An object is thrown vertically upward and has a speed of 32.6 m/s when it reaches two thirds of its maximum height above the launch point. Determine its maximum height.

Answer:

a ) 2.13 X 10⁶ m/s .

b ) 1.28 X 10⁶ m/s

Explanation:

When electrons are repelled by negative plates and attracted by positive plates , it will increase their kinetic energy.

Increase in their energy = 4.15 eV

= 4.5 X 1.6 X 10⁻¹⁹ J

= 6.64 x 10⁻¹⁹ J

Initial kinetic energy

= 1/2 mv²

= 1/2 x 9.1 x 10⁻³¹ x ( 1.76 x 10⁶)²

= 14.09 X 10⁻¹⁹ J

Total energy

= 6.64 x 10⁻¹⁹+14.09 X 10⁻¹⁹

= 20.73 x 10⁻¹⁹

If V be the increased velocity

1/2 m V² = 20.73 X 10⁻¹⁹

.5 X 9.1 X 10⁻³¹ V² = 20.73 X 10⁻¹⁹

V = 2.13 X 10⁶ m/s .

b ) When electrons are released from positive plate , their speed are reduced because of attraction between electrons and positively charge plates.

Initial kinetic energy

= 14.09 x 10⁻¹⁹ J (see above )

reduction in kinetic energy  

= 6.64 x 10⁻¹⁹ J ( See above )

Total energy with electron -

= 14.09 x 10⁻¹⁹  - 6.64 x 10⁻¹⁹

= 7. 45 x 10⁻¹⁹ J .

If V be the energy of electrons reaching the negative plate,

1/2 m V² =7. 45 x 10⁻¹⁹

V = 1.28 X 10⁶ ms⁻¹

Answer:

Maximum current in the inductor will be 3.824 A

Explanation:

In first case inductive reactance = 59.2 ohm

Frequency = 60 Hz

We know that inductive reactance is given by [tex]X_L=\omega L[/tex]

[tex]59.2=2\pi f\times L[/tex]

[tex]59.2=2\times 3.14\times  60\times L[/tex]

[tex]L=0.157H[/tex]

In second case frequency f = 45 Hz

Now inductive reactance [tex]X_L=\omega L =2\times 3.14\times 45\times .157=44.368ohm[/tex]

Now current [tex]i=\frac{V}{X_L}=\frac{120}{44.368}=2.70A[/tex]

Maximum current [tex]i_{max}=\sqrt{2}i=1.414\times 2.70=3.824A[/tex]

To find the maximum current when the inductor is connected to a 45Hz source, one needs to first calculate the inductance of the inductor using the provided inductive reactance at 60Hz. Then, with the inductance determined, the inductive reactance at 45Hz can be computed. Finally, Arnold's law (I = V/X₁) is used to determine the current.

In this problem, the concept of inductive reactance and Ohm's law are involved. First, we know that the inductive reactance (X₁) is directly proportional to the frequency, and it is given by the formula X₁ = 2πfL. From the information provided, we can find the inductance (L) of the inductor by using this formula at 60Hz, then looking at how X₁ changes when we use a 45Hz source. Once we determine L, we can find the new inductive reactance at 45Hz.

Next, we use Ohm's Law (I = V/X₁) to find the current at 45Hz, where V is the rms voltage across the inductor and X₁ is the inductive reactance. This would give us the maximum current in the inductor.

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Answer: Ok, so we know that them start on the same place.

First, the first one rides 1290 to the east, and 1410 to the north.

So if we put the 0 at the place where they start, north is the Y axis and east the X axis, we could describe the campground is in (1290,1410)

a) when the second cyclist turns, he traveled 1890m to the north, se the point where he is, is described by (0, 1890). We want to know how far is from the campground which we already know that is located in (1290,1410), so if we subtract we get distance = (0,1890) - (1290,1410) = (-1290, 1890 - 1410) = (1290, 480). The total distance will be D = [tex]\sqrt{1290^{2} + 480^{2}  }[/tex] = 1376.4 meters.

b) You want to know in what angle should he turn now.

So again, the cyclist is on the point (0,1890) and the campground is on the point (1290,1410) so he want to go 1290 meters to east, and -480 meters to north. If we define this two amounts as the cathetus of triangle rectangle, the route that he must follow is the hypotenuse of said triangle.

So the angle that he must turn (counting from the east, or the +x axis, counterclockwise) is defined by Tg(A) = -480/1290 = -0.372.

A = aTg(-0.372) = -20° aprox.

Answer:

[tex]v=\frac{v_{1}v_{2}}{\left ( v_{1}+v_{2} \right )}[/tex]

Explanation:

Let the distance traveled in first half is d and then in the next half is also d.

Let the time taken in first half is t1 and the time taken  in the second half is t2.

[tex]t_{1}=\frac{d}{v_{1}}[/tex]

[tex]t_{2}=\frac{d}{v_{2}}[/tex]

Total time taken

[tex]t=t_{1}+t_{2}=\frac{d}{v_{1}}+\frac{d}{v_{2}}=\frac{d\left ( v_{1}+v_{2} \right )}{v_{1}v_{2}}[/tex]

The average speed of a body is defined as the total distance traveled by teh body to the total time taken.

[tex]Average speed = \frac{total distance}{total time}[/tex]

[tex]v=\frac{2d}{t}[/tex]

[tex]v=\frac{2d}{\frac{2d\left ( v_{1}+v_{2} \right )}{v_{1}v_{2}}}[/tex]

[tex]v=\frac{v_{1}v_{2}}{\left ( v_{1}+v_{2} \right )}[/tex]

The average speed of a runner who ran at different speeds for two equal halves of a race is calculated as 2V1V2 / (V1+V2), where V1 and V2 are the speeds for the first and second halves respectively.

The question pertains to calculating average speed in a two-part race which essentially is a concept in Mathematics. We'll start by understanding the definition of average speed. It is calculated as the total distance travelled divided by the total time taken.

In this case, as the race is divided into two halves, the distance covered in the first half is equal to the distance covered in the second half. Hence, the total distance covered by the runner is 2D (where D is the distance of one half of the race).

However, the time to cover each half of the race is different. The time taken to cover the first and second halves would be D/V1 and D/V2 respectively. Therefore, the total time taken for the whole race is D/V1+D/V2.

Now we can calculate the average speed by dividing the total distance by the total time, which gives: Average Speed = 2D / (D/V1+D/V2). This can be simplified to: Average Speed = 2V1V2 / (V1+V2).

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Answer:

speed when the block had slid 3.40 m is 2.68 m/s

Explanation:

given data

distance = 6.80 m

speed = 3.80 m/s

to find out

speed when the block had slid 3.40 m

solution

we will apply here equation of motion that is

v²-u² = 2×a×s   ..............1

here s is distance, a is acceleration and v is speed and u is initial speed that is 0

so put here all value in equation 1 to get a

v²-u² = 2×a×s

3.80²-0 = 2×a×6.80

a = 1.06 m/s²

so

speed when distance 3.40 m

from equation 1 put value

v²-u² = 2×a×s

v²-0 = 2×1.06×3.40

v² = 7.208

v = 2.68

so speed when the block had slid 3.40 m is 2.68 m/s

Answer:

a)  radio < microwaves < visible < gamma

Explanation:

Electromagnetic waves are transverse waves produced by oscillations of the electric and magnetic fields, so light speeds and must comply with the relationship of the wave speed

      v = λ f

  Name             Wavelength              Frequency

  Radio waves        Km                          10⁵

 Microwaves          cm                           10¹⁰

 Visible light           μm                          10¹⁴

 Gamma rays         pm                           10¹⁹

All waves also meet the Planck energy equation

       E = h f

h is the constant of Planck 6.6 10 -34 J s (4.136 10-15 eV-s), the unit of electron volts is very like for the waves of greater energy we substitute the frequencies and we calculate

Name            Frequency             Energy( J)    Energy (eV)

Gamma ray      10¹⁹                        10⁻¹⁵                10⁺⁴

Visible              10¹⁴                        10⁻²⁰               10⁻¹

Microwave       10¹⁰                        10⁻¹⁴                10⁻⁵

Radio waves    10⁵                         10⁻²⁹               10⁻¹⁰

You can see that frequency and energy are proportional

Answer:

In increasing order of frequency:

radio waves < microwaves< visible light < gamma rays

In increasing order of energy per photon:

radio waves < microwaves< visible light < gamma rays

Explanation:

Electromagnetic waves can transfer energy in the space as well. Electromagnetic spectrum is a range of radiation having different frequency.  The radiation which has more frequency has more energy but smaller wavelength. Gamma rays have maximum frequency where as radio waves have smaller  frequency and radiation.

Answer:

The final velocity of the ball just hits the ground is [tex]\sqrt{2gh}[/tex].

Explanation:

A ball is dropped from the height h. As the ball is dropped means there is no initial velocity given to the ball.

GIVEN:

The distance traveled by the ball is equal to h.

Initial velocity of the ball is zero.

Concept:

Two type of force comes into play in dropping of ball.

1. Gravitational force

Due to gravitational force ball forced to fall downward towards the surface.

2. Air resistance

Air develops drag force on the ball in opposite direction of the gravitational force.

Assumption:

There is no air resistance in falling of ball. So, there will be no drag force on the ball.

Calculation:

It is given that a ball is dropped from a building of height h.

The Newton’s equations of motion are as follows:

[tex]V^{2}-u^{2}=2as[/tex]

[tex]s=ut+\frac{1}{2}at^{2}[/tex]

[tex]v=u+at[/tex]

STEP 1

Newton’s equation of motion is expressed as follows:

[tex]V^{2}-u^{2}=2as[/tex]

Here, V is final velocity of the ball just before the ball hits the ground, a is acceleration due to gravity, u is the initial velocity of the ball and s is the distance traveled by the ball.

Here, initial velocity is zero. So, u = 0 m/s.

Vertical distance traveled by the ball is h. So, s= h.

Only gravitational force is responsible for motion as well as for acceleration. Thus, the acceleration due to gravity acts on the ball. Acceleration due to gravity is denoted by small g.

Acceleration due to gravity is constant for different planate. The acceleration due to gravity for the earth is 9.8 m/s². Here, no planet is defined. So the acceleration due to gravity is a = g.

Substitute 0 for m/s, g for a and h for s in above equation as follows:

[tex]V^{2}-u^{2}=2as[/tex]

[tex]V^{2}-(0)^{2}=2gh[/tex]

[tex]V^{2}=2gh[/tex]

[tex]v=\sqrt{2gh}[/tex]

Thus, the final velocity of the ball just hits the ground is [tex]\sqrt{2gh}[/tex].

The coefficient of friction in this scenario is calculated as the ratio of the frictional force to the normal force. After calculating the frictional force to be 50N and the normal force to be 300N, the coefficient of friction is 0.167.

Dana is pulling a box on a level surface with a rope that makes an angle of 30 degrees. The box has a weight of 300 N and the tension in the rope is 100 N. We first must understand that the force of friction is balancing the force that Dana is applying to move the box.

The vertical component of the tension (Tsin30) should be equal to the force of friction since they are balancing each other. Therefore, Tsin30 = Friction. Substituting the given values (100N*sin30 = Friction), we can calculate friction as 50 N.

The coefficient of friction can be calculated as the ratio of the frictional force to the normal force. As there is no vertical acceleration, the normal force is equal to the weight of the box which is 300N. Therefore, the coefficient of friction (mu) equals friction/normal force is equal to 50N/300N which is 0.167.

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The coefficient of friction is approximately 0.247.

The coefficient of friction between the box and the surface is given by the ratio of the frictional force to the normal force. The frictional force is the horizontal component of the tension in the rope, and the normal force is the vertical component of the tension in the rope plus the weight of the box acting downward.

First, let's calculate the horizontal and vertical components of the tension in the rope. The horizontal component can be found using the cosine of the angle, and the vertical component can be found using the sine of the angle:

[tex]\[ T_{\text{horizontal}} = T \cdot \cos(\theta) \] \[ T_{\text{vertical}} = T \cdot \sin(\theta) \][/tex]

Given that the tension in the rope (T) is 100 N and the angle [tex](\(\theta\))[/tex] is 30 degrees, we have:

[tex]\[ T_{\text{horizontal}} = 100 \text{ N} \cdot \cos(30^\circ) \] \\T_{\text{vertical}} = 100 \text{ N} \cdot \sin(30^\circ) \][/tex]

Using the values of cosine and sine for 30 degrees:

[tex]\[ T_{\text{horizontal}} = 100 \text{ N} \cdot \frac{\sqrt{3}}{2} \approx 86.6 \text{ N} \]\\T_{\text{vertical}} = 100 \text{ N} \cdot \frac{1}{2} = 50 \text{ N} \][/tex]

The normal force (N) is the sum of the weight of the box (W) and the vertical component of the tension:

[tex]\[ N = W + T_{\text{vertical}} \] \[ N = 300 \text{ N} + 50 \text{ N} \] \[ N = 350 \text{ N} \][/tex]

The frictional force (f) is equal to the horizontal component of the tension since the box is moving at constant velocity (implying that the net force in the horizontal direction is zero):

[tex]\[ f = T_{\text{horizontal}} \] \[ f = 86.6 \text{ N} \][/tex]

Now, the coefficient of friction is the ratio of the frictional force to the normal force:

[tex]\[ \mu = \frac{f}{N} \] \[ \mu = \frac{86.6 \text{ N}}{350 \text{ N}} \] \[ \mu \approx 0.247 \][/tex]

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

[tex]R = u² sin2θ/g[/tex]

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

[tex]u² = \frac{Rg}{sin2\theta}[/tex]

[tex]u^2 = \frac{9.1 x 9.80}{sin26}[/tex]

[tex] u^2 = 113.17[/tex]

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

[tex]v^2 = (v_osin\theta)^2 -2gh[/tex]

[tex]h =\frac{(v_o^2sin\theta)^2}{2g}[/tex]

[tex]h  =  \frac{(113.17)(sin26)^2}{(2 x 9.80)}}[/tex]

h = 1.10 m

We have that for the takeoff speed and maximum height above ground is

From the question we are told

A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo leaves the ground at a 26° angle. If this is so:

1) What is its takeoff speed

2) What is its maximum height above the ground?

Generally the equation for the Projectile motion   is mathematically given as

[tex]R=\frac{u^2sin 2\theta}{g}\\\\Therefore\\\\9.1=\frac{u^2sin 52}{9.8}[/tex]

u=10.69m/s

And

Generally the equation for the Projectile Height   is mathematically given as

[tex]Hmax=\frac{u^2sin^2 2\theta}{2g}\\\\Therefore\\Hmax=\frac{114.27*0.184}{19.6}[/tex]

Hmax=1.072m

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Therefore, the stone takes 3.06 seconds to reach the bottom, has a speed of 29.8 m/s downwards just before hitting, and travels a total distance of 150.0 meters.

Here's how to solve the problems about the falling stone:

1. Time to reach the bottom:

First half of the journey (upward):

Use the equation v = u - gt, where v is the final velocity (0 m/s at the top), u is the initial velocity (15.0 m/s), g is the acceleration due to gravity (9.81 m/s^2), and t is the time.

Solving for t, we get t = u/g = 15.0 m/s / 9.81 m/s^2 = 1.53 seconds.

Second half of the journey (downward):

The stone falls freely again with an initial velocity of 0 m/s. The time taken will be the same as the upward journey, 1.53 seconds.

Total time to reach the bottom:

Add the times for both halves: 1.53 seconds + 1.53 seconds = 3.06 seconds.

2. Speed just before hitting:

Use the same equation v = u - gt, but this time u is 0 m/s (at the top) and t is the total time (3.06 seconds).

Solving for v, we get v = 0 m/s - 9.81 m/s^2 * 3.06 seconds = -29.8 m/s (negative sign indicates downward direction).

Therefore, the stone's speed just before hitting is 29.8 m/s downwards.

3. Total distance traveled:

The stone travels twice the height of the cliff: once going up and once going down.

Total distance = 2 * cliff height = 2 * 75.0 m = 150.0 m.

Therefore, the stone takes 3.06 seconds to reach the bottom, has a speed of 29.8 m/s downwards just before hitting, and travels a total distance of 150.0 meters.

Answer:

c.m=(1/∑mi)*∑mi*yi

[tex]c.m.=\frac{m*d+2m*3d}{m+2m}=7d/3[/tex]

Explanation:

We solve this problem, with the equation for the center of mass:

c.m=(1/∑mi)*∑mi*yi

[tex]m_i:[/tex]   represent the different mass of the system

[tex]y_i:[/tex]    represent the position of the different mass of the system

c.m=(1/∑mi)*∑mi*yi=(1/(m+2m))*(m*d+2m*3d)=7d/3

Answer:[tex]\Delta E=2.0715\times 10^{-21} J[/tex]

Explanation:

Given

Temperature of the gas is increased from 100 to 200

Also we know that average kinetic energy of the molecules is

[tex]E=\frac{3}{2}\cdot \frac{R}{N_A}T[/tex]

Where

R=Gas constant

[tex]N_A[/tex]=Avogadro's number

T=Temperature in kelvin

[tex]\frac{R}{N_A}=1.381\times 10^{-23}[/tex]

So kinetic energy increases by

[tex]\Delta E=\frac{3}{2}\times 1.381\times 10^{-23}\left ( 200-100\right )[/tex]

[tex]\Delta E=2.0715\times 10^{-21} J[/tex]

Answer:

a) 919 mts

b) 392 mts

Explanation:

In order to solve this, we will use the formulas of acceletared motion problems:

[tex](1)Y=Yo+Vo*t+\frac{1}{2}*a*t^2\\(2)Vf^2=Vo^2+2*a*y[/tex]

We are looking to obtain the initial velocity of the canister right after it was relased, we will use formula (2):

[tex]Vf^2=(0)^2+2*(3.10)*(240)\\Vf=38.6 m/s[/tex]

we need to calculate the time the canister takes to reach the ground, we will use formula (1):

[tex]0-240m=38.6*t+\frac{1}{2}(-9.8m/s^2)*t^2\\\\-4.9*t^2+38.6*t+240=0\\t=11.9seconds[/tex]

in order to know the new height of the rocket we have to use the formula (1) again:

[tex]Y=240+38.6*(11.9)+\frac{1}{2}*(3.10)*(11.9)^2\\Y=919mts[/tex]

We can calulate the total distance the canister traveled before reach the ground by (2):

[tex](0)^2=(38.6)^2+2*(-9.8)*y\\Y=76m[/tex]

So the canister will go up another 76m, so the total distance will be:

[tex]Yc=Yup+Ydown+240m\\Yc=76+76+240\\Yc=392 mts[/tex]

Answer:

I think D is the answer.

D. The development of reading and writing in young children

Answer:

51.94 ft/s²

257.63 ft/s

Explanation:

t = Time taken = 4 s

u = Initial velocity = 34 mi/h

v = Final velocity

s = Displacement = 615 ft

a = Acceleration

Converting velocity to ft/s

[tex]34\ mi/h=\frac{34\times 5280}{3600}=49.87\ ft/s[/tex]

Equation of motion

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow a=2\frac{s-ut}{t^2}\\\Rightarrow a=2\left(\frac{615-49.87\times 4}{4^2}\right)\\\Rightarrow a=51.94\ ft/s^2[/tex]

Acceleration is 51.94 ft/s²

[tex]v=u+at\\\Rightarrow v=49.87+51.94\times 4\\\Rightarrow v=257.63\ ft/s[/tex]

Final velocity at this time is 257.63 ft/s

Final answer:

The glass globe with a net electric charge of 5.31 × 10^-6 C has 3.32 × 10^13 fewer electrons than it did in its neutral state, because positive net charge indicates electron removal.

Explanation:

When a glittering glass globe is given a net electric charge of 5.31 × 10^-6 C, it means that electrons have been added or removed to achieve this net charge. The charge on an electron is known to be approximately -1.6 × 10^-19 C. To determine if the globe has more or fewer electrons than in its neutral state, we compare the sign of the given charge. Here, the positive charge indicates that electrons have been removed. To calculate the quantity of electrons removed, we use the formula:

Number of electrons (n) = Net charge (Q) / Elementary charge (e).

Plugging in the values:

n = 5.31 × 10^-6 C / 1.6 × 10^-19 C/e-

n = 3.32 × 10^13 electrons.

Therefore, the globe now has 3.32 × 10^13 fewer electrons than it did in its neutral state.

Answer:

Explanation:

Total momentum of the system before the collision

.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left

If v be the velocity of the stuck pucks

momentum after the collision = 2 v

Applying conservation of momentum

2 v = -  .75

v =  - .375 m /s

Let after the collision v be the velocity of .5 kg puck

total momentum after the collision

.5 v + 1.5 x .231 = .5v +.3465

Applying conservation of momentum law

.5 v +.3465 = - .75

v = - 2.193 m/s

2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.

Kinetic energy before the collision

= 2.25 + 1.6875

=3.9375 J

kinetic energy after the collision

= .04 + 1.2 =1.24 J

So kinetic energy is not conserved . Hence collision is not elastic.

3 ) Change in the momentum of .5 kg

1.5 - (-1.0965 )

= 2.5965

Average force applied = change in momentum / time

= 2.5965 / 25 x 10⁻³

= 103.86 N

Answer:

The momentum is not conserved.

Explanation:

When the superhero throws the piano, he exerts a force on the piano. Newton's third law says that each action has an opposite and equal reaction. In other words, this means that when the superhero throws the piano, the piano also exerts a force on the superhero, therefore, the superhero should recoil. You can visualize a gun being shot to visualize this phenomenon, when the gun is shot, it will tend to go backwards, trying to conserve momentum.

The scenario in the movie is not physically possible and violates the principles of Newton's laws of motion and conservation of momentum.

The scenario in the movie where the superhero hovers in the air and throws a piano while remaining stationary is not physically possible. It violates the principles of Newton's laws of motion and the conservation of momentum. In order for the superhero to hover and throw a piano, they would need some external force or propulsion, like jetpack or levitation powers, which are not commonly associated with superheroes.

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Answer:

Their kinetic energy will increase but potential energy will decrease.

Explanation:

Given that

Initial velocities of electron and proton is zero.

We know that ,electron have negative charge and proton have positive charge it means that they will attract to each other.We know that opposite charge attract to each other and same charge repels to each other.

It means that the velocities of proton and electron will increase and that leads to increase in the kinetic energy of proton and electron.

We know that potential energy U

[tex]U\alpha -\dfrac{1}{r}[/tex]

So when r will decrease then U will increase but in negative direction it means that U will decrease.

So we can say that their kinetic energy will increase but potential energy will decrease.

Answer:

40m (40m east)

Explanation:

The fist move is 70m east, and east is the positive direction so the truck initially is at +70m.

The second move is 120m to the west, since east is the positive direction, west must be the negative direction, this means the truck now is at:

[tex]+70m -120m = -50m[/tex]

The third move is 90m to the east, again, this is the positive direction, so the new position is:

[tex]-50m + 90m = +40m.[/tex]

The truck is at + 40m, so it ended up 40m away from its initial position and this is the resultant dispacement.

Explanation:

Given that,

Initial speed of the sports car, u = 80 km/h = 22.22 m/s

Final speed of the runner, v = 0

Distance covered by the sports car, d = 80 km = 80000 m

Let a is the acceleration of the sports car.  It can be calculated using third equation of motion as :

[tex]v^2-u^2=2ad[/tex]

[tex]a=\dfrac{v^2-u^2}{2d}[/tex]

[tex]a=\dfrac{0-(22.22)^2}{2\times 80000}[/tex]

[tex]a=-0.00308\ m/s^2[/tex]

Value of g, [tex]g=9.8\ m/s^2[/tex]

[tex]a=\dfrac{-0.00308}{9.8}\ m/s^2[/tex]

[tex]a=(-0.000314)\ g\ m/s^2[/tex]

Hence, this is required solution.

Answer:

An electric field has a direction, +ve to -ve. This is the trajectory in which the electric field will bring about to accelerate a positive charge.

Explanation:

An electric field has a direction, +ve to -ve. This is the trajectory in which the electric field will bring about to accelerate a positive charge.

if the +ve charge moves in the very same direction as the electrical field vector then the velocity of the particle will increase. It will slow down if it moves in the reverse direction.

if the adverse charge moves in the very same direction as the electrical field vector, the particle velocity will decelerate. If it moves in the reverse direction, it will speed up.

Answer and Explanation:

When a particle is placed in an Electric field having a charge Q and mass M, then it experiences an electric force on it. and it is given by:

[tex]F = QE[/tex]

This force is the overall force which is responsible for the acceleration of the particle.

We know, from Newton's second law:

F = Ma

and the force on a particle in electric field:

[tex]F_{E} = QE[/tex]

Therefore, from these two forces:

Ma = QE

a = [tex]\frac{QE}{M}[/tex]

In case of uniform Electric field, the particle moves with constant acceleration.

This acceleration is also capable of bending the object and to move it in a circle.

Answer:

Q = - 256 X 10⁻⁷ C .

Explanation:

Electric field due to a charge Q at a distance d from the center is given by the expression

E = k Q /d² Where k is a constant and it is equal to 9 x 10⁹

Put the given value in the equation

9 x 10² = [tex]\frac{9\times10^9\times Q}{(14+2)^2\times10^{-4}}[/tex]

Q = [tex]\frac{9\times16^2\times10^{-2}}{9\times10^9}[/tex]

Q = - 256 X 10⁻⁷ C .

It will be negative in nature as the field is directed towards the center.

Final answer:

To find the charge on the surface of the copper ball, we use Coulomb's law and the provided electric field magnitude, resulting in an approximate charge of 2.56 x 10⁻⁹ C or 2.56 nC, indicating a positive charge.

Explanation:

The question involves calculating the amount of charge on the surface of a copper ball using the given electric field intensity. The electric field due to a charged sphere at a distance from its surface can be calculated using Coulomb's law and the principle of superposition, which for a sphere of charge translates to E = kQ/r², where E is the electric field, k is Coulomb's constant (8.99 x 10⁹ Nm²/C²), Q is the charge, and r is the distance from the center of the sphere to the point of interest (in this case, 14 cm from the surface or 16 cm from the center considering the radius of the copper ball is 2 cm).

Substituting the given values into the equation, we have 9.0 x 10² = (8.99 x 10⁹)Q/(0.16)². Solving for Q gives us Q = 9.0 x 10² x (0.16)² / (8.99 x 10⁹) C, which yields approximately 2.56 x 10⁻⁹ C or 2.56 nC. The charge is positive, as the electric field is directed towards the center of the ball, indicating a positive source.

Answer: 950 Kg/m^3

Explanation: We can deduce from the Archimedes principle that there is a relation between the density and the volumes displaced, as follows:

Density*Volume= Mass

So for equilibrium Density of body= Density of water *Vw/Vb

Being Vw/Vb the relation between  the displaced water and the body volume, and given the water density as 1000 Kg/m^3 we got:

Density(B)= 0.95 * 1000 Kg/m^3.

Answer:

Explanation:

Conduction of heat per second Q  through a metal sheet having thermal conductivity K , thickness d , area A and temperature difference between cold and hot surface ( T₂ -T₁ ) is given by the following equation

Q = [tex]\frac{KA(T_2-T_1)}{d}[/tex]

Area A = π R² = 3.14 X 10000 X 10⁻⁶

= 3.14 X 10⁻2 m²

For aluminium plate :-

T₂ - T₁ = [tex]\frac{Q\times d}{KA}[/tex]

= [tex]\frac{600\times5\times10^{-3}}{240\times3.14\times10^{-2}}[/tex]

T₂ - T₁ = .4

T₂ = 110.4 ° C

For copper plate :

[tex][tex]\frac{600\times5\times10^{-3}}{390\times3.14\times10^{-2}}[/tex][/tex]

T₂ - T₁ = .25

T₂ = 110.25 ° C

The temperature of the surface in contact with the stove is approximately 113.0 °C for aluminum and 111.3 °C for copper, considering the thermal conductivity and dimensions of the pan.

This is a heat transfer problem involving conduction through different materials. The heat conduction through a flat wall, like the bottom of a pan, can be calculated using Fourier's law of heat conduction: q = -kA(T₂ - T₁)/d, where q is the heat transfer rate (600 W in this case), 'k' is the thermal conductivity, 'A' is the area across which heat is transferring, 'T₁' and 'T₂' represent the temperatures on either side of the wall, and 'd' represents the wall thickness.

First, convert the diameter of the pan to the radius in meters, which is (200mm/2)/1000 = 0.1 m. The area, A, is then π*(0.1 m)² = 0.0314 m². The thickness d given is already in meters: 5 mm = 0.005 m.

For aluminum, rearrange Fourier's law to find T₂, the temperature on the stove side of the pan: T₂ = T₁ + qd/(kA), where T₁ = 110°C and k for aluminum = 240 W/m K. Plugging in all values we get: T₂ = 110 + (600*0.005)/(240*0.0314) ≈ 113.0 °C.

For copper, using k for copper = 390 W/m K in the above formula will give: T₂ = 110 + (600*0.005)/(390*0.0314) ≈ 111.3 °C.

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Answer:

The height of the cliff is 39.655 m

Given:

Height at which the arrow was shot, h = 1.3 m

Velocity of the arrow, u = 34 m/s

Angle, [tex]\theta' = 58.1^{\circ}[/tex]

Time of the fight, t = 4.1 s

Solution:

Let the Height of the cliff be H

Since, the motion of the object is projectile motion and the direction of motion  is vertical at some angle

Therefore, we consider the vertical component of velocity, [tex]u_{y} = usin\theta[/tex].

Now,

The Height of the cliff is given by applying the second equation of motion in the projectile:

Thus

[tex]s = u_{y}t - \frac{1}{2}gt^{2}[/tex]

[tex]s = 34sin60^{\circ}\times 4.1 - \frac{1}{2}\times 9.8\times 4.1^{2}[/tex]

s = 38.355 m

Now, the height of the cliff, H:

H = s + h = 38.355 + 1.3 = 39.655 m

Answer:

The blue boat traveled 6.1 km farther west than the green boat

The green boat traveled 10.7 km farther south than the blue boat

Explanation:

The equation for linear uniform speed movement is

X(t) = X0 + v * t

Since we have two coordinates (X, Y) we use

X(t) = X0 + vx * t

Y(t) = Y0 + vy * t

The dock will be the origin of coordinates so X0 and Y0 will be zero. The X axis will be pointing west and the Y axis south.

The blue boat moves with a direction 24° south of west, so it will have speeds:

vxb = 104 * cos(24) = 95 km/h

vyb = 104 * sin(24) = 42.3 km/h

And the green boat:

vxg = 104 * cos(37.7) = 82.3 km/h

vyg = 104 * sin(37.7) = 63.6 km/h

After half hour the boats will have arrived at positions

Xb = 95 * 0.5 = 47.5 km

Yb = 42.3 * 0.5 = 21.1 km

And

Xg = 82.3 * 0.5 = 41.4 km

Yg = 63.6 * 0.5 = 31.8 km

The difference in positions of the boats

47.5 - 41.4 = 6.1 km

31.8 - 21.1 = 10.7 km

An equipotential line is like the contour line of a map that had lines of equal altitude. In this case the "altitude"is the electrical potential or voltage. Equipotential lines are always perpendicular to the electric field. In three dimensions these lines form equipotential surfaces. The movement along an equipotential surface does not perform work, because that movement is always perpendicular to the electric field.

Answer:

Time taken by the coin to reach the ground is 1.69 s

Given:

Initial speed, v = 11.8 m/s

Height of the building, h = 34.0 m

Solution:

Now, from the third eqn of motion:

[tex]v'^{2} = v^{2} + 2gh[/tex]

[tex]v'^{2} = 11.8^{2} + 2\times 9.8\times 34.0 = 805.64[/tex]

[tex]v' = \sqrt{805.64} = 28.38 m/s[/tex]

Now, time taken by the coin to reach the ground is given by eqn (1):

v' = v + gt

[tex]t = \frac{v' - v}{g} = \frac{28.38 - 11.8}{9.8} = 1.69 s[/tex]

Answer:

circumference of earth = 24000 miles

Explanation:

given data

New York and Los Angeles distance = 3000 mi

time difference = 3 h

to find out

circumference of the Earth

solution

we find speed first that is

speed = [tex]\frac{distance }{time}[/tex]

speed = [tex]\frac{3000}{3}[/tex]

speed = 1000 mph

now we know speed of earth is

speed = [tex]\frac{circumference of earth}{time of rotation}[/tex]

so

circumference = speed × time of rotation

and we know earth take 24 hours to rotate at its axis

so

circumference of earth = 1000 × 24

circumference of earth = 24000 miles

Final answer:

To calculate the Earth's circumference, we establish that a 3-hour difference in time between New York and Los Angeles represents a 45° longitudinal difference. Since 3000 miles correspond to this 45°, we extend this to a full 360° rotation to find that the Earth's estimated circumference is about 24,000 miles.

Explanation:

To calculate the circumference of the Earth using the time difference between New York and Los Angeles, we use the relationship between time and longitude. Since the Earth rotates 360° in 24 hours, it rotates 15° per hour. Thus, a 3 hour time difference corresponds to a 45° difference in longitude (because 3 hours x 15° per hour = 45°).

If New York and Los Angeles are approximately 3000 miles apart, and this represents 45° of the Earth's rotation, we can find the total circumference of the Earth by setting up the proportion:

3000 miles / 45° = Circumference / 360°

Circumference = (3000 miles / 45°) x 360°

Circumference = 24,000 miles

Therefore, the estimated circumference of the Earth is approximately 24,000 miles.