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3 years ago

What are the main subatomic particles that make up the atom?

Physics

1 answer:

Sergeeva-Olga [200]3 years ago

3 0

Answer:

Protons, Neutron and Electrons.

Explanation:

An atom comprises of three main subatomic particles, the protons (define the type of element), the neutrons and the electrons.

Protons and neutrons are clustered together in the centre with the electrons orbiting around the cluster.

Protons have a positive charge, Neutrons have no charge and Electrons have a negative charge.

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Ben walks 500 meters from his house to the corner store. He then walks back toward his house but stops to talk to a neighbor whe

spin [16.1K]

Average velocity =

(displacement) / (time for the displacement)
and
(direction of the displacement) .

Displacement =

(distance from the start-point to the end-point)
and
    (direction from the start-point to the end-point) .

When Ben is 200 meters from the corner store,
he is (500 - 200) = 300 meters from his house.

His displacement is

   300 meters in the direction
   from his house to the neighbor .

His average velocity is

   (300/910) = 0.33 meters per second, in the
   direction from his house to the neighbor .

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3 years ago

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Identify one organism in the chart that is a decomposer

FromTheMoon [43]

Answer:

fungi

Explanation:

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3 years ago

What are important things about mental health

Paul [167]

Answer:

Mental health includes our emotional, psychological, and social well-being. It affects how we think, feel, and act. It also helps determine how we handle stress, relate to others, and make healthy choices. Mental health is important at every stage of life, from childhood and adolescence through adulthood

Explanation:

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3 years ago

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Three charges are located at a different position in a plane: q1= 10μC at →r1=(5,6)cm q2=−27μC at →r2=(−6,10)cm and q3=−12μC at

sasho [114]

Answer:

E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

E_{total} = 2,467 10⁶ N / A θ = -21.8

Explanation:

For this exercise we will use that the electric field is a vector quantity, so the total field is

E_total = E₁₃ + E₂₃

bold font vectors . We can work with the components of the electric field in each axis

X- axis

E_ total x = E₁₃ₓ + E_{23x}

y-axis

E_{total y} = E_{13y} + E_{23y}

the expression for the electric field is

E = k q / r²

where r is the distance between the charge and the positive test charge

in this exercise

Let's find the field created by charge 1

q₁ = 10 μC = 10 10⁻⁶ C

x₁ = 5 cm = 0.05 m

x₃ = 21 cm = 0.21 m

E_{13x} = 9 10⁹ 10 10⁻⁶ / (0.21 -0.05)²

E_{13x} = 3.516 10⁶ N / C

y₁ = 6 cm = 0.06 cm

y₃ = -12 cm = -0.12 m

E_{13y} = 9 10⁹ 10 10⁻⁶ / (-0.12 - 0.06)²

E_{13y} = 2,777 10⁶ N / C

let's find the field produced by charge 2

q₂ = -27 μC = - 27 10⁻⁶ C

x₂ = -6 cm = -0.06 m

x₃ = 0.21 m

E_{23x} = 9 10⁹ 27 10⁻⁶ / (0.21 + 0.06)²

E_{23x} = 1.23 10⁶ N / A

y₂ = 10 cm = 0.10 m

y₃ = -0.12 m

E_{23y} = 9 10⁹ 27 10⁻⁶ / (-0.12 - 0.10)²

E_{23y} = 1.86 10⁶ N / C

Taking the components we can calculate the total electric field, we must use that charge of the same sign repel and attract the opposite sign, remember that the test charge is always considered positive.

E_{total x} = E_{13x} - E_{23x}

E_{total x} = (3.516 - 1.23) 10⁶

E_{total x} = 2.29 10⁶ N / A

E_{total y} = -E_{13y} + E_{23y}

E_{total y} = (-2.777 +1.86) 10⁶ N / A

E_{total y} = -0.917 10⁶ N / A

we can give the result in two ways

E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

or in the form of modulus and angle, let's use the Pythagorean theorem to find the modulus

E_{total} = √ (E_{total x}^2 + E_{total y}^ 2)

E_{total} = √ (2.29² + 0.917²) 10⁶

E_{total} = 2,467 10⁶ N / A

let's use trigonometry for the angle

tan θ = E_total and / E_totalx

θ = tan⁻¹ E_{total y} / E_{total x}

θ = tan⁻¹ (-0.917 / 2.29)

θ = -21.8

The negative sign indicates that the angle is measured with respect to the x-axis in a clockwise direction.

7 0

3 years ago

The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T

zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

f) The two trains avoid collision if the acceleration of the freight train is at least $0.35 m/s^2$

Explanation:

We can describe the position of the passenger train at time t with the equation

$x_p(t)=u_p t + \frac{1}{2}at^2$

where

$u_p = 25.0 m/s$ is the initial velocity of the passenger train

$a=-0.100 m/s^2$ is the deceleration of the train

On the other hand, the position of the freight train is given by

$x_f(t)=x_0 + v_f t$

where

$x_0=200 m$ is the initial position of the freight train

$v_f = 15.0 m/s$ is the constant velocity of the train

The collision occurs if the two trains meet, so

$x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0$

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

$x_f(t)=x_0 +v_ft$

And substituting t = 22.5 s, we find:

$x_f(22.5)=200+(15)(22.5)=537.5 m$

We can verify that the passenger train is at the same position at the time of the collision:

$x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m$

So, the two trains collide at x = 537.5 m.

In the graph in attachment, the position-time graph of each train is represented. We have:

The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

In order to avoid the collision, the freight train must have a initial position of

$x_0'$

such that the two trains never meet.

We said that the two trains meet if:

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t$

Re-arranging,

$\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0$

Substituting the values for the acceleration and the velocity,

$0.05t^2-10t+x_0'=0$

The solution of this equation is given by the formula

$t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}$

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

$10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m$

Therefore, the freight train must have a head start of 500 m.

In this case, we want to find the acceleration $a'$ of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t$

Re-arranging

$\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0$

Substituting,

$-0.5at^2-10t+200=0$

The solution to this equation is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}$

Again, the two trains never meet if the discriminant is negative, so

$10^2-4\cdot (-0.5a') \cdot (200)$

So, the deceleration must be smaller (towards negative value) than $-0.25 m/s^2$

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

$x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2$

where $a_f$ is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

$x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2$

Re-arranging,

$\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0$

And the solution is

$t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}$

Again, the two trains avoid collision if the discriminant is negative, so

$10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2$

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

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3 years ago