fertilization that takes place when the union of the sex cells happens outside the body​

Answer:

the radius of curvature of the track for this instant is 266 m

Explanation:

Given that;

The Initial Velocity u = 100 km/h = 100 × [tex]\frac{5}{18}[/tex] = 27.77 m/s

velocity of the train at t=12 s is;

[tex]V_{t=12}[/tex] = 50 km/h = 50 × [tex]\frac{5}{18}[/tex] = 13.89 m/s

now, we calculate the deceleration of the train

[tex]V_{t=12}[/tex]  = u + at

13.89 = 27.77 + [tex]a_{t}[/tex]12

[tex]a_{t}[/tex] = (13.89 - 27.77) / 12

[tex]a_{t}[/tex] = -13.88 / 12

[tex]a_{t}[/tex] = - 1.1566 m/s²

Now, the velocity of the train at 6 seconds is;

[tex]V_{t=6}[/tex]  = u + at

[tex]V_{t=6}[/tex]  = 27.77 + ( - 1.1566 m/s²)6

[tex]V_{t=6}[/tex]  = 27.77 - 6.9396

[tex]V_{t=6}[/tex]  = 20.83 m/s

The acceleration at t=6 s is;

a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]

a = √[ ([tex]a_{t}[/tex] )² + ([tex]a_{n}[/tex])²]

we substitute

2m/s² = √[ (- 1.15 )² + ([tex]a_{n}[/tex])²]

4 = (- 1.1566 )² + ([tex]a_{n}[/tex])²

4 = 1.3377 +  ([tex]a_{n}[/tex])²

([tex]a_{n}[/tex])² = 4 - 1.3377

([tex]a_{n}[/tex])² = 2.6623

[tex]a_{n}[/tex] = √2.6623

[tex]a_{n}[/tex]  = 1.6316 m/s²

Now the radius of curve is;

a = V² / p

[tex]p_{t=6}[/tex] = ( [tex]V_{t=6}[/tex] )² /  [tex]a_{n}[/tex]

[tex]p_{t=6}[/tex] = ( 20.83 m/s )² /  1.6316 m/s²

[tex]p_{t=6}[/tex] = 433.8889 / 1.6316

[tex]p_{t=6}[/tex] = 265.9 m ≈ 266 m

Therefore;  the radius of curvature of the track for this instant is 266 m

Answer:

c

Explanation:

it is stored energy because it is built up in said object

Answer:

i think its ancestor

Explanation:

sry if im wrong

Answer:

scientists compare organisms DNA to support the theory that all species share a common Ancestor.

Answer:

Hello the diagram related to your question is attached below

answer: a) 851 m/s

             b)  8506.1 secs

Explanation:

calculate the periodic time of the satellite using the equation below

t = [tex]\frac{2\pi }{R} \sqrt{\frac{(R+h)^{3} }{g} }[/tex]  --  ( 1 )

where ; R = 6370 km

h = 500 km

g = 9.81 m/s^2

input given values into equation 1

t = 5670.75 secs

next calculate the periodic time taken by the space craft  

a) determine the increase in speed

V = v - [tex]\sqrt{\frac{gR^2}{R + h} }[/tex]  

where ; v = 8463 m/s , R = 6370 km, h = 500 km

V = 851 m/s

b) Determine the periodic time for the elliptic orbit

τ = [tex]\frac{3t}{2}[/tex]

 = [tex]\frac{3*5670.76}{2}[/tex]  =  8506.1 secs

attached below is the remaining part of the detailed solution

Answer:

1.40 m/s^2

Explanation:

Given data

Velocity= 17.4 m/s

time= 12.4 seconds

We want to find the acceleration of the rock

We know that

acceleration = velocity/time

Substitute

acceleration= 17.4/12.4

acceleration=1.40 m/s^2

Hence the acceleration is 1.40 m/s^2

Answer:

a)  [volts] = [N m / C],

b) The lines or surface that has the same potential are called equipotential

c) the equipotential lines must also be perpendicular to the electric field lines

Explanation:

a) find the units of the volt

the electric potential energy is

             V = k q / r

             V = [N m² / C²] C / m

              V = [N m / C]

The electric potential is defined as

             V = E .s

             V = [N / C] [m]

             V = [N m / C] = [volt]

we see that in the two expressions the same result is obtained therefore the volt is

            [volts] = [N m / C]

b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.

c) The electric potential is defined as the gradient of the electric field

             v =  [tex]- \frac{dE}{dx} i^[/tex]

therefore the equipotential lines must also be perpendicular to the electric field lines

(a) The unit of electric potential is volts.

(b) Equipotential lines are lines along which the electric potential is constant.

(c) Equipotential lines are obtained by drawing a line perpendicular to electric field lines.

Electric potential is the work done in moving a unit positive from infinite to a point in the electric filed.

Electric potential = EQ

Electric potential = (V/C) x (C)

Electric potential = Volts.

This is a line along which the electric potential is constant.

Equipotential lines are obtained by drawing a line perpendicular to electric field lines.

Learn more about equipotential lines here: https://brainly.com/question/12074544

Answer:

* displacement from equilibrium position

* The spring constant,

Explanation:

The work done by a spring is two times the expressions

         W = ∫ F. dx

in a spring the force is given by hooke's law

          F = - k Δx

we substitute

         W = - k ∫ x dx cos θ

In the case of the spring, the displacement is in the same direction of work, therefore the angle is zero and the cosine is equal to 1

we integrate

           W = ½ k Δx²

we can see that we need two parts to calculate the work

* displacement from equilibrium position

          Δx = 12 -8 = 4 cm

* The spring constant, this constant can be found from the displacement measurements as a function of the applied force.

Answer:

B

Explanation:

Friction is a force that opposes motion between any surfaces that are touching. Friction occurs because no surface is perfectly smooth Friction produces heat because it causes the molecules on rubbing surfaces to move faster and have more energy.

Answer:

t = 12.6 seconds

Explanation:

Given that,

Initial velocity, u = 0

Acceleration of an electric vehicle, a = 2.3 m/s²

Final velocity, v = 29 m/s

We need to find the time elapses from start to stop. The acceleration of an object is given by the relation as follows :

[tex]a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{29-0}{2.3}\\\\t=12.6\ s[/tex]

So, 12.6 seconds is elapsed from start to stop.

Answer:

Vf = 29.4 m/s

h = 44.1 m

Explanation:

Data:

==================================================================

Final Velocity

Use formula:

Replace:

Multiply:

==================================================================

Height

Use formula:

Replace:

Multiply time squared:

Simplify the s², and multiply in the numerator:

It divides:

What is the velocity when falling to the ground?

The final velocity is 29.4 meters per seconds.

How high is the building?

The height of the building is 44.1 meters.

Answer:

the answer is transmitted

Explanation:

Answer:

PRECAUTIONS

-The refracting faces of the glass prism should be smooth, transparent and without any air bubble or broken edge. ...

-Use a sharp pencil to draw boundary of the prism and rays of light.

-The alpins should have sharp tip and should be fixed exactly vertical to the plane of the paper.

Explanation:

Please give thanks to all my answers and please mark as brilliant and please follow me

net force = 30 N

mass = 8.16 kg

acceleration = 3.68 m/s²

Given

80 N force applied

mass of object = 10 kg

Friction force = 50 N

Required

Net force

mass

acceleration

Solution

Net force = force applied(to the right) - friction force(to the left)

Net force = 80 - 50 = 30 N

Gravitational force(downward) : F = mg

m = F : g

m = 80 : 9.8

m = 8.16 kg

a = F net / m

a = 30 / 8.16

a = 3.68 m/s²

Answer:

t = 5.88 10⁻² s

Explanation:

The speed of the sound wave after it is emitted by the bat is constant, so we can use the uniform motion relationships

          v = [tex]\frac{x}{t}[/tex]

          t = [tex]\frac{x}{v}[/tex]

in this case the distance is that of the sound in going from the bat to the insect and back

         x = 2d

         x = 2 10

         x = 20 m

let's calculate

        t = 20/340

        t = 5.88 10⁻² s

We can see that the time is very short, so the distance traveled by the two animals has little influence on the result.

Answer:

A) 1268 MeV

B)  299MeV/c

C) 1268 MeV

Explanation:

Given :

π meson rest energy = 139.6 MeV

Speed = 0.921c

proton at rest energy = 938.3 MeV

a) Find the total relativistic energy of resulting composite particle

E = E(meson) + E(proton)

  = [tex]\frac{(mc^2)_{meson} }{\sqrt{1-\frac{v^2}{c^2} } } + (mc^2)_{proton}[/tex]

  = [tex]\frac{139.6MeV}{\sqrt{1-\frac{(0.906c)^2}{c^2} } } + 938.3[/tex]

E = 1268 MeV

B) determine the total linear momentum of the composite particle

= 299MeV/c

attached below is the detailed solution

C) Determine the rest energy of the composite particle

E = 1268 MeV

Answer:

A

Explanation:

Answer:

it 1.5 meters

Explanation:

if u could put the option number it will be cool and hope it help and if it doesnt am really sorry ;)

Answer:

Well, you said the answer is A, so it’s A!

Answer:

4

2

1

3

Explanation:

Be safe, lovelies <3

Answer:

d = 29.89 m

Explanation:

To solve this, we need to separate this problem in two parts.

One part would be the the time taken by the rock to actually hit the bottom, and the other part would be the time taken by the sound to reach the inspector.

Joining these two times we have:

t = t₁ + t₂    (1)

This time is 2.56 s.

Now, as we are asked to determine the distance from the top floor to the bottom, and we have two times taken in different ways, one by sound and the other the actual, we can say the same thing on distance, we need a distance relationed to the time taken by rock to hit the bottom, and the other distance relationet to the time taken by sound to reach the inspector.

Doing this we have that the distance traveled by the rock is:

y₁ = gt²/2

y₁ = 9.8t²/2 = 4.9t₁²    (2)

Now, the distance traveled by sound would be:

y₂ = v * t₂ = 336t₂   (3)

Remember that the speed of the sound is 336 m/s

From this last expression (3), we can actually write t₂ in function of t₁, using (1):

2.56 = t₁ + t₂

t₂ = 2.56 - t₁    (4)

Replacing (4) in (3):

y₂ = 336(2.56 - t₁)    (5)

Now that we have y₁ and y₂, we can equal (2) and (5), both expressions to get the value of t₁, and then, calculate the distance:

4.9t₁² = 336(2.56 - t₁)

4.9t₁² = 860.16 - 336t₁

4.9t₁² + 336t₁ - 860.16 = 0

Using the quadractic formula, we can calculate t₁:

t₁ = -336 ±√(336)² + 4*4.9*860.16 / (2*4.9)

t₁ = -336 ±√129,7555.136 / 9.8

t₁ = -336 ± 360.21 / 9.8     Using only the positive value we have:

t₁ = 2.47 s

This means that the rock hits the bottom in 2.47 s, and the remaining 0.09 s belongs to the time taken by sound. (2.47 + 0.09 = 2.56 s)

With this, we can calculate the distance of the rock using expression (2):

y₁ = 4.9 * (2.47)²

Hope this helps

Answer:

What was it

Explanation:

It would increase; g is inversely proportional to the square of the radius of the Earth. The correct option is A.

A current idea in geology that describes how the earth's crust is made up of a few big, hard plates that move independently of one another, causing deformation, volcanism, and seismic activity along their boundaries.

Because it explains how mountain ranges, earthquakes, volcanoes, shorelines, and other features often emerge where the moving plates contact along their boundaries, plate tectonics provides "the overall picture" of geology.

The Earth has been shrinking as it gradually cools, according to some geological hypotheses from the nineteenth century that have now been completely debunked.

If that were the case, it would rise since g is inversely proportional to the square of the Earth's radius.

Thus, the correct option is A.

For more details regarding geological theory, visit:

https://brainly.com/question/10134253

#SPJ2

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

Answer:

convection currents

Explanation:

Answer:

(a) The negative charge on one of the charges is -8.79630245 × 10⁻⁷C

(b) The positive charge on one of the other charges is 8.79630245 × 10⁻⁷C

Explanation:

The given parameters are;

The force of attraction between the two spheres = 0.0988 N

The distance between their centers = 44.5 cm = 0.445 m

Therefore, we have;

[tex]F = \dfrac{k \cdot q_1 \cdot q_2}{d^2}[/tex]

Therefore, we have;

[tex]0.0988 \ N = -\dfrac{8.99 \times 10^9 N\cdot m^2 \cdot C^{-2}\cdot q_1 \cdot q_2}{(0.445 \ m)^2}[/tex]

Therefore, we have;

q₁·q₂ = -0.0988 N × (0.445 m)²/(8.99 × 10⁹ N·m²·C⁻²) = -2.17629255 × 10⁻¹² C²

q₁·q₂ = -2.17629255 × 10⁻¹² C²...(1)

When the two charges are connected, we get;

[tex]F_2 = \dfrac{k \cdot \left (\dfrac{q_1 + q_2}{2} \right) ^2}{d^2}[/tex]

Therefore, we have;

[tex]q_1 + q_2 = \sqrt{\dfrac{4 \cdot F_2 \cdot d^2}{k} }[/tex]

[tex]q_1 + q_2 = \sqrt{\dfrac{4 \times 0.0276 \ N \times(0.445 \ m)^2}{8.99 \times 10^9 N\cdot m^2 \cdot C^{-2}} } = 1.59446902743 \times 10^{-6} \ C[/tex]

q₁ + q₂ = 1.59446902743 × 10⁻⁶ C...(2)

From, equation (2), we have;

q₁ = 1.59446902743 × 10⁻⁶ C -  q₂

Plugging in the value of q₁ in equation (1) givens;

q₁·q₂ = -2.17629255 × 10⁻¹²

Therefore, we have;

(1.59446902743 × 10⁻⁶ -  q₂) × q₂ = -2.17629255 × 10⁻¹²

Which gives;

-q₂² + 1.59446902743 × 10⁻⁶·q₂+2.17629255 × 10⁻¹² = 0

Solving, with a graphing calculator, we get;

q₂ = 2.4741×10⁻⁶ C, or -8.79630245 × 10⁻⁷C

q₁ = 8.79630245 × 10⁻⁷C or -2.4741×10⁻⁶ C

Therefore, we have;

(a) The negative charge on one of the charges = -8.79630245 × 10⁻⁷C

(b) The positive charge on one of the other charges = 8.79630245 × 10⁻⁷C

Answer:

Acceleration

Explanation:

Answer: Acceleration

Explanation:

Answer:

the required mass flow rate is 49484.37 kg/s

Explanation:

Given the data in the question;

we first determine the relation for mass flow rate of water that passes through the turbine;

so the relation for net work on the turbine due to the change in potential energy considering 100% efficiency is;

[tex]W_{net}[/tex] = m ( Δ P.E )

so we substitute (gh) for ( Δ P.E );

[tex]W_{net}[/tex] = m (gh)

m = [tex]W_{net}[/tex] / gh

so we substitute our given values into the equation

m = 100 MW / ( 9.81 m/s²) × 206 m

m = ( 100 MW × 10⁶W/MW) / ( 9.81 m/s²) × 206 m

m = 10 × 10⁷ / 2020.86

m = 49484.37 kg/s

Therefore, the required mass flow rate is 49484.37 kg/s

Answer:lever

Explanation:

Answer:

The ribbon will move to the right.

Explanation:

To know the the correct answer to the question, we shall determine the net force and direction. This can be obtained as follow:

Force to the right (Fᵣ) = 120 N

Force to the left (Fₗ) = 80 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 120 – 80

Fₙ = 40 N to the right.

From the calculation made above, the net force is 40 N to the right. Thus, the ribbon will move to the right.

Answer:

     l = 0.548 m

Explanation:

For this exercise we compensate by finding the speed of the car

         p = m v

         v = p / m

         v = 0.58 / 0.2

         v = 2.9 m / s

this is how fast you get to the ramp, let's use conservation of energy

starting point. Lowest point

         Em₀ = K = ½ m v²

final point. Point where it stops on the ramp

         [tex]Em_{f}[/tex] = U = m g h

  mechanical energy is conserved

          Em₀ = Em_{f}

          ½ m v² = m g h

           h = [tex]\frac{m v^2}{2 g}[/tex]

let's calculate

          h = [tex]\frac{0.2 \ 2.9^2}{2 \ 9.8}[/tex]

          h = 0.0858 m

to find the distance that e travels on the ramp let's use trigonometry, we look for the angle

          tan θ = y / x

          tan θ = 12/75 = 0.16

          θ = tan⁻¹ 0.16

          θ = 9º

therefore

           sin 9 = h / l

           l = h / sin 9

           l = 0.0858 / sin 9

           l = 0.548 m