Please help !!!!!!!!!!!!

Answer: a= 2 m/s²

Explanation: acceleration = change of speed/ time = 20 m/s / 10 s

Answer:

his distance is too small (r = 0.01 mm), therefore the cut can be at any distance

Explanation:

For this exercise let's use the ampere law.

Let's use a cylinder as the circulating surface

          ∫ B. ds = μ₀ I

in this case the field is circular and ds is circular therefore the angle between them is zero and cos 0 = 1

          B 2π r =  μ₀ I

          r =  [tex]\frac{\mu_o I}{2\pi B}[/tex]

The field needed to demagnetize the card is B = 1000 gauss = 0.1 T

          r = [tex]\frac{4\pi 10^{-7} 5.0 }{2\pi \ 0.1}[/tex]

           r = 2 10⁻⁷  5.0/0.1

          r = 1  10⁻⁵ m

this distance is too small (r = 0.01 mm), therefore the cut can be at any distance

Answer:

A Data is transmitted in the form of electromagnetic radiation.

B Signals travel faster through space than through fiber-optic cables.

C No ground equipment is needed to access the internet.

D The network is made of many satellites organized in a grid pattern.

Explanation:

THose are the options

Answer:

B

Explanation:

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{x_2^2}[/tex]

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = [tex]k \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}[/tex]

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = [tex]k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )[/tex]

            E_ {total} = [tex]k \frac{Q}{x_1^2}[/tex]   [tex]( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )[/tex]

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = [tex]-k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}[/tex]

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = [tex]k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}[/tex]

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = [tex]+ k \frac{Q}{(d-x_1)^2}[/tex]+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

5) Right side point

            E_ {total} = [tex]k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}[/tex]

             E_ {total} = [tex]- k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )[/tex]- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

Answer: A tsunami is a very long-wavelength wave of water that is generated by sudden displacement of the seafloor or disruption of any body of standing water.  Tsunami are sometimes called "seismic sea waves", although they can be generated by mechanisms other than earthquakes.  Tsunami have also been called "tidal waves", but this term should not be used because they are not in any way related to the tides of the Earth.  Because tsunami occur suddenly, often without warning, they are extremely dangerous to coastal communities. Ocean trenches are steep depressions in the deepest parts of the ocean [where old ocean crust from one tectonic plate is pushed beneath another plate, raising mountains, causing earthquakes, and forming volcanoes on the seafloor and on land.  

Explanation:

Answer:Air bags can leave you in even more injury, From the impact they give

You could end up with a broken nose,arm

concussion

Answer:

a) 6636 km

b) 0.0154

Explanation:

The height above the earth at its furthest point is 368 km

The height above the earth at its closest point is 164 km

Radius of the Earth is 6370 km

The distance of the satellite from the center of the earth to the furthest point is 6370 + 368 km = 6738 km

The distance of the satellite from the center of the earth to the closest point is 6370 + 164 = 6534 km

If we add together the sum of the distance of the satellite from the furthest and its closest distance, it is equal to the 2 major semi axis.

Basically,

2a = R + r

a = (R + r) / 2

a = (6738 + 6534) / 2

a = 13272 / 2

a = 6636 km

Eccentricity, e = (a - r) / a

Eccentricity, e = (6636 - 6534) / 6636

Eccentricity, e = 102 / 6636

Eccentricity, e = 0.0154

Answer:

72 J

Explanation:

Use the Work formula

W= F x d

Given:

F - 48 N

d - 1.5 m

Solution:

W= F x d

W=  48 N x 1.5 m

W= 72 J

Answer:

A

Explanation:

The angular momentum of a system of particles around a point in a fixed inertial reference frame is conserved if there is no net external torque around that point:

d

→

L

d

t

=

0

or

→

L

=

→

l

1

+

→

l

2

+

⋯

+

→

l

N

=

constant

.

Note that the total angular momentum

→

L

is conserved. Any of the individual angular momenta can change as long as their sum remains constant. This law is analogous to linear momentum being conserved when the external force on a system is zero.

Answer:

They're going to come home as soon as the movie is over.

Answer:

B: They're going to come home as soon as the movie is over.

Explanation:

Answer:

speed =distance /time

speed =4.82/0.00360

speed =1338.8m/s

Answer:

comparing the beam waist for both lasers ( ratio of the beam waists )

4.536 μm / 2.117 μm = 2.14

Explanation:

Nd-YAG laser system : emits at 532 nm , beam diameter = 8 mm

Ti-sapphire laser system : emits at 855 nm , Beam diameter = 6mm

Comparing the beam waist for both lase systems using a focusing lens

Focal length = 10 mm

assumption : light fills lenses in each laser system

Beam waist radius ( W ) =  [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex]

β = wavelength , D = diameter illuminated  , F = focal length

For

Nd-YAG laser system

β = 532 mm , D = 8 mm

hence ( Wn ) =   [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex] =  ( 2*532 / π ) ( 10 / 8 )   = 2.117 μm

For

Ti-sapphire laser

β = 855 nm ,  D = 6 mm

hence ( Wt )  [tex](\frac{2\beta }{\pi } )(\frac{F}{D} )[/tex] =  ( 2* 855 ) / π ) ( 50 / 6 ) = 4.536 μm

comparing the beam waist for both lasers ( ratio of the beam waists )

4.536 μm / 2.117 μm = 2.14

La erosión costera es la pérdida o desplazamiento de tierra, o la remoción a largo plazo de sedimentos y rocas a lo largo de la costa debido a la acción de olas, corrientes, mareas, agua impulsada por el viento, hielo transportado por el agua u otros impactos de marejadas ciclónicas.

This problem involved half projectile.

initial velocity, vo = 20 m/s

time of flight, t = 7 s

(a) Simply use the formula to get the height, h:

h = vo*t - (1/2)gt^2

(b) To get the final vertical velocity or terminal velocity (vf), use the formula:

(vf)^2 - (vo)^2 = 2gh

(c) Use the formula find the horizontal distance traveled, R:

R = vo * cos(θ) * t

But since the angle involved with respect to horizontal is zero, and cos(0) = 1, we have

R = vo * t

Hope this helps~ `u`

Jai

Answer: kooi kooi

how r u and thanks for the free points :)

Answer:2.23 m/s

Explanation:

Given

Mass of person and vehicle is [tex]m=1600\ kg[/tex]

Total work done is [tex]W_t=6000\ J[/tex]

Friction consumes one-third of the energy

The remaining two-third is consumed to increase the kinetic energy

[tex]\Rightarrow \dfrac{2}{3}\times 6000=\dfrac{1}{2}\times 1600\times v^2\\\\\Rightarrow 4000\times 2=1600\times v^2\\\\\Rightarrow v=\sqrt{5}\ \approx 2.23\ m/s[/tex]

Answer:

false im just trying to get it if you'd like to give it to me

The answer is false :)

Answer:

OK we appreciate your concern.

Explanation:

force=35

distance=18

time=30

power=f×d/t

p=35×18/30

p=21

Answer: Molecules of H202, H20 and 02 are still forming. ( A P E X )

Explanation: I know this is late but for anyone looking at this later

The above question is incomplete, I think the original  question is

Belinda observes that the reaction 2H20+02 2H2O2 appears to have stopped because the concentrations of reactants and products are not changing. Which statement is true at this point in the reaction?

A. Molecules of H₂O₂ are formed only when the reaction vessel is disturbed.

B. Molecules of H2O2 have stopped forming.

C. Molecules of H20 and O₂ have stopped forming.

D. Molecules of H2O2, H2O, and O2 are still forming.

At this point in the reaction, the system has reached a state of equilibrium i.e there is no net change in the concentrations of reactants and products. So the Molecules of H2O2 have stopped forming. The correct option is B.

In chemistry, a state of equilibrium refers to a situation where the concentrations of reactants and products in a chemical reaction are no longer changing over time.

This occurs when the rates of the forward and reverse reactions are equal, and there is no net change in the concentrations of reactants and products.

At equilibrium, the system is in a dynamic state, where the forward and reverse reactions continue to occur, but at the same rate, resulting in no overall change in the concentrations of the species involved in the reaction. The equilibrium state is typically characterized by the equilibrium constant, which is a measure of the relative concentrations of reactants and products at equilibrium for a given reaction under specific conditions of temperature, pressure, and concentration.

Here in this question,

The molecules of H2O2 have stopped forming. This means that the forward reaction (2H2O + O2 → 2H2O2) and the reverse reaction (2H2O2 → 2H2O + O2) are occurring at the same rate, resulting in no net change in the concentration of H2O2. However, molecules of H2O and O2 are still being consumed and produced as they are involved in the reaction, but their concentrations are not changing because they are being replenished at the same rate that they are being consumed.

Therefore, the correct answer is B i.e Molecules of H2O2 have stopped forming.

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Answer:

correct ooooooooooooo

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

[tex]m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}[/tex] (1)

From the bottom to the top of the second hill

[tex]\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}[/tex] (2)

Where:

[tex]m[/tex] - Mass of the roller coaster car, in kilograms.

[tex]v_{1}[/tex] - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

[tex]h_{1}[/tex] - Height of the first top of the hill with respect to the bottom, in meters.

[tex]W_{1, loss}[/tex] - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

[tex]v_{2}[/tex] - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

[tex]h_{2}[/tex] - Height of the second top of the hill with respect to the bottom, in meters.

[tex]W_{2, loss}[/tex] - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

[tex]m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}[/tex] (3)

Where [tex]W_{loss}[/tex] is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that [tex]m = 175\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]h_{1} = 18\,m[/tex], [tex]h_{2} = 8\,m[/tex] and [tex]v_{2} = 11\,\frac{m}{s}[/tex], then the work done by non-conservative force is:

[tex]W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right][/tex]

[tex]W_{loss} = 6574.75\,J[/tex]

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Answer:

2) a_y= -g  3) vₓ=constant v_y = v_{oy} - g t, 4)  vₓ = v₀ₓ - ax t

5)  changes the horizontal speed, should change range

7) changes the vertical speed change the maximum height

Explanation:

1) After reading your long writing, we are going to solve the exercise, in the attachment you can see the different vectors.

2) The acceleration vectors are vertical and directed downwards due to the attraction of the Earth (gravity force) this force is constant, on the x axis there is no acceleration

3) the velocity vectors on the x-axis are constant because there are no relationships and the y-axis changes value according to the expression

           v_y = v_{oy} - gt

at the point of maximum height, vy = 0 is equal to the maximum height

4) For someone to change the horizontal acceleration we must assume a friction with the air, in this case they relate it would be in the opposite direction to the horizontal speed

In the graph it would be directed to the left, therefore the velocity would be

           vₓ = v₀ₓ - ax t

5 and 6) If someone changes the horizontal speed, they should change the range of the shot for greater horizontal speed, the rock goes further.

the equations of motion are

           x = v₀ₓ t

           y = v_{oy} t - ½ g t²

7) If someone changes the vertical speed change the maximum height, but not the scope of the shot, for higher speed higher maximum height,

the equations of motion are the same.

Imagine you have three flashlights. One has a blue light, one has a green light and one is red. If you were to shine each of those lights in the same spot, I think i would see white color.

At one end of the electromagnetic wave line, which has the group of the visible spectrum. This has been referred to as the visible light of the spectrum. The visible light with the shortest of the wavelength has the blue light and the one with the longest has the red light.

The primary colors of the light has been the spectrum are red, blue, and green. The combination of these colors will form other as well as the colors which has been referred to as the secondary colors. The combination of the beam of the red light and the beam of green light will form yellow color.

Therefore, Imagine you have three flashlights. One has a blue light, one has a green light and one is red. If you were to shine each of those lights in the same spot, I think i would see white color.

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Answer:

.lyrjhg.

Explanation:

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Answer: I think the answer is b

Explanation: