Use the drop-down menus to complete the sentences.
A concave lens is
in the middle than at the edges, and causes light rays to
A convex lens is
in the middle than at the edges, and causes light rays to

Answer:

I think it's because the light waves travel faster than the sound waves.

The speed of light is far greater than the speed of sound hence, sound waves always reach the observer after the light waves.

Light occurs in the electromagnetic spectrum. Recall that light can be transmitted through vaccuum unlike sound.

The speed of light is far greater than the speed of sound hence, sound waves always reach the observer after the light waves.

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Answer: 5 N

Explanation:

Given

Mass of box [tex]m=4\ kg[/tex]

acceleration of box [tex]a=2\ m/s^2[/tex]

Force applied by Manual [tex]F_m=8\ N[/tex]

Friction force [tex]f=5\ N[/tex]

The net force on the block is [tex]F_{net}=m\times a=4\times 2=8 \N[/tex]

Suppose [tex]F_c[/tex] is the force applied by Carlos

[tex]\Rightarrow F_m+F_c-f=F_{net}\\\Rightarrow F_c=F_{net}-F_m+f\\\Rightarrow F_c=8-8+5=5\ N[/tex]

The strength of the force that Carlos pushes on the box will be [tex]F_c=5\ N[/tex]

The force is defined as the external effort applied on any object to move it or to restrict it.

Here following information is given in the question:

Mass of the box m=4 kg

The acceleration of the box [tex]a=2\ \frac{m}{s^2}[/tex]

The force applied by the Manual [tex]F_m=8\ N[/tex]

The frictional force is [tex]F_f=5\ N[/tex]

The net force on the body will be calculated by

[tex]F_{net}=m\times a=4\times 2=8\ N[/tex]

The force that Carlos pushes on the box will be calculated by the equilibrium of the forces:

[tex]F_m+F_c-F_f=F_{net}[/tex]

[tex]8+F_c-5=8[/tex]

[tex]F_c=5\ N[/tex]

Thus the strength of the force that Carlos pushes on the box will be [tex]F_c=5\ N[/tex]

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Answer:

hi..bro

make pinky sis answer as brinlist ok

have a good day

Answer:

23 asden

Explanation:

first clothes and shoes

Answer:

Explanation:

In an elliptical orbit , the angular momentum of the planet remains constant because velocity of planet makes a right angle to the direction to the star .

mvr = constant .

mv₁r₁ = mv₂r₂

v₁r₁ = v₂r₂

v₁ x 21 = 38 x 75

v₁ = 135.71 km/s .

Answer:

= 0.517 m

Explanation:

This is a resonance exercise where the ends of the string are fixed, therefore it has a node of them, the fundamental (longest) wavelength created has the form

                λ = 2L

wave speed is related to wavelength and frequency

               v = λ f

               v = 2L f

let's calculate

                v = 2 0.50 440

                v = 440 m / s

since they indicate that the tension of the string does not change and the linear density of the string is constant, the speed of the wave also remains constant

               f =[tex]\frac{v}{2L}[/tex]

let's find the length for the new resonance frequency

               L = [tex]\ \frac{v}{2f}[/tex]

let's calculate

               L = [tex]\frac{440}{2 \ 528}[/tex]

               L = 0.5166 m

Answer:

the speed of the ball just after the collision is 1.5 m/s.

Explanation:

Given;

mass of the ball, m₁ = 1.6 kg

initial velocity of the ball, u₁ = 0

mass of the block, m₂ = 0.8 kg

initial velocity of the block, u₂ = 0

final velocity of the block, v₂ = 3 m/s

let the final velocity of the ball after collision = v₁

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁ + m₂u₂ = m₁v₁  +  m₂v₂

1.6 x 0   +    0.8 x 0       =   1.6 x v₁     +  0.8 x 3

0 = 1.6v₁  + 2.4

-1.6v₁ = 2.4

v₁  = -2.4 / 1.6

v₁ = - 1.5 m/s

v₁ = 1.5 m/s (in opposite direction of the block)

Therefore, the speed of the ball just after the collision is 1.5 m/s.

Answer:

b) N = 560 N, c)  fr = 138.56 N, d)  μ = 0.247

Explanation:

a) In the attachment we can see the free body diagram of the system

b) Let's write Newton's second law on the y-axis

              N + T_y -W = 0

              N = W -T_y

let's use trigonometry for tension

             sin θ = T_y / T

             cos θ = Tₓ / T

             T_y = T sin θ

             Tₓ = T cos θ

we substitute

              N = W - T sin 30

we calculate

              N = 640 - 160 sin 30

              N = 560 N

c) as the system goes at constant speed the acceleration is zero

X axis

              Tₓ - fr = 0

               Tₓ = fr

we substitute and calculate

              fr = 160 cos 30

              fr = 138.56 N

d) the friction force has the formula

             fr = μ N

             μ = fr / N

we calculate

             μ = 138.56 / 560

             μ = 0.247

Uhm.. based on what? Can you please specify/add an image or some more text and elaborate?

Answer:

a) [tex] E_{p} = 0 [/tex]

[tex] E_{k} = 168.7 J [/tex]

[tex] E_{m} = 168.7 J [/tex]

b) [tex] E_{p} = 73.6 J [/tex]

[tex] E_{k} = 95.8 J [/tex]

[tex] E_{m} = 169.4 J [/tex]

c) [tex] E_{p} = 169.2 J [/tex]

[tex] E_{k} = 0 [/tex]

[tex] E_{m} = 169.2 J [/tex]

Explanation:

We have:

m: is the ball's mass = 1.5 kg

v₀: is the initial speed = 15 m/s

g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

[tex] E_{p} = mgh = 0 [/tex]

The kinetic energy is:

[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J [/tex]

And the mechanical energies:

[tex] E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J [/tex]

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

[tex] E_{p} = mgh = 1.5*9.81*5 = 73.6 J [/tex]

Now, to find the kinetic energy we need to calculate the speed at 5 m:

[tex] v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9 [/tex]

[tex] v_{f} = \sqrt{126.9} = 11.3 m/s [/tex]

[tex] E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(11.3)^{2} = 95.8 J [/tex]

And the mechanical energies:

[tex] E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J [/tex]

c) At its maximum height:

[tex] v_{f}[/tex]: is the final speed = 0

[tex] h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m [/tex]

Now, the potential, kinetic and mechanical energies are:

[tex] E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J [/tex]

[tex] E_{k} = \frac{1}{2}mv^{2} = 0 [/tex]

[tex] E_{m} = 169.2 J + 0 = 169.2 J [/tex]

I hope it helps you!    

a metal spike in an airport runway

A metal spike in an airport runway  is NOT a way to stay safe from static electricity. Hence option C is correct.

Electricity is a collection of physical phenomena related with the presence and motion of matter having an electric charge. Both electricity and magnetism are connected to the phenomena of electromagnetic, as defined by Maxwell's equations. Lightning, static electricity, electric heating, electric discharges, and other frequent occurrences are all connected to electricity.

An electric field is created when either a positive or negative electric charge is present. The movement of electric charges results in an electric current and a magnetic field. In most applications, a force of magnitude determined by Coulomb's law operates on a charge. Volts are commonly used to measure electric potential.

Hence option C is correct.

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Answer:

3. they can travel through solids

4. they move rock at right angles to the direction of wave travel

Explanation:

Answer:

1, 3, 4

Explanation:

S waves are transverse waves, so they move rock at right angles to the direction of wave travel. Also, S waves can only travel through solids. S waves and P waves both begin at an earthquake's focus.

Answer:

the net energy Gained per hour equals 30Kcal/h

Answer:

Q = 282,000 J

Explanation:

Given that,

The mass of liquid water, m = 125 g

Temperature, T = 100°C

The latent heat of vaporization, Hv = 2258 J/g.

We need to find the amount of heat needed to vaporize 125 g of liquid water. We can find it as follows :

[tex]Q=mH_v\\\\Q=125\ g\times 2285\ J/g\\\\Q=282250\ J[/tex]

or

Q = 282,000 J

So, the required heat is 282,000 J .

Answer:

 y = 4.666 10⁻² m

Explanation:

The constructive interference experiment for the double slit

         d sin sin θ = m λ

Let's use trigonometry to find a sine relationship.

         Tan θ = y / L

          tan θ = sin θ/ cos θ

in these experiments the angles are very small

           tan θ = sin θ

           sin θ = y / L

            [tex]d \frac{y}{L}[/tex] = m λ

              y = [tex]\frac{ m \lambda \ L}{d}[/tex]

we replace the values

              y = 3  560 10⁻⁹ 3.0 / 0.108 10⁻³

              y = 4.666 10⁻² m

Answer:

[tex]69.06\ \text{N/m}[/tex]

Explanation:

m = Mass of bungee jumper = 55 kg

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

h = Height at the bottom = 40 m

k = Spring constant

x = Displacement of bungee cord = 25 m

Applying the principle of conservation of energy to the system we have

[tex]mgh=\dfrac{1}{2}kx^2\\\Rightarrow k=\dfrac{2mgh}{x^2}\\\Rightarrow k=\dfrac{2\times 55\times 9.81\times 40}{25^2}\\\Rightarrow k=69.06\ \text{N/m}[/tex]

The spring constant of the bungee cord is [tex]69.06\ \text{N/m}[/tex].

Answer:

--------------------------------

Explanation:

Answer:

9.6

Hope this helps.

Answer:

Ur answer is 12 sir!

Explanation:

HAve a good day sir

Explanation:

The lower the level of resistivity the more electrical conductivity a metal has. Copper has low resistivity, and therefore is an excellent conductor. Copper is also less oxidative than other metals

Answer:

Explanation:

For equilibrium ,

weight of mass 470 gram = restoring force in spring

If x be extension in spring and k be spring constant

restoring force = kx

weight of mass = kx

.47 x 9.8 = k x .36

k = 12.8 N /m.

Answer:

It is A. Light blue

_____________

Hope this helps!

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Answer:

These are the 4  correct answers that should be on your test.

A. A model can represent something very large like the solar system.

B. A model can represent something very small like a single cell.

C. A model can represent something very complex like the flow of energy in an ecosystem.

D. A model can represent something that cannot be seen or touched.

Explanation:

Answer:

d

Explanation:

rzp-yyib-oiv

plèase jóin ón góógle mèèt

Answer:

a. a problem

Explanation:

The engineering process begins by defining a problem and completing background research on the problem.

Answer: Increase the charge of one particle by a factor of 16

Explanation:Just took the quiz

Answer:

C. Increase the charge of one particle by a factor of 16.