Ask question

Ask question

All categories

3 years ago

11

Does the Earth stay in one place throughout the year? If it doesn't, describe its motion and where it's located in the solar syt

em.

1 answer:

stiv31 [10]3 years ago

6 0

the earth moves throughout the year such as rotate around the sun, so yes the it does move and it sits roughly at 93.048 million miles away from the sun. I hope this helps you out! :)

You might be interested in

Velocity and Time

Murrr4er [49]

Answer:

h=112.35

Explanation:

becuase it had to

3 0

3 years ago

A quarter is flipped from a height of 1.45 m above the ground. How much time will it take to reach the ground if the person flip

Artemon [7]

It will take the quarter 0.151 seconds to reach the ground.

<u>Given the following data:</u>

Height = 1.45 meters

Initial velocity = 10.32 m/s

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find how much time it will take the quarter to reach the ground, we would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

$S = ut + \frac{1}{2} at^2$

Where:

S is the height or distance covered.

u is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the values into the formula, we have;

$1.45 = 10.32(t) + \frac{1}{2} (9.8)t^2\\\\1.45 = 10.32t + 4.9t^2\\\\4.9t^2 + 10.32t - 1.45 = 0$

The standard form of a quadratic equation is:

$ax^2 + bx + c = 0$

a = 4.9, b = 10.32 and c = 1.45

We would solve the above quadratic equation by using the quadratic equation formula;

$x = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}$

Substituting the values, we have;

$t = \frac{-10.32\; \pm \;\sqrt{10.32^2\; - \;4(4.9)(1.45)}}{2(4.9)}\\\\t = \frac{-10.32\; \pm \;\sqrt{106.5024\; - \;28.42}}{9.8}\\\\t = \frac{-10.32\; \pm \;\sqrt{78.0824}}{9.8}\\\\t = \frac{-10.32\; \pm \;8.84}{9.8}\\\\t = \frac{-10.32\; + \;8.84}{9.8}\\\\t = \frac{1.48}{9.8}$

<em>Time, t = 0.151 seconds.</em>

Therefore, it will take the quarter 0.151 seconds to reach the ground.

Read more: brainly.com/question/8898885

3 0

2 years ago

The force P is applied to the 45-kg block when it is at rest. Determine the magnitude and direction of the friction force exerte

cluponka [151]

Answer:

Check attachment for free body diagram of the question.

I used the free body diagram and the angles given in the diagram missing in the question above, but I used the data given in the above question.

Explanation:

Let frictional force be Fr acting down the plane

Let analyze the structure before inserting values

Using Newton's second law along the y-axis

ΣFy = Fnet = m•ay

Since the body is not moving in the y-direction, then ay = 0

N+PSinβ — WCosθ = 0

N+PSin20—441.45Cos15 = 0

N+PSin20—426.41 = 0

N = 426.41 — PSin20 , equation 1

The maximum Frictional force to be overcome is given as

Fr(max) = μsN

Fr(max) = 0.25(426.41 — PSin20)

Fr(max)= 106.6 —0.25•PSin20

Fr(max) = 106.6 — 0.08551P, equation 2

This is the maximum force that must be overcome before the body starts to move

Using Newton's law of motion in the x direction

Note, we took the upward direction up the plane as the direction of motion since the force want to move the block upward

Fnetx = ΣFx

Fnetx = P•Cosβ —W•Sinθ — Fr

Fnetx = P•Cos20—441.45•Sin15—Fr

Fnetx = 0.9397P — 114.256 — Fr

Equation 3

When Fnetx is positive, then, the body is moving up the plane, if Fnetx is negative, then, the maximum frictional force has not yet being overcome and the object is still i.e. not moving

a. When P = 0

From equation 2

Fr(max) = 106.6 — 0.08551P

Fr(max) = 106.6 — 0.08551(0)

Fr(max)= 106.6 N

So, 106.6N is the maximum force to be overcome

So, here the only force acting on the body is the weight and it acting down the plane, trying to move the body downward.

Wx = WSinθ

Wx = 441.45× Sin15

Wx = 114.256 N.

Since the force trying to move the body downward is greater than the maximum static frictional force, then the body is not in equilibrium, it is moving downward.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — PSin20 , equation 1

N = 426.41 N, since P=0

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 426.41

Fr = 93.81 N.

b. Now, when P = 190N

From equation 2

Fr(max) = 106.6 — 0.08551(190)

Fr(max) = 106.6 —16.2469

Fr(max)= 90.353 N

So, 90.353 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 190Cos20 — 441.45Sin15

Fnetx = 64.29N

So the force moving the body up the incline plane is 64.29N

Fnetx < Fr(max)

Then, the frictional force has not being overcome yet.

Then, the body is in equilibrium.

Then, applying equation 3.

Fnetx = 0.9397P — 114.256 — Fr

Fnetx = 0, since the body is not moving

0 = 0.9397(190) —114.246 — Fr

Fr = 64.297 N

Fr ≈ 64.3N

c. When, P = 268N

From equation 2

Fr(max) = 106.6 — 0.08551(268)

Fr(max) = 106.6 —16.2469

Fr(max)= 83.68 N

So, 83.68 N is the maximum force to be overcome

Now the force acting on the x axis is the horizontal component of P and the horizontal component of the weight

Fnetx = P•Cosβ —W•Sinθ

Fnetx = 268Cos20 — 441.45Sin15

Fnetx = 137.58 N

So the force moving the body up the incline plane is 137.58 N

Fnetx > Fr(max)

Then, the frictional force has being overcome.

Then, the body is not equilibrium.

So, finding the magnitude of frictional force

From equation 1

N = 426.41 — 268Sin20 , equation 1

N = 334.75 N, since P=268N

Then, using law of kinetic friction

Fr = μk • N

Fr = 0.22 × 334.75

Fr = 73.64 N

d. The required force to initiate motion is the force when the block want to overcome maximum frictional force.

So, Fnetx = Fr(max)

Px — Wx = Fr(max)

From equation 1

Fr(max) = 106.6 — 0.08551P,

P•Cosβ-W•Sinθ = 106.6 — 0.08551P

P•Cos20 — 441.45•Sin15 = 106.6 — 0.08551P

P•Cos20—114.256=106.6 - 0.08551P

PCos20+0.08551P =106.6 + 114.256

1.025P=220.856

P = 220.856/1.025

P = 215.43 N

3 0

3 years ago

An electron with velocity v = 1.0 ´ 106 m/s is sent between the plates of a capacitor where the electric field is E = 500 V/m. I

oksano4ka [1.4K]

Answer:

The deviation in path is $4.39 \times 10^{-3}$

Explanation:

Given:

Velocity $v = 1 \times 10^{6}$ $\frac{m}{s}$

Electric field $E = 500$ $\frac{V}{m}$

Distance $x = 1 \times 10^{-2}$ m

Mass of electron $m = 9.1 \times 10^{-31}$ kg

Charge of electron $q = 1.6 \times 10^{-19}$ C

Time taken to travel distance,

$t = \frac{x}{v}$

$t = \frac{1 \times 10^{-2} }{1 \times 10^{6} }$

$t = 10^{-8}$ sec

Acceleration is given by,

$F = qE$

$ma = qE$

$a = \frac{qE}{m}$

$a = \frac{1.6 \times 10^{-19} \times 500}{9.1 \times 10^{-31} }$

$a = 8.77 \times 10^{13}$ $\frac{m}{s^{2} }$

For finding the distance, we use kinematics equations.

$y = vt + \frac{1}{2} at^{2}$

Where $v = 0$ because here initial velocity zero

$y = \frac{1}{2} at^{2}$

$y = \frac{1}{2} \times 8.77 \times 10^{13 } \times (10^{-2} )^{2}$

$y = 4.39 \times 10^{-3}$ m

Therefore, the deviation in path is $4.39 \times 10^{-3}$

6 0

2 years ago

un columpio de balancin tiene una barra de 6m de longitud y en ella se sientan 2 personas,una de 60kg y otra de 40kg, calcular e

Kitty [74]

Answer:

<em>El punto de apoyo debe colocarse a 3.6 metros de la persona de 40 Kg</em>

<em>La ventaja mecánica es 1.5</em>

Explanation:

<u>Máquinas Simples</u>

Un balancín es un ejemplo de máquina simple, donde se aplica una fuerza física y ésta puede amplificarse o reducirse a voluntad cambiando la configuración física de la máquina.

En nuestro caso, el punto de apoyo o fulcro se coloca entre las dos fuerzas constituyendo una máquina de primer grado.

La situación planteada se muesta en la figura anexa. Debemos averiguar el valor de x para que las dos personas sentadas en el balancín puedan estar en equilibrio.

Para determinar el valor de x, se establece la condición de equilibrio de torques mecánicos. Ya que el balancín se asume en reposo, los torques aplicados de cada lado del mismo deben ser iguales, haciendo que el torque neto sea cero.

El torque es el producto de la fuerza por la distancia:

T = F.d

De cada extremo del balancín, se aplica una fuerza igual al peso de cada persona, es decir, llamando F1 al peso de la persona de 40 Kg y F2 al peso de la persona de 60 Kg:

$F_1 = 40 kg * 9.8 m/s^2=392N\\\\F_2 = 60 kg * 9.8 m/s^2=588 N$

El torque neto del balancín debe ser cero, es decir (refiérase a la figura):

$F_1*x=F_2*(6-x)$

Reemplazando los valores obtenidos:

$392*x=588*(6-x)$

Operando:

$392*x+588*x=588*6$

Simplificando

$980*x=3528$

Resolviendo

$\displaystyle x=\frac{3528}{980}=3.6\ m$

El punto de apoyo debe colocarse a 3.6 metros de la persona de 40 Kg, es decir, a (6 - 3.6) = 2.4 metros de la persona de 60 Kg

La ventaja mecánica se calcula como el cociente de ambas distancias

$\displaystyle VM=\frac{3.6}{2.4}=1.5$

La ventaja mecánica es 1.5, es decir, se amplifica la fuerza vez y media

3 0

3 years ago