Pelican hunts fish in the water as th pelican the waters surface why must it aim for the fish in a slightly different place than where the fish appears to be located​

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

distance ; time

Explanation:

wavelength is in metres [m]

period is in seconds [s]

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:

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

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}  }

Explanation:

m means mass and v means velocity.

now, you can calculate

The change in potential energy in moving a 10kg box from the floor to a table 1 m high is 98 Joules.

To find the potential energy, the given values are,

Mass of the box = 10Kg

Height h = 1 m

The potential energy can be defined as the energy that is stored and that can be determined through various parts in a particular system.

Also, the energy will gets stored when the object is not moving.

For example: Spring. Even when the spring is not stretched or contracted it stores energy.

The formula for Potential energy PE= mgh joules.

m is the Mass,

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

h is the height.

Substituting all the given values,

Potential Energy= 10 (9.8)(1)

PE= 98 J

Thus the potential energy of the box= 98 J.

To learn more about the potential energy,

https://brainly.com/question/13548111

#SPJ2

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:

The Rutherford model was made by Ernest Rutherford, to describe a atom. That is a brief explanation

Explanation:

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Answer:

The very small particles that make up matter are I) Atoms

Matter - Anything that have mass and occupies space is called matter . it is made up of atoms and molecules

Atoms - The smallest part of matter is called atom.

Molecule - Group of atoms combine together to form a molecule.

Answer:

C. A. B.

Explanation:

Credit to Lainey in the comments for the answer!

Answer:

1] C

2] A

3] B

Explanation:

Answer:

The value is  [tex]a_t = 2.42 \ m/s^2[/tex]

Explanation:

From the question we are told that

  The radius of the tires is  [tex]r = 0.22 \ m[/tex]

   The  angular acceleration is  [tex]\alpha = 11.0 \ rad/s^2[/tex]

Generally the linear acceleration is mathematically represented as

     [tex]a_t = r * \alpha[/tex]

=>  [tex]a_t = 0.22 * 11[/tex]

=>  [tex]a_t = 2.42 \ m/s^2[/tex]

Answer:

The value is  [tex]m = 69.24 \ kg[/tex]

Explanation:

From the question we are told that

   The value of the external force is  [tex]F = 59.0 \ N[/tex]

    The magnitude of the astronaut's acceleration is  [tex]a = 0.852 \ m/s[/tex]

Generally Newton's Second Law of Motion from the mass of the astronauts is mathematically represented as

            [tex]m = \frac{F}{a}[/tex]

=>         [tex]m = \frac{59 }{0.852 }[/tex]

=>         [tex]m = 69.24 \ kg[/tex]  

Answer: ur mom

Explanation:

Answer:

i think the red runner travels greater distance ie.40 m

Explanation:

although ,the blue one travels 21 m but in the backward direction .so the correct ans is "the red runner travels 40 m"

Answer:

no .the blue runner began 16 m ahead of the red runner

Answer:

[tex]Relative\ Velocity = 105m/s[/tex]

Explanation:

Given

[tex]V_A = 45m/s[/tex]

[tex]V_B = 60m/s[/tex]

Required

Determine the speed of B w.r.t A

The question implies that, we determine the relative velocity of B w.r.t A

Because both trains are moving towards one another, the required velocity is a [tex]sum\ of\ velocities\ of[/tex] both trains:

This is shown below:

[tex]Relative\ Velocity = V_A + V_B[/tex]

[tex]Relative\ Velocity = 45m/s + 60m/s[/tex]

[tex]Relative\ Velocity = 105m/s[/tex]

Answer:

A: Perpendicular

Explanation:

The question is incomplete as it lacks the image of the rope wave motion.

However, as found on "estudyassistant", the options are;

A) Perpendicular

B) Circular

C) Longitudinal

D) Parallel

From all that, we can say that;

The rope's are moving simultaneously in the same pattern without touching each other.

This is therefore a mechanical wave being created with the motion having oscillations that are perpendicular to the direction of energy transfer of the ropes.

This is a definition of transverse waves because the rope particle motion is perpendicular to the wave motion.

Answer:

A: Perpendicular

Explanation:

Read above explanation.

Answer:

[tex]74.31\ \text{N}[/tex]

[tex]16.59^{\circ}[/tex]

Explanation:

P = 50 N

Q = 30 N

[tex]\theta[/tex] = Angle between the vectors = [tex]45^{\circ}[/tex]

Resultant is given by

[tex]R=\sqrt{P^2+Q^2+2PQ\cos\theta}\\\Rightarrow R=\sqrt{50^2+30^2+2\times 50\times 30\times \cos45^{\circ}}\\\Rightarrow R=74.31\ \text{N}[/tex]

Angle of resultant

[tex]\phi=\tan^{-1}\dfrac{Q\sin\theta}{P+Q\cos\theta}\\\Rightarrow \phi=\tan^{-1}\dfrac{30\times \sin45^{\circ}}{50+30\cos45^{\circ}}\\\Rightarrow \phi=16.59^{\circ}[/tex]

Magnitude of the resultant is [tex]74.31\ \text{N}[/tex]

Direction of the resultant is [tex]16.59^{\circ}[/tex]

Answer:

Electrical to Thermal

We are given:

Initial Velocity(u) = 15 m/s

Time interval(t) = 20s

Solving for the distance covered:

Since the girl keeps riding her bike at 15 m/s, her speed is constant and hence, acceleration of the bike is 0 m/s²

acceleration(a) = 0 m/s²

We know that:

s = ut + 1/2(at²)                [second equation of motion]

s = (15)(20) + 1/2(0)(400)  [plugging the values]

s = 150 + 0

s = 300 m

Hence, the girl covered 300 m in 20 seconds

Answer:

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

Complete Question

1 a A little girl pushes a 5.0 kg toy baby stroller at constant speed 7.0 m across the floor. She pushes on the handle with a force of 40 N at an angle of 30o with the horizontal.  How much work is done by the girl on the wagon?

1b  A farmhand pushes 20 ㎏ bale of hay 4m across the floor of the barn if she exerts a horizontal force of 60 N on the hay, how much work is done? (5 pts)

All parts are 4 points each

Answer:

1a

  [tex]W = 242.5 \ J[/tex]

1b

    [tex]W = 240 \ J[/tex]

Explanation:

Considering question a

From the question we are told that

       The mass of the toy baby stroller is  [tex]m = 5.0 \ kg[/tex]

        The distance covered is    [tex]d = 7.0\ m[/tex]

          The force the girl applies on the handle  is  [tex]F = 40 \ N[/tex]

          The angle at which this force is applied is  [tex]\theta = 30^o[/tex]

Generally the workdone is mathematically represented as

          [tex]W = F_x * d[/tex]

Here  [tex]F_x[/tex] is the force along the horizontal axis , this is mathematically represented as

           [tex]F_x = F cos (\theta )[/tex]

=>         [tex]F_x = 40 * cos(30 )[/tex]

=>         [tex]F_x = 34.64 \ N[/tex]

So

              [tex]W = 34.64 * 7[/tex]

=>           [tex]W = 242.5 \ J[/tex]

Considering question b

From the question we are told that

       The mass of the toy baby stroller is  [tex]m = 20 \ kg[/tex]

        The distance covered is    [tex]d = 4 \ m[/tex]

          The force the girl applies on the handle  is  [tex]F = 60 \ N[/tex]      

Generally the workdone is mathematically represented as

          [tex]W = F * d[/tex]

=>     [tex]W = 60 * 4[/tex]

=>     [tex]W = 240 \ J[/tex]

Answer:

277.78 hours

Explanation:

The formula for calculating the amount of charge is expressed as;

Q = It

I is the current

t is the time

Given

I =0.05A

Q = 50,000C

Required

Time t

Recall that: Q = It

t = Q/I

t = 50,000/0.05

t = 1,000,000secs

Convert to hours

1,000,000secs  = 1,000,000/3600

1,000,000secs = 277.78 hours

Hence it will take 277.78 hours for the charge to flow through the diode

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

Explanation:

By Hooke's Law, Fe = kx.

Since Fe = 1.6N and x = 9.2cm - 8cm = 1.2cm,

k = Fe/x = 1.6N/1.2cm = 1.33N/cm.

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

Answer:

Radient to ElEcTrIcAAl

Explanation:

The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.

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.