Name the part of the human eye which acts as a protective layer for the eye​

Answer:

2.096m/s

Explanation:

The speed of this soccer ball can be calculated using the formula;

Speed = distance/time

According to this question, the distance of the ball before it hits the tree is 6.5m, the time it takes is 3.1s, hence;

Speed = 6.5/3.1

Speed of the ball = 2.096m/s

Therefore, the speed of the ball before hitting the tree is 2.096m/s

Answer:

Thermal Energy

Explanation:

Answer:

d = 150 km

Explanation:

Speed of an airplane is 600 km/h

We need to find how far it travel in 1/4 hour.

We know that the speed of an object is given by distance travelled divided by the time taken. Let d is the distance covered in 1/4 hour. So,

[tex]v=\dfrac{d}{t}\\\\d=v\times t\\\\d=600\ km/h \times \dfrac{1}{4}\ h\\\\d=150\ km[/tex]

So, it will cover 150 km.

Answer:

D

Explanation:

Answer:

[tex]A. \: A \: series \: circuit[/tex]

Explanation:

♨Rage♨

Answer:

  C.  A parallel circuit

Explanation:

Adding a bulb in parallel with the existing bulb will apply the same voltage to both bulbs. They will light equally bright.

You would design a parallel circuit.

_____

In a series circuit the same current would flow in both bulbs, but that current would be at half the original current level. Both bulbs would be dimmer than the first bulb was.

It is difficult to create a "complex" circuit with only two components. An "incomplete" circuit would result in no light at all.

Answer:

According to the University of Iowa, the average length of a stride is 5ft.

Now, the total distance of the marathon is:

26mi and 385yd.

Let's transform that distance into ft.

1mi = 5280ft

Then:

26mi = 26*5280ft = 137,280ft

1yd = 3ft

then:

385yd = 385*3ft = 1,155ft.

Then the total distance of the marathon, in ft, is:

D = 137,280ft + 1,155ft = 138,435 ft.

Now the average number of strides needed will be equal to the quotient between the total distance of the marathon and the distance traveled in each stride.

N = 138,435ft/5ft = 27,687.

Answer:

Option C.

Explanation:

To know which velocity-time graph best describes the motion of the ball, let us calculate the velocity of the ball and the time taken for the ball to get the ground. This can be obtained as follow:

1. Determination of the velocity.

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Final velocity (v) =.?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 100)

v² = 0 + 1960

v² = 1960

Take the square root of both side.

v = √(1960)

v = 44.27 m/s

2. Determination of the time taken.

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 100 m

Time (t) =.?

h = ½gt²

100 = ½ × 9.8 × t²

100 = 4.9 × t²

Divide both side by 4.9

t² = 100 / 4.9

Take the square root of both side

t = √(100 / 4.9)

t = 4.52 s

From the above illustration,

Initial time (t1) = 0 s

Final time (t2) = 4.52 s

Initial velocity (u) = 0 m/s

Final velocity (v) = 44.27 m/s

Thus, we can see that as the time increase, the velocity also increase. Therefore, option C gives the correct answer to the question.

Density ρ is mass m per unit volume v, or

ρ = m / v

Solving for v gives

v = m / ρ

So the given object has a volume of

v = (130 g) / (65 g/cm³) = 2 cm³

Answer:

Explanation:

Given parameters:

Initial velocity = 0

Acceleration = 11m/s²

Time  = 6s

Unknown:

Final velocity  = ?

Solution:

 From the given parameters, we use one of the appropriate equations of motion to solve this problem.

     V = U + at

V is the final velocity

U is the initial velocity

a is the acceleration due to gravity

t is the time taken

Input the parameters and solve;

     V  = 0 + 11 x6

     V  = 66m/s

The final velocity is 66m/s

Answer:

A. 8.64 secs.

B. 14.58 secs.

C. 26.002 secs.

D. 33.46secs.

Explanation:

A. P wave would travel 7000km

p-wave travels on a speed of 13.5km/s

= 7000km/13.5km/s

= 8.64 secs.

B. S-wave time to travel 7000km

s-wave travels on a speed of 8km/s

= 7000km/8km/s

= 14.58 secs.

C Love wave travels at a speed of 10,000m/s ( 2.7778 m/s ).

= 7000km to miles

= 4349.598m/2.788m/s

= 26.002 secs.

D. Rayleigh wave to travel 7,000 km

10,000m/s ( 2.1667 m/s ).

= 7000km to miles

= 4349.598m/2.1667m/s

= 33.46secs.

Answer:

ns for high-speed rail in the United States date back to the High Speed Ground Transportation Act of 1965. Various state and federal proposals have followed. Despite being one of the world's first countries to get high-speed trains (the Metroliner service in 1969), it failed to spread. Definitions of what constitutes high-speed rail vary, including a range of speeds over 110 mph (180 km/h) and dedicated rail lines. Inter-city railwith top speeds between 90 and 125 mph (140 and 200 km/h) is sometimes referred to in the United States as higher-speed rail.[1]

Amtrak's Acela Express (reaching 150 mph, 240 km/h), Silver Star, Northeast Regional, Keystone Service, Vermonter and certain MARC Penn Line express trains (all five reaching 125 mph, 201 km/h) are the only high-speed services in the country.

As of 2020, the California High-Speed Rail Authority is working on the California High-Speed Rail project and construction is under way on sections traversing the Central Valley. The Central Valley section is planned to open in 2029 and Phase I is planned for completion in 2031.[2]

Contents

1 Definitions in American context

2 History

2.1 Faster inter-city trains: 1920–1941

2.2 Post-war period: 1945–1960

2.3 First attempts: 1960–1992

2.4 Renewed interest: 1993–2008

2.5 Plans for 2008–2013

3 Current state and regional efforts

3.1 The Northeast

3.1.1 Northeast Corridor: Next Generation High-Speed Rail

3.1.1.1 Proposed routes

3.1.2 Northeast Maglev proposal

3.1.3 New Jersey–New York City upgrades

3.1.4 New York

3.1.5 Pennsylvania

3.2 Western States

3.2.1 California

3.2.2 Pacific Northwest

3.2.3 Arizona

3.3 Mid-Atlantic and the South

3.3.1 Florida

3.3.2 Southeast

3.3.3 Texas

3.4 Midwest

3.4.1 Illinois and the Midwest

3.5 The Southwest

4 Federal high-speed rail initiatives

4.1 American Recovery and Reinvestment Act of 2009

4.1.1 Strategic plan

4.2 2009 federal grant funding

4.3 2010 allocation

4.3.1 Cancellation of funds for Wisconsin, Ohio, and Florida

4.4 2011 and 2012 proposals and rejections of funding

5 See also

6 Notes

7 Further reading

8 External links

Explanation:

Answer: It will be about 44.1m

Explanation:

Answer:

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

[tex]volume = \frac{mass}{density} \\[/tex]

From the question

mass = 10 g

density = 87 g/ml

We have

[tex]volume = \frac{10}{87} \\ = 0.114942528...[/tex]

We have the final answer as

Hope this helps you

Answer:

I chose c because it is the greater slope at point c

Answer:

The value is    [tex]t = 56.68 \  s  [/tex]

Explanation:

From the question we are told that

   The rating of the hoist motor is  [tex]k  =  159hp = 159 *746 =118614 \ W[/tex]

    The  percentage of it power used is  [tex]z = 0.72 * 118614=85402.08 \ W[/tex]

      The  mass of the load is m  = 5550 kg

      The distance is  h = 89.0 m

The potential  energy required to lift the load through that distance is

     [tex]E =  m *  g * h[/tex]

=>    [tex]E =  5550 *  9.8 *  89.0[/tex]

=>   [tex]E =  4840710 \ J[/tex]

Generally the time taken is mathematically represented as

       [tex]t = \frac{E}{ z}[/tex]

=>    [tex]t = \frac{4840710}{ 85402.08}[/tex]

=>    [tex]t = 56.68 \  s  [/tex]

Answer:

* when L → H    chord too long

in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross

* when L → 0 very short string

         the speed of the platform is very small, so we do not have the minimum required value

        vox = √ (g / (2 (H)) D

Explanation:

For this exercise we are going to solve it using conservation of energy to find the velocity of the body and the launch of projectiles to find the velocity to cross the well.

Let's start with the projectile launch

as the body leaves the vertical its velocity must be horizontal

         x = v₀ₓ t

         y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²

when reaching the ground its height of zero (y = 0) and the initial vertical velocity is zero

         t = √ 2 y₀ / g

we substitute

        x = vox √2y₀ / g

        v₀ₓ = √(g / 2y₀)     x

In the exercise, it tells us that the width of the well is D (x = D) and the initial height is the height of the platform minus the length of the rope (I = H - L)

       v₀ₓ = √(g /(2 (H -L))    D

this is the minimum speed to cross the well.

Now let's use conservation of energy

starting point. On the platform

      [tex]Em_{o}[/tex] = U = m g H

final point. At the bottom of the swing

      Em_{f} = K + U = 1 / 2m v² + m g (H -L)

as there is no friction the mechanical energy is conserved

        Em_{o} = Em_{f}

       m g H = 1 / 2m v² + m g (H -L)

        v = √ (2gL)

let's write our two equations

the minimum speed to cross the well

       v₀ₓ = √ (g /(2 (H -L))    D

the speed at the bottom of the oscillatory motion

       v = √ (2g L)

we analyze the extreme cases

* when L → H    chord too long

in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross

* when L → 0 very short string

         the speed of the platform is very small, so we do not have the minimum required value

        vox = √ (g / (2 (H)) D

From this analysis we see that there is a range of lengths that allows us to have the necessary speeds to cross the well

      V₀ₓ = v

      g / (2 (H -L) D² = 2g L

       4 L (H- L) = D²

        4 H L - 4 L2 - D² = 0

        L² - H L - D² / 4 = 0

let's solve the quadratic equation

      L = [H ± √ (H2-D2)] / 2

we assume that H> D

       L = ½ H [1 + - RA (1 - (D / H) 2)]

The two values ​​of La give the range of values ​​for which the two speeds are equal

A) The person lands in the moat if the rope's length is very short because :

B) The person lands in the moat if the rope length is similar to the height of the platform because :

Following the assumptions;

size of the person is much smaller than L and H

D = horizontal distance

Hence we can conclude that The person lands in the moat if the rope's length is very short because The speed of the platform is less than the required minimum speed  and  The person lands in the moat if the rope length is similar to the height of the platform because,the speed required to cross the moat approaches infinity.

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Answer:

The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]

Explanation:

From Rotation Physics we know that linear velocity of a point moving with uniform circular motion is:

[tex]v = r\cdot \omega[/tex] (Eq. 1)

Where:

[tex]r[/tex] - Radius of rotation of the particle, measured in meters.

[tex]\omega[/tex] - Angular velocity, measured in radians per second.

[tex]v[/tex] - Linear velocity of the point, measured in meters per second.

But we know that angular velocity is also equal to:

[tex]\omega = \frac{\theta}{t}[/tex] (Eq. 2)

Where:

[tex]\theta[/tex] - Angular displacement, measured in radians.

[tex]t[/tex] - Time, measured in seconds.

By applying (Eq. 2) in (Eq. 1) we get that:

[tex]v = \frac{r\cdot \theta}{t}[/tex] (Eq. 3)

From Geometry we must remember that circular arc ([tex]s[/tex]), measured in meters, is represented by:

[tex]s = r\cdot \theta[/tex]

[tex]v = \frac{s}{t}[/tex]

The linear velocity is represented by the following expression: [tex]v = \frac{s}{t}[/tex]

Answer:

The value  is  [tex]  \frac{dR}{dt} =  -286.2 \  km/hr [/tex]

Explanation:

From the question we are told that  

   The speed of the airplane from the north is [tex]\frac{dN}{dt}  =  -181 \  km /hr[/tex]

The negative sign is because the direction is towards the south

  The speed of the airplane from the east is  [tex]\frac{dE}{dt}  =  -278 \  km/hr[/tex]

The negative sign is because the direction is towards the west

   The distance of the southbound plane from the airport is  [tex]N  =  30 \  km[/tex]

   The distance of the westbound plane is  [tex]E =  15 \  km[/tex]

Generally the distance between the plane is mathematically represented using Pythagoras theorem  as

    [tex]R^2  = N^2 + E^2[/tex]

Next differentiate implicitly this equation to obtain the rate at which the distance between the planes changes

So

     [tex]2R\frac{dR}{dt} =  2N \frac{dN}{dt} +   2E\frac{dE}{dt}[/tex]

Here

     [tex]R = \sqrt{N^2 + E^2}[/tex]

=>    [tex]R = \sqrt{30^2 + 15^2}[/tex]

=>    [tex]R = \sqrt{30^2 + 15^2}[/tex]

=>    [tex]R =33.54 \ m [/tex]

    [tex]2(33.54) * \frac{dR}{dt} =  2( 30)*(-181)  +   2*15*(-278)[/tex]

=>   [tex] 67.08 * \frac{dR}{dt} =  -19200[/tex]

=>   [tex]  \frac{dR}{dt} =  -286.2 \  km/hr [/tex]

The rate of change of the distance between the planes is 286.23 km/hr.

The given parameters;

The distance between the two planes is calculated by applying Pythagoras theorem as shown below;

[tex]d^2 = n^2 + e^2\\\\d = \sqrt{n^2 + e^2} \\\\d = \sqrt{30^2 + 15^2} \\\\d = 33.54 \ km[/tex]

The rate of change of the distance between the planes is calculated as follows;

[tex]d^2 = e^2 + n^2\\\\2\frac{dd}{dt} = 2e\frac{de}{dt} + 2n\frac{dn}{dt} \\\\d\frac{dd}{dt} = e\frac{de}{dt} + n\frac{dn}{dt}\\\\(33.54) \frac{dd}{dt} = (15)(278) \ + (30)(181)\\\\(33.54) \frac{dd}{dt} = 9600\\\\\frac{dd}{dt} = \frac{9600}{33.54} \\\\\frac{dd}{dt} = 286.23 \ km/hr[/tex]

Thus, the rate of change of the distance between the planes is 286.23 km/hr.

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Answer:

they don't strech so they tear a muscle when they perform

Explanation:

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas  = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 240 K

= initial volume of gas  =  

= final volume of gas  =  

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

Answer:

the scientific study of natural forces such as light, sound, heat, electricity, pressure, etc.

Explanation:

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Answer:

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

Explanation:

The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:

[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]

[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)

Where:

[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.

[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.

And we expand the equation above by definitions of elastic potential energy and kinetic energy:

[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]

[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)

Where:

[tex]m[/tex] - Mass of the block, measured in kilograms.

[tex]k[/tex] - Spring constant, measured in newtons per meter.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.

[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.

If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:

[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]

[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]

The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.

Answer:

kenietic and potential i guess

Explanation:

Answer:

The maximum torque on the loop is 395.80 N.m.

Explanation:

Given;

number of turns of the wire, N = 150 turns

length of the square loop, L = 18.0 cm = 0.18 m

current in the wire, I = 50.9 A

Magnetic field, B = 1.6 T

Maximum torque on the loop is given by;

τ = NIAB

τ = (150)(50.9)(0.18²)(1.6)

τ = 395.80 N.m

Therefore, the maximum torque on the loop is 395.80 N.m.

Answer:

The speed as it reaches the bottom is  24m/s

Explanation:

Given parameters:

   Height of hill  = (3 x 9.8)m = 29.4m

Unknown:

Final speed as it reaches the bottom = ?

Solution:

 To solve this problem, we apply;

    v² = u² + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity = 9.8m/s²

h is the height

note, the initial speed of the body is 0;

Input the parameters and solve;

    v² = 0² + 2 x 9.8 x 29.4

   v² = 576.24

   v = √576.24  = 24m/s

The speed as it reaches the bottom is  24m/s

Answer:

2 and 8

Explanation:

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