What is the lithosphere?
A. the outer layer of the Earth's crust
B. the inner core
C. the middle portion of the mantle
D. the outer core

Answer:

A. The sum of all the forces acting on an object.

Answer:

The maximum elongation of the spring is 0.5 meters.

Explanation:

The statement is incorrect. The correct form is:

A 2-kg block is attached to a horizontal ideal spring with a spring constant of 200 newton per meter. When the spring has its equillibrium length, the block has a speed of 5 meters per second. What is the maximum elongation of the spring?

The block experiments a simple harmonic motion, where there are no non-conservative forces and the total energy is the sum of translational kinetic energy of the mass and the elastic potential energy of the spring. The maximum elongation of the spring is done when elastic potential energy reach its maximum. By the Principle of Energy Conservation, the maximum elastic potential energy is equal to the maximum translational kinetic energy, which corresponds to the instant when the mass reaches the equilibrium position. Then, the equation modelling the system is:

[tex]U_{max} = K_{max}[/tex] (1)

Where:

[tex]U_{max}[/tex] - Maximum elastic potential energy of the spring, measured in joules.

[tex]K_{max}[/tex] - Maximum translational kinetic energy of the mass, measured in joules.

By definitions of the maximum elastic potential energy of the spring and the maximum translational kinetic energy of the mass, the expression above is expanded and simplified:

[tex]\frac{1}{2}\cdot k\cdot x_{max}^{2} = \frac{1}{2}\cdot m \cdot v_{max}^{2}[/tex]

[tex]x_{max} = \sqrt{\frac{m}{k} }\cdot v_{max}[/tex] (2)

Where:

[tex]x_{max}[/tex] - Maximum elongation of the spring, measured in meters.

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

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

[tex]v_{max}[/tex] - Maximum speed of the mass, measured in meters per second.

If we know that [tex]m = 2\,kg[/tex], [tex]k = 200\,\frac{N}{m}[/tex] and [tex]v_{max} = 5\,\frac{m}{s}[/tex], then the maximum elongation of the spring is:

[tex]x_{max} = \sqrt{\frac{2\,kg}{200\,\frac{N}{m} } }\cdot \left(5\,\frac{m}{s} \right)[/tex]

[tex]x_{max} = 0.5\,m[/tex]

The maximum elongation of the spring is 0.5 meters.

Answer:

[tex]\lambda =533.6 nm [/tex]

Explanation:

the slits spacing, d = 0.21 mm

distance of screen, D = 61 cm

The condition for minima is given as

[tex]dsin(\theta) = \left ( n+\frac{1}{2} \right )\lambda[/tex]

So, first minima, n = 0

[tex]dsin(\theta_1) = \frac{1}{2}\lambda[/tex]

fifth minima, n = 4

[tex]dsin(\theta_5) = \frac{9}{2}\lambda[/tex]

[tex]d(sin(\theta_5) -sin(\theta_1))= 4\lambda[/tex]

For small angle

[tex]d(tan(\theta_5) -tan(\theta_1))= 4\lambda[/tex]

From the figure:

[tex]d(\frac{y_5}{D}-\frac{y_1}{D})= 4\lambda[/tex]

[tex]\frac{d}{D} (y_5-y_1) = 4\lambda[/tex]

[tex]\lambda = \frac{d}{4D} (y_5-y_1)[/tex]

[tex]\lambda = \frac{0.021}{4(61)} (6.2 \times 10^{-3})[/tex]

[tex]\lambda =533.6 nm[/tex]

"533.5 nm" would be the wavelength of the light used in this experiment.

According to the question,

Slits spacing,

Distance of screen,

The condition for minima is given as:

→ [tex]d (sin \Theta) = (n+\frac{1}{2} ) \lambda[/tex]

So,

1st minima, n = 0

→ [tex]d sin(\Theta_1)= \frac{1}{2} \lambda[/tex]

5th minima, n = 4

→ [tex]d (sin (\Theta_5) - sin (\Theta_1) = 4 \lambda[/tex]

For small angle,

→ [tex]d (tan (\Theta_5) - tan (\Theta_1) = 4 \lambda[/tex]

From the figure, we get

→ [tex]d(\frac{y_5}{D} - \frac{y_1}{D} )= 4 \lambda[/tex]

  [tex]\frac{d}{D} (y_5 -y_1) = 4 \lambda[/tex]

                 [tex]\lambda = \frac{d}{4D}(y_5-y_1)[/tex]

                    [tex]= \frac{0.021}{4\times 61} (6.2\times 10^{-3})[/tex]

                    [tex]= 533.6 \ nm[/tex]

Thus the above answer is correct.  

Learn more:

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Answer:

The value is [tex]u_x = 597 \ m/s[/tex]

Explanation:

From the question we are told that

    The distance of the target from the riffle is  [tex]d = 40 \ m[/tex]

    The height at which the bullet hit the target is  [tex]y = 2.2 \ cm = 0.022 \ m[/tex]

Considering the vertical motion  

Generally from kinematic equations we have that

        [tex]y = u_y t + \frac{1}{2} * gt^2[/tex]

At the initial stage the velocity is  zero  i.e [tex]u_y = 0 \ m/s[/tex]

=>     [tex]0.022 = 0 * t + \frac{1}{2} * 9.8 t^2[/tex]

=>    [tex]t = 0.067 \ s[/tex]

Generally the velocity of the bullet when it leaves the riffle is mathematically represented as

         [tex]u_x = \frac{ d}{t}[/tex]

=>     [tex]u_x = \frac{40 }{ 0.067 }[/tex]

=>     [tex]u_x = 597 \ m/s[/tex]

Answer:

Power_input = 85.71 [W]

Explanation:

To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.

 [tex]W =F*d[/tex]

where:

W = work [J] (units of Joules)

F = force [N] (units of Newton)

d = distance [m]

We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.

Now replacing:

 [tex]W=m*g*d[/tex]

[tex]W=80*10*3\\W=2400[J][/tex]

Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.

 [tex]P=W/t[/tex]

where:

P = power [W] (units of watts)

W = work [J]

t = time = 40 [s]

[tex]P=W/t\\P=2400/40\\P=60 [W][/tex]

The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.

 [tex]Effic=0.7\\Effic=P_{required}/P_{introduced}\\P_{introduced}= 60/0.7\\P_{introduced}=85.71[W][/tex]

Answer:

4500W

Explanation:

[tex]P = {I}^{2} R[/tex]

where P = Power consumed , I = Current & R = Resistance.

In the question it's given that

I = 15A ; R = 20Ω

So ,

[tex]P = {15}^{2}\times 20 = 225 \times 20 = 4500W[/tex]

Answer:

S = 2461.53 [kJ]

Explanation:

The change in entropy in a process such as melting can be calculated by means of the following expression.

[tex]S=\frac{H*m}{T}[/tex]

where:

S = entropy [kJ]

H = fusion heat = 3.36*10¹ [J/kg]

m = mass = 0.02 [kg]

T = temperature in kelvin = 273 [K]

[tex]S = \frac{0.02*3.36*10^{1} }{(273+0)}\\S = 2461.53 [kJ][/tex]

Answer:

Magnesium is the correct answer

Explanation:

It is magnesium because as we move down the periodic table group, atomic sizes increase which is as a result of lesser forces of attraction between nucleus and valence of an atom which makes the atom to be prone to leave as electrons so easily, elements become more reactive.

But as we move along the period , atomic sizes decrease because there is a more force of attraction between nucleus of an atom and it valence electrons making it less reactive just as the case in magnesium. Other elements are more reactive than it across the group.

Answer:

Magnesium

Explanation: Correct on edge!

Hope this Helps! :)

Answer:

Electrical to Thermal

Answer:

Resistance = 2.2 Ohms.

Explanation:

Given the following data;

Resistivity, P = 0.25

Length, L = 11.2

Area, A = 1.26

To find the resistance.

Resistance is given by the formula below;

Resistance = PL/A

Substituting into the equation, we have;

Resistance = (0.25*11.2)1.26

Resistance = 2.8/1.26

Resistance = 2.2 Ohms.

Therefore, the resistance of the wire is 2.2 Ohms.

Answer:

In a step-up transformer, the secondary coil is made of same wire as the primary coil but it is rolled with the presence of the insulated which never conduct the opposite current to the coil.

Hence it function property without meeting any fire and other problem.

Even if you want to get clear ideas, then you have to go with official website.

Answer:

0.0158m/s

Explanation:

Using the law of conservation of energy which states that the sum of momentum before collision is equal to the sum after collision. It is expressed mathematically as;

m1u1 + m2u2 =  m1v1 + m2v1

m1 and m2 are the masses of the object

u1 and u2 are the initial velocities

v1 and v2 are the final velocities

Given

m1 = 300g  = 0.3kg

u1 = 1.10m/s

m2 = 4.00kg

u2 = 0m/s (at rest)

v1 = 0.890

v2 = ?

Substitute the given values into the formula;

0.3(1.10) + 0 = 0.3(0.89) + 4v2

0.33 = 0.267 + 4v2

0.33-0.267 = 4v2

0.063 = 4v2

v2 = 0.063/4

v2 = 0.0158m/s

Hence the speed of the large cart after the collision is 0.0158m/s

According to the question:

Mass,

Final velocity,

Initial velocity,

By using the law of conservation, we get

→ [tex]m_1 u_1 +m_2 u_2 =m_1 v_1 +m_2 v_1[/tex]

By substituting the values, we get

→ [tex]0.3(1.10)+0 = 0.3(0.89)+4v_2[/tex]

→              [tex]0.33=0.267+4 v_2[/tex]

   [tex]0.33-0.267 = 4 v_2[/tex]

              [tex]0.063 =4v_2[/tex]

                   [tex]v_2 = \frac{0.063}{4}[/tex]

                       [tex]= 0.0158 \ m/s[/tex]

Thus the response above is appropriate.

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Answer:

F = 404.6 N

Explanation:

Mechanical Force

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

[tex]v_f=v_o+a.t[/tex]

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

Another useful equation for speeds and distance X is:

[tex]V_f^2=V_o^2+2aX[/tex]

To calculate the acceleration, we solve for a:

[tex]\displaystyle a=\frac{V_f^2-V_o^2}{2X}[/tex]

The SUV starts from rest (vo=0) to vf=10.4 m/s traveling a distance of X=211 m, thus the acceleration is:

[tex]\displaystyle a=\frac{10.4^2-0^2}{2(212)}[/tex]

[tex]\displaystyle a=\frac{108.16}{424}[/tex]

[tex]a=0.26 \ m/s^2[/tex]

The force needed to accelerate the SUV of mass m=1,586 Kg is:

F = 1,586*0.26

F = 404.6 N

force between the two charges (+q1 and +q2),if they are at a distance 'a' is

F1=1/4pieEo q1q2/d^2................ (1)

when the metal spheres are in contact the charge flow from one sphere to another till both the sides acquires the same charge. here q1 and q2 are of same sign,hence after contact each sphere will have a charge

[tex] \binom{q1 + q2 }{2} [/tex]

now,the force between them,

f2=1/4pieEo

(q1+q2/2)^2/d^2

from eq (1)and eq (2)

f2=f1 (q1+q2)^2/4q1q2

Answer:

A lands first becuase its heavier

Answer:

x = A cos (w \sqrt{2y_{o}/g})

a) maximun  Ф= \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) minimun     Ф = [tex]\frac{\pi }{2}[/tex] - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

Explanation:

For this exercise let's use kinematics to find the time it takes for the mass to reach the floor

         y = y₀ + v₀ t - ½ g t²

as the mass is released from rest, its initial velocity is zero (vo = 0) and its height upon reaching the ground is zero (y = 0)

      0 = y₀ - ½ g t²

      t = [tex]\sqrt{2y_{o}/g}[/tex]

The bucket-spring system has a simple harmonic motion, which is described by

     x = A cos wt

in this expression we assumed that the phase constant (Ф) is zero

let's replace the time

     x = A cos (w \sqrt{2y_{o}/g})

this is the distance where the system must be for the mass to fall into it.

a) The new system has a total mass of m ’= 3.0 kg, so its angular velocity changes

          w = [tex]\sqrt{k/m}[/tex]

In the initial state

         w = \sqrt{k/2}

When the mass changes

         w ’= \sqrt{k/3}

the displacement in each case is

         x = A cos (wt)

for the new case

        x ’= A cos (w’t + Ф)

the phase constant is included to take into account possible changes due to the collision of the mass.

we see that this maximum expressions when the cosine is maximum

        cos (w´t + Ф) = 1

         w’t + Ф = 0

        Ф = -w ’t

        Ф = - [tex]\sqrt{k/3}[/tex] [tex]\sqrt{2y_{o}/g}[/tex]

       \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

b) the function is minimun if

        cos (w’t + fi) = 0

        w’t + Ф = π / 2

        Ф = π / 2 - w ’t

        Ф = [tex]\frac{\pi }{2}[/tex] - \sqrt{\frac{2}{3}  \frac{2 y_{o} }{g}  }

Answer:

[tex]l_o=550.055\ cm[/tex]

Explanation:

Given that,

Length of a street bridge, l = 5.5 m

The coefficient of bridge, [tex]\alpha =0.00001 ^0 C[/tex]

We need to find how much will it expand when temperatures is by 10°C.

The change in length per unit original length is given by :

[tex]\dfrac{\Delta l}{l}=\alpha \Delta T\\\\\Delta l = l\alpha \Delta T\\\\=5.5\times 0.00001 \times 10\\\\\Delta l=0.00055\\\\(l_o-l)=0.00055\\\\l_o=0.00055+5.5\\\\=5.50055\ m\\\\l_o=550.055\ cm[/tex]

Hence, the length will expanded 550.055 cm.

Answer:

1. it explains what's happening im the universe now

Answer:

distance ; time

Explanation:

wavelength is in metres [m]

period is in seconds [s]

Answer:

Elastic potential energy is stored when materials stretch or compress. Examples of it include springs, rubber bands, and slingshots.

Explanation:

Answer:

Time

An explanation of basic pond measurements that are vital to basic pond management practices.

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Pond Measurements: Area, Volume and Residence Time - Articles UPDATED: OCTOBER 29, 2015

Pond Measurements: Area, Volume and Residence Time

The importance of getting an accurate estimation of your pond surface area cannot be overestimated. The majority of pond owners visually estimate their pond area, which usually results in an overestimate of the true pond surface area. Pond area and water volume should be calculated based on some simple measurements. The effort necessary to estimate pond surface area is directly related to your pond's shape and uniformity. The simplest method--using basic equations for common shapes--can be applied if your pond closely resembles a circle, square, rectangle, or trapezoid in shape.

Pond Shapes

Circular

pond shape can be estimated by measuring the distance around the pond shoreline in feet. Square the shoreline distance and divide by 547,390 to get the pond area in acres. For example, a pond that is 450 feet around the shoreline would have an area = (450 feet)2 / 547,390 or 0.37 acres.

Explanation:

i looked this up

ANSWER: The first thing to learn is how to convert numbers back and forth between scientific notation and ordinary decimal notation. The expression "10n", where n is a whole number, simply means "10 raised to the nth power," or in other words, a number gotten by using 10 as a factor n times:

105 = 10 x 10 x 10 x 10 x 10 = 100,000 (5 zeros)

108 = 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 100,000,000 (8 zeros)

Notice that the number of zeros in the ordinary decimal expression is exactly equal to the power to which 10 is raised.

If the number is expressed in words, first write it down as an ordinary decimal number and then convert. Thus, "ten million" becomes 10,000,000. There are seven zeros, so in powers of ten notation ten million is written 107.

A number which is some power of 1/10 can also be expressed easily in scientific notation. By definition,

1/10 = 10-1     ("ten to the minus one power")

More generally, the expression "10-n" (where n is a whole number) means ( 1/10 )n. Thus

10-3 = ( 1 / 10 )3 = 1 / ( 10 x 10 x 10) = 1/1000

10-8 = ( 1 / 10 )8 = 1/100,000,000

Scientific notation was invented to help scientists (and science students!)deal with very large and very small numbers, without getting lost in all the zeros. Now answer the following on a separate sheet of paper and check your answers by clicking on "Answers":

Explanation:

Answer:

the cooler the wire, the less the resistance

Answer: ur mom

Explanation:

Answer:

1.3

Explanation:

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