=ma), B is the correct description of an application.
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Consider the rocket in the image. Which unbalanced force accounts for the direction of the net force of the rocket ?
Wittaler [7]

Answer:

unbalenced force

Explanation:

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3 years ago
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A linear accelerator uses alternating electric fields to accelerate electrons to close to the speed of light. A small number of
kobusy [5.1K]

Answer:

8.1 x 10^13 electrons passed through the accelerator over 1.8 hours.

Explanation:

The total charge accumulated in 1.8 hours will be:

Total Charge = I x t = (-2.0 nC/s)(1.8 hrs)(3600 s/ 1 hr)

Total Charge = - 12960 nC = - 12.96 x 10^(-6) C

Since, the charge on one electron is e = - 1.6 x 10^(-19) C

Therefore, no. of electrons will be:

No. of electrons = Total Charge/Charge on one electron

No. of electrons = [- 12.96 x 10^(-6) C]/[- 1.6 x 10^(-19) C]

<u>No. of electrons = 8.1 x 10^13 electrons</u>

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3 years ago
The fluid inside the hydraulic jack has a pressure of 30,000 Pa. If the surface of
aleksley [76]

Explanation:

p = F /A

F = P×A

F = 30,000 Pa / 0.1 m²

F = 300,000 N

5 0
2 years ago
CAN SOMEONE PLS HELP ME!!!
scZoUnD [109]

Answer:

c

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2 years ago
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A5 cm object is 18.0 cm from a convex lens, which has a focal length of 10.0 cm.
Masja [62]

Explanation:

We have,

Height of object is 5 cm

Object distance from a convex lens is 18 cm

Focal length of convex lens is 10 cm

i.e. h = 5 cm

u = -18 cm

f = +10 cm

Let v is distance of the image from the lens. Using lens formula :

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{10}+\dfrac{1}{(-18)}\\\\v=22.5\ cm

The magnification of lens is :

m=\dfrac{v}{u}=\dfrac{h'}{h}, h' is height of the image

\dfrac{v}{u}=\dfrac{h'}{h}\\\\h'=\dfrac{vh}{u}\\\\h'=\dfrac{22.5\times 4}{(-18)}\\\\h'=-5\ cm

h' = -5.00 cm (in three significant figures)

6 0
3 years ago
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