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2 years ago

Definition of displacement

1 answer:

5 0

The moving of something from its place or position.
"vertical displacement of the shoreline"
the removal of someone or something by someone or something else that takes their place.
"males may be able to resist displacement by other males"

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Question 25 of 30

VARVARA [1.3K]

Answer:

it's B. circuit a and b are series circuit while c is parallel

7 0

2 years ago

A wall has a negative charge distribution producing a uniformhorizontal electric field. A small plastic ball of mass .01kg carry

Sauron [17]

Answer:

a) E = -4 10² N / C , b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰

Explanation:

For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball

X axis

$F_{e}$ - $T_{x}$ = m a

Axis y

$T_{y}$ - W = 0

Initially the system is in equilibrium, so zero acceleration

Fe = $T_{x}$

T_{y} = W

Let us search with trigonometry the components of the tendency

cos θ = T_{y} / T

sin θ = $T_{x}$ / T

T_{y} = cos θ

$T_{x}$ = T sin θ

We replace

q E = T sin θ

mg = T cosθ

a) the electric force is

$F_{e}$ = q E

E = $F_{e}$ / q

E = -0.032 / 80 10⁻⁶

E = -4 10² N / C

b) the distance to this point can be found by dividing the two equations

q E / mg = tan θ

θ = tan⁻¹ qE / mg

Let's calculate

θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)

θ = tan⁻¹ 0.3265

θ = 18 ⁰

sin 18 = x/0.30

x =0.30 sin 18

x = 0.093 m

c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations

X axis

$F_{e}$ = m aₓ

aₓ = q E / m

aₓ = 80 10⁻⁶ 4 10² / 0.01

aₓ = 3.2 m / s²

Axis y

W = m $a_{y}$

a_{y} = g

a_{y} = 9.8 m/s²

The total acceleration is can be found using Pythagoras' theorem

a = √ aₓ² + a_{y}²

a = √ 3.2² + 9.8²

a = 10.31 m / s²

The Angle meet him with trigonometry

tan θ = a_{y} / aₓ

θ = tan⁻¹ a_{y} / aₓ

θ = tan⁻¹ (-9.8) / 3.2

θ = -71.9⁰

Movement is two-dimensional type with acceleration in both axes

8 0

2 years ago

Which graph below represents how the velocity of the sphere changes over time when falling with constant acceleration?

Amiraneli [1.4K]

Graph B represents the velocity of the sphere changes over time when falling with constant acceleration.

Acceleration is the measure of how quickly a body's velocity varies with regard to time, and constant acceleration occurs when a body's velocity changes proportionately over a period of time, or at a constant rate. It measures in m/s2.

It is claimed that a body has continual positive acceleration when it begins to move with an initial velocity of zero and gradually increases to a positive value over time.

Constant positive acceleration is demonstrated by a ball falling freely in a vertical direction.

To know more about constant acceleration. visit : brainly.com/question/9754169

#SPJ1

4 0

1 year ago

Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t

Readme [11.4K]

Answer: $3.66(10)^{33}kg$

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity $\omega$ of the planet P1 with a period $T=750years=2.36(10)^{10}s$ :

$\omega=\frac{2\pi}{T}=\frac{V_{1}}{R}$ (1)

Where:

$V_{1}=40.2km/s=40200m/s$ is the velocity of planet P1

$R$ is the radius of the orbit of planet P1

Finding $R$ :

$R=\frac{V_{1}}{2\pi}T$ (2)

$R=\frac{40200m/s}{2\pi}2.36(10)^{10}s$ (3)

$R=1.5132(10)^{14}m$ (4)

On the other hand, we know the gravitational force $F$ between the star S with mass $M$ and the planet P1 with mass $m$ is:

$F=G\frac{Mm}{R^{2}}$ (5)

Where $G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

In addition, the centripetal force $F_{c}$ exerted on the planet is:

$F_{c}=\frac{m{V_{1}}^{2}}{R^{2}}$ (6)

Assuming this system is in equilibrium:

$F=F_{c}$ (7)

Substituting (5) and (6) in (7):

$G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}}$ (8)

Finding $M$ :

$M=\frac{V^{2}R}{G}$ (9)

$M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}}$ (10)

Finally:

$M=3.66(10)^{33}kg$ (11) This is the mass of the star S

4 0

3 years ago

A rock falls from rest off a cliff and hits the ground in 2 seconds. If there is no air resistance, determine the rock's velocit

mihalych1998 [28]

Answer:

19.6m/s

Explanation:

A Rock falling off a cliff can be modeled as an object starting with zero velocity moves with constant acceleration for certain period of time, for such motion following equation of motion can be used.

$v(t) = v_{0} +at$

here in our case $v_{0} =0$ because object starts off from rest and $a = g =9.8m/s^2$ is acceleration because of gravity ( Motion under gravity).

and of course t = 2 second.

Now by substituting all this information in equation of motion we get.

$v(2s) = 0+9.8m/s^2 *2s = 19.6m/s$

that would be the velocity of rock as it would hit the ground.

Note! We have assumed that there is no air resistance.

A rock falling off a cliff can be modeled as an object starting with zero velocity moves with constant acceleration for a certain period of time, for such motion following equation of motion can be used.

here in our case because object starts off from rest and is acceleration because of gravity ( Motion under gravity).

and of course t = 2 seconds.

Now by substituting all this information in equation of motion we get.

V = 19.6m/s

that would be the velocity of rock as it would hit the ground.

Note! We have assumed that there is no air resistance.

5 0

2 years ago