All categories

3 years ago

Which type of friction keeps a mound of rocks from falling away from each other

2 answers:

6 0

Hello Lunaerica62ovuaaf, Which type of friction keeps a mound of rocks from falling away from each other,
<span>B) static friction.</span>

nirvana33 [79]3 years ago

6 0

b) static friction

Static means "not moving." So friction that is static is friction that keeps things from moving.

You might be interested in

A simple series circuit consists of a 120 Ω resistor, a 21.0 V battery, a switch, and a 3.50 pF parallel-plate capacitor (initia

slega [8]

Answer

Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)

Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.

ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)

Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC

30.7pC/εo = 3.47 V∙m <----- C)

ic(0.5ns) = 29.7ma <----- D)

8 0

3 years ago

Q11) If you were standing at the top of a building and you dropped a rock.

Dafna1 [17]

Answer:

Part A

The distance travel by the rock is approximately 132.496 m

Part B

The speed when the rock hits the ground is approximately 50.96 m/s

Explanation:

Part A

The question is focused on the kinetics equation of a free falling object

The given parameter is the time it takes the rock to hit the ground, t = 5.2 s

For an object in free fall, we have;

h = 1/2·g·t²

Where;

h = The height from which the object is dropped

g = The acceleration due to gravity ≈ 9.8 m/s²

t = The time taken to travel the distance, h = 5.2 s

∴ h = 1/2 × 9.8 m/s² × (5.2 s)² ≈ 132.496 m

The distance travel by the rock, h ≈ 132.496 m

Part B

The speed, 'v', when the rock hits the ground, is given by the following kinematic equation,

v = g·t

∴ v = 9.8 m/s² × 5.2 s = 50.96 m/s

The speed when the rock hits the ground, v ≈ 50.96 m/s.

8 0

2 years ago

A charged particle accelerated to a velocity v enters the chamber of a mass spectrometer. The particle's velocity is perpendicul

gladu [14]

Answer:

Circle

Explanation:

When a charged particle is in motion in a region with magnetic field, the particle experiences a force whose magnitude is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

$\theta$ is the angle between the directions of v and B

In this problem, the velocity of the particle is perpendicular to the magnetic field, so

$\theta=90^{\circ}$

and the formula reduces to

$F=qvB$

Also, the direction of this force is perpendicular to the direction of motion of the particle. This means that as the charge moves in the region of the magnetic field, the force acting on it acts as a centripetal force: therefore, the particle will start moving by unifom circular motion, with constant speed (because the magnetic force does no work on the particle, since it is perpendicular to the direction of motion).

So, the path of the particle will be a circle.

4 0

3 years ago

How far away is the sun from earth in AU?

julia-pushkina [17]

That distance is the definition of 1 AU.

5 0

2 years ago

A biker pedals hard to ride his bike to the top of a 44 m hill. He decides to let his bike coast down the hill, and is having so

Dafna11 [192]

Answer:

The bikers speed at the top of other hill is <u>25.82 m/s.</u>