A walker covers a disatance of 4.0 km in a time of 53 minutes. (a) What is the average speed of the walker for this distance in km/hr?

The car strikes the tree at approximately 10.7 m/s after slowing down with a uniform acceleration of -5.75 m/s^2 for 60.0 m.

To calculate the speed with which the car strikes the tree, we'll use the kinematic equations for uniformly accelerated motion. Given the uniform acceleration of -5.75 m/s2 and the time of 4.40 s, we can use the following equation to find the initial velocity (vi) before the car started braking:

v = vi + at
Where:

By rearranging the equation to solve for the initial velocity (vi), we get:

vi = v - at

Substituting the known values:
vi = 0 - (-5.75 m/s2 * 4.40 s)
vi = 25.3 m/s

This is the speed with which the car was travelling before it hit the brakes. However, to find the speed at which the car strikes the tree, we must consider the distance of the skid marks. Using the kinematic equation for distance (d), where d equals the initial velocity times time plus half the acceleration times time squared:

d = vit +  rac{1}{2}at2

Since the distance to the tree is 60.0 m and we're looking for the speed at the end of this distance, we rearrange the equation to solve for final velocity (v). But first, we need to calculate the time it takes to stop over this distance. We can use the formula:

d =  rac{vi2 - v2}{2a}

By solving for v we find:

v = [tex]\sqrt{x}[/tex]{vi2 - 2ad}

v = [tex]\sqrt{x}[/tex]{(25.3 m/s)2 - 2(-5.75 m/s2)(60.0 m)}

v = [tex]\sqrt{x}[/tex]{(640.09) - (-690)}

v = 10.7 m/s

Therefore, the car strikes the tree at approximately 10.7 m/s.

The car strikes the tree at a speed of 0.98 m/s.

To determine the speed at which the car strikes the tree, we can utilize the kinematic equations of motion. We are given the following data:

First, we calculate the initial velocity using the equation:

Substituting the given values:

Solving this:

Now, to find the final velocity when the car strikes the tree, we use the kinematic equation:

Substituting the values:

So, the car strikes the tree at a speed of 0.98 m/s.

Answer:

Explanation:

Given:

Length of each side of the cube, [tex]L=10\ cm[/tex]

The Elecric flux through one of the side of the cube is, [tex]\phi =-1 kNm^2/C.[/tex]

The net flux through a closed surface is defined as the total charge that lie inside the closed surface divided by [tex]\epsilon_0[/tex]

Since Flux is a scalar quantity. It can added to get total flux through the surface.

[tex]\phi_{total}=\dfrac{Q_{in}}{\epsilon_0}\\6\times {-1}\times 10^3=\dfrac{Q_{in}}{\epsilon_0}\\\\Q_{in}=-6}\times 10^3\epsilon_0\\Q_{in}=-6}\times 10^3\times8.85\times 10^{-12}\\Q_{in}=-53.1\times10^{-9}\ C[/tex]

So the the charge at the centre is calculated.

Answer:

a) Focal length of the lens is 8 cm which is a convex lens

b) 6 cm

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

Explanation:

u = Object distance =  4 cm

v = Image distance = -8 cm

f = Focal length

Lens Equation

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{4}+\frac{1}{-8}\\\Rightarrow \frac{1}{f}=\frac{1}{8}\\\Rightarrow f=\frac{8}{1}=-8\ cm[/tex]

a) Focal length of the lens is 8 cm which is a convex lens

Magnification

[tex]m=-\frac{v}{u}\\\Rightarrow m=-\frac{-8}{4}\\\Rightarrow m=2[/tex]

b) Height of image is 2×3 = 6 cm

Since magnification is positive the image upright

c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.

(a) The focal length of the lens is -2.67 cm.

(b) The magnification of the image is 2 and the height is 6 cm.

(c) The imaged formed by the lens is upright, virtual and magnified.

The focal length of the lens is determined by using lens formulas as given below;

[tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \\\\\frac{1}{f} = \frac{-1}{8} - \frac{1}{4} \\\\\frac{1}{f} = \frac{-1 -2}{8} = \frac{-3}{8} \\\\f = -2.67 \ cm[/tex]

The magnification of the image is calculated as follows;

[tex]m = \frac{v}{u} \\\\m = \frac{8}{4} \\\\m = 2\\\\[/tex]

The height of the image is calculated as follows;

[tex]H = mu\\\\H = 2(3cm) = 6\ cm[/tex]

The imaged formed by the lens is;

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Answer:

[tex]v=0[/tex]

Explanation:

Knowing that the formula for average velocity is:

[tex]v=\frac{x_{2}-x_{1}}{t_{2}-t_{1}}[/tex]

Being said that, we know that the person's displacement is zero because it returns to its starting point

[tex]x_{2}=x_{1}[/tex]

That means [tex]x_{2}-x_{1}=0[/tex]

[tex]v=\frac{0}{t_{2}-t_{1}}=0[/tex]

Answer:

[tex]\theta=145[/tex]

Explanation:

The amplitude of he combined wave is:

[tex]B=2Acos(\theta/2)\\[/tex]

A, is the amplitude from the identical harmonic waves

B, is the amplitude of the resultant wave

θ, is the phase, between the waves

The amplitude of the combined wave must be 0.6A:

[tex]0.6A=2Acos(\theta/2)\\ cos(\theta/2)=0.3\\\theta/2=72.5\\\theta=145[/tex]

Answer:

Distance, d = 112.5 meters

Explanation:

Initially, the bicyclist is at rest, u = 0

Final speed of the bicyclist, v = 30 m/s

Acceleration of the bicycle, [tex]a=4\ m/s^2[/tex]

Let s is the distance travelled by the bicyclist. The third equation of motion is given as :

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

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

[tex]s=\dfrac{(30)^2}{2\times 4}[/tex]

s = 112.5 meters

So, the distance travelled by the bicyclist is 112.5 meters. Hence, this is the required solution.

Answer:

The fraction of time for turn on is 0.3852

Solution:

As per the question:

Temperature at which oil bath is maintained, [tex]T_{o} = 50.5^{\circ}[/tex]

Heat loss at rate, q = 4.68 kJ/min

Resistance, R = [tex]60\Omega[/tex]

Operating Voltage, [tex]V_{o} = 110 V[/tex]

Now,

Power that the resistor releases, [tex]P_{R} = \frac{V_{o}^{2}}{R}[/tex]

[tex]P_{R} = \frac{110^{2}}{60} = 201.67 W = 12.148 J/min[/tex]

The fraction of time for the current to be turned on:

[tex]P_{R} = \frac{q}{t}[/tex]

[tex]12.148 = \frac{4.68}{t}[/tex]

t = 0.3852

The acceleration experienced by the space travelers during their launch would be considered unrealistically large. It would exceed both the limits of human endurance and the acceleration experienced during free fall.

To calculate the acceleration experienced by the space travelers during their launch, we can use the formula:

Acceleration = (Final Velocity - Initial Velocity) / Time

In this case, the final velocity is given as 10.97 km/s, and the initial velocity is 0 m/s (since the spaceship starts from rest). The time is not provided, so we cannot calculate the exact acceleration. However, we can compare it to the acceleration that a human can withstand, which is 15g for a short time. One g is equivalent to the acceleration due to gravity, which is approximately 9.8 m/s².

So, 15g is equal to 15 * 9.8 m/s² = 147 m/s². Therefore, any acceleration larger than 147 m/s² would be unrealistically large for the space travelers during their launch.

Comparing this with the free-fall acceleration, which is approximately 9.8 m/s², we can see that the acceleration proposed by Jules Verne would be much larger than both the limits of human endurance and the acceleration experienced during free fall.

Answer:

The resultant field will have a magnitude of 241.71 V/m, 30.28° to the left of E1.

Explanation:

To find the resultant electric fields, you simply need to add the vectors representing both electric field E1 and electric field E2. You can do this by using the component method, where you add the x-component and y-component of each vector:

E1 = 99 V/m, 0° from the y-axis

E1x = 0 V/m  

E1y = 99 V/m, up

E2 = 164 V/m, 48° from y-axis

E2x = 164*sin(48°) V/m, to the left

E2y = 164*cos(48°) V/m, up

[tex]Ex: E_{1_{x}} + E_{2_{x}} = 0 V/m - 164 *sin(48) V/m= -121.875 V/m\\Ey: E_{1_{y}} + E_{2_{y}} = 99 V/m + 164 *cos(48) V/m = 208.74 V/m\\[/tex]

To find the magnitude of the resultant vector, we use the pythagorean theorem. To find the direction, we use trigonometry.

[tex]E_r = \sqrt{E_x^2 + E_y^2}= \sqrt{(-121.875V/m)^2 + (208.74V/m)^2} = 241.71 V/m[/tex]

The direction from the y-axis will be:

[tex]\beta = arctan(\frac{-121.875 V/m}{208.74 V/m}) = 30.28[/tex]° to the left of E1.

the resultant electric field has a magnitude of approximately 247.8 V/m and is directed at an angle of approximately 63.5 degrees above the horizontal, not the vertical.

To find the resultant electric field, you should add the horizontal and vertical components of E1 and E2 separately:

Vertical Component:

E1y = 99 V/m (vertical component of E1)

E2y = 164 V/m * sin(48°) ≈ 123.6 V/m (vertical component of E2)

Horizontal Component:

E2x = 164 V/m * cos(48°) ≈ 109.8 V/m (horizontal component of E2)

Add the vertical components:

Ey = E1y + E2y

Ey = 99 V/m + 123.6 V/m

Ey ≈ 222.6 V/m

Add the horizontal components:

Ex = E2x

Ex ≈ 109.8 V/m

Calculate the magnitude of the resultant electric field (E) using the Pythagorean theorem:

E = √(Ex² + Ey²)

E = √((109.8 V/m)² + (222.6 V/m)²)

E ≈ 247.8 V/m

So, the resultant electric field has a magnitude of approximately 247.8 V/m and is directed at an angle of approximately 63.5 degrees above the horizontal, not the vertical, as previously stated.

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Answer:

Explanation:

The total energy of an electron in an orbit consists of two components

1 ) Potential energy which is - ve because the field is attractive

2) Kinetic energy which represents moving electron having some velocity.

Kinetic is always positive.

3 ) In an orbit , The magnitude of potential energy is twice that of kinetic energy. So if -2E is the value of potential energy E wil be the value of kinetic energy.

4 ) Total energy will become some of potential energy and kinetic energy

-2E + E = -E

5 ) So total energy becomes equal to kinetic energy with only sign reversed.

In the given case total energy is -0.28 eV . Hence kinetic energy will be +0.28 eV.

When kinetic energy is calculated as +.28 eV , the potential energy will be

- 2 x .28 or - 0.56  eV .

Answer:

[tex]-1.144\ \mu C[/tex]

Explanation:

Given:

Assume:

We know that the electric field is the electric force applied on a unit positive charge i.e.,

[tex]\vec{E}=\dfrac{\vec{F}}{Q}[/tex]

This means the electric force applied on this additional charge placed in the field is given by:

[tex]\vec{F}=Q\vec{E}\\\Rightarrow \vec{F} =  1.04\times 10^{-8}\ n C\times (148.0\ \hat{i}-110.0\ \hat{j})\ N/C\\\Rightarrow \vec{F} = (1.539\ \hat{i}-1.144\ \hat{j})\ \mu N\\[/tex]

From the above expression of force, we have the following y-component of force on this additional charge.

[tex]F_y = -1.144\ \mu N[/tex]

Hence, the y-component of the electric force on the this charge is [tex]-1.144\ \mu N[/tex].

By installing motion sensors, the storage room would save $1.27 per day in energy costs. The simple payback period for installing motion sensors is approximately 77.17 days.

To determine the amount of energy and money that will be saved as a result of installing motion sensors, we need to compare the energy consumption and cost of the current lighting system with that of the motion sensor system. Currently, the storage room uses six fluorescent light fixtures, each containing four lamps rated at 60 W each. The lamps are on for 12 hours a day. So, the total energy consumed by the current system per day is 6 * 4 * 60 W * 12 hours = 17,280 W. With an electricity price of $0.11/kWh, the cost of energy consumed per day is 17,280 W / 1000 * $0.11 = $1.90.

Now, let's consider the motion sensor system. If the room is only used for an average of 3 hours a day, the total energy consumed by the motion sensor system per day would be 6 * 4 * 60 W * 3 hours = 5,760 W. This results in a cost of 5,760 W / 1000 * $0.11 = $0.63 per day. Therefore, by installing motion sensors, the storage room would save $1.90 - $0.63 = $1.27 per day in energy costs.

To determine the simple payback period, we need to calculate the cost of the motion sensor system and the savings achieved per day. The total cost of the motion sensor system is $32 for the purchase price + $66 for installation, which equals $98. With daily savings of $1.27, the payback period can be calculated as $98 / $1.27 = 77.17 days. Therefore, the simple payback period for installing motion sensors in the storage room is approximately 77.17 days.

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Answer:

Explanation:

mass, m = 1 kg

Position (2, 3 ) m

height, h = 2 m

acceleration due to gravity, g = 9.8 m/s^2

Here, no force is acting in horizontal direction, the force of gravity is acting in vertical direction, so the work done by the gravitational force is to be calculated.

Force  mass x acceleration due to gravity

F = 1 x 9.8 = 9.8 N

Work = force x displacement x CosФ

Where, Ф be the angle between force vector and the displacement vector.

Here the value of Ф is 180° as the force acting vertically downward and the displacement is upward

So, W = 9.8 x 2 x Cos 180°

W = - 19.6 J

Thus, option (A) is correct.

Answer:

(a) Maximum height = 53.88 meters

(b) Range of the ball = 924.36 meters

Explanation:

The ball has been launched at a speed = 44.4 meters per second

Angle of the ball with the horizontal = 25°

Horizontal component of the speed of the ball = 44.4cos25° = 40.24 meters per second

Vertical component = 44.4sin25° = 18.76 meters per second

We know vertical component of the speed decides the height of the ball so by the law of motion,

v² = u² - 2gh

where v = velocity at the maximum height = 0

u = initial velocity = 18.76 meter per second

g = gravitational force = 9.8 meter per second²

Now we plug in the values in the given equation

0 = (18.76)² - 2(9.8)(h)

19.6h = 352.10

h = [tex]\frac{352.10}{19.6}[/tex]

h = 17.96 meters

By another equation,

[tex]v=ut-\frac{1}{2}gt^{2}[/tex]

Now we plug in the values again

[tex]0=(18.76)t-\frac{1}{2}(9.8)t^{2}[/tex]

18.76t = 4.9t²

18.76 = 4.9t

t = [tex]\frac{18.76}{4.9}=3.83[/tex]seconds

Since time t is the time to cover half of the range.

Therefore, time taken by the ball to cover the complete range = 2×3.83 = 7.66 seconds

  So the range of the ball = Horizontal component of the velocity × time

                                           = 40.24 × 7.66

                                           = 308.12 meters

This we have calculated all for our planet.

Now we take other planet.

(a) Since the golfer drives the ball 3 times as far as he would have on earth then maximum height achieved by the ball = 17.96 × 3 = 53.88 meters

(b) Range of the ball = 3×308.12 = 924.36 meters

Answer:

The answer is [tex] 1.956 \times 10^7\ m/s[/tex]

Explanation:

The amount of energy is not enough to apply the relativistic formula of energy [tex]E = mc^2[/tex], so the definition of energy in this case is

[tex]E = \frac{1}{2}m v^2[/tex].

From the last equation,

[tex]v= \sqrt{2E/m}[/tex]

where

[tex]E = 2 MeV = 3.204 \times 10^{-13} J[/tex]

and the mass of the neutron is

[tex]m = 1.675\times 10^{-27}\ Kg[/tex].

Then

[tex]v = 1.956 \times 10^7\ m/s[/tex]

the equivalent of [tex]0.065[/tex] the speed of light.

Answer:

The vector has a magnitude of 33.86 units and a direction of 121.64°.

Explanation:

To find the magnitude, you use the Pitagorean theorem:

[tex]||V|| = \sqrt{x^2 + y^2}= \sqrt{(-26.5)^2 + (43)^2} = 33.86 units[/tex]

In order to find the direction, you can use trigonometry. You have to keep in mind, that as the y component of the vector is positive and the x component is negative, the vector must have an angle between 90 and 180°, or in the second quadrant of the plane.:

[tex]tan(\alpha) = \frac{y}{x} \\\alpha = tan^{-1}(\frac{43}{-26.5})= 180 - tan^{-1}(\frac{43}{26.5}) = 121.64[/tex]° or, 58.36° in the second quadrant

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁=  m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

[tex]v_{f1} = \sqrt{2*a_{1}*d_{1}  }[/tex]  Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁  :Newton's second law

-Ff₁+Wx₁ = ma₁   , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁      Equation (2)

∑Fy₁=0   : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁  =  m*g*cos30°

We replace   N₁  =  m*g*cos30 and  Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁   :  We divide by m

-μ₁*g*cos30°+g*sin30° = a₁  

g*(-μ₁*cos30°+sin30°) = a₁  

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

[tex]v_{f1} = \sqrt{2*3.2*3}  }[/tex]

[tex]v_{f1} =\sqrt{2*3.2*3}[/tex]

[tex]v_{f1} = 4.38 m/s[/tex]

Rough surface  kinematics

vf₂²=v₀₂²+2*a₂*d₂   v₀₂=vf₁=4.38 m/s

0   =4.38²+2*a₂*d₂  Equation (3)

Rough surface  kinetics (μ= 0.3)

∑Fx₂=ma₂  :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂   Equation (4)

∑Fy₂= 0  :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂   We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0   =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= -  6.79*3 = 20.4 N*m

Wf=  -20.4 J    is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

The work done by the friction force is negative, and the work done by the gravitational force is positive as the object slides down the inclined plane.

The work done by the friction force while the object slides down the inclined plane is negative. The work done by the gravitational force while the object slides down the inclined plane is positive.

When the object slides down the inclined plane, the friction force acts in the opposite direction to its motion. Since friction always opposes the motion, the work done by friction is negative.

The gravitational force, on the other hand, acts in the same direction as the object's motion. Therefore, the work done by the gravitational force is positive.

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Answer:

0.02442 × 10⁻⁹

Explanation:

Given:

Diameter of copper ball = 2.00 mm = 0.002 m

Charge on ball = 40 nC = 40 × 10⁻⁹ C

Density of copper = 8900 Kg/m³

Now,

The number of electrons removed, n = [tex]\frac{\textup{Charge on ball}}{\textup{Charge of an electron}}[/tex]

also, charge on electron = 1.6 × 10⁻¹⁹ C

Thus,

n = [tex]\frac{40\times10^{-9}}{1.6\times10^{-19}}[/tex]

or

n = 25 × 10¹⁰ Electrons

Now,

Mass of copper ball = volume × density

Or

Mass of copper ball =  [tex]\frac{4}{3}\pi(\frac{d}{2})^3[/tex]  × 8900

or

Mass of copper ball =  [tex]\frac{4}{3}\pi(\frac{0.002}{2})^3[/tex]  × 8900

or

Mass of copper ball = 0.03726 grams

Also,

molar mass of copper = 63.546 g/mol

Therefore,

Number of mol of copper in  0.03726 grams = [tex]\frac{ 0.03726}{63.546}[/tex]

or

Number of mol of copper in  0.03726 grams = 5.86 × 10⁻⁴ mol

and,

1 mol of a substance contains = 6.022 × 10²³ atoms

Therefore,

5.86 × 10⁻⁴ mol of copper contains = 5.86 × 10⁻⁴ × 6.022 × 10²³ atoms.

or

5.86 × 10⁻⁴ mol of copper contains = 35.88 × 10¹⁹ atoms

Now,

A neutral copper atom has 29 electrons.  

Therefore,

Number of electrons in ball = 29 × 35.88 × 10¹⁹ = 1023.37 × 10¹⁹ electrons.

Hence,

The fraction of electrons removed = [tex]\frac{25\times10^{10}}{1023.37\times10^{19}}[/tex]

or

The fraction of electrons removed = 0.02442 × 10⁻⁹

The fraction of electrons removed from the copper ball can be calculated by comparing its net charge to the charge of a single electron. Using the provided information, we can find that approximately 2.5 x 10^10 electrons have been removed from the ball. To determine the fraction, we need to compare this number to the total number of electrons in the ball, which can be calculated using the mass, density, and atomic mass of copper. By plugging in the values, we can find the fraction of electrons removed.

To determine the fraction of electrons that have been removed from the copper ball, we need to compare the net charge of the ball to the charge of a single electron. The net charge of the ball is 40 nC, which is equivalent to 40 x 10^-9 C. The charge of a single electron is 1.60 x 10^-19 C.

We can calculate the number of electrons that have been removed using the formula:

Number of electrons removed = Net charge of the ball / Charge of a single electron

Number of electrons removed = (40 x 10^-9 C) / (1.60 x 10^-19 C) = 2.5 x 10^10 electrons

To find the fraction of electrons removed, we need to compare the number of electrons removed to the total number of electrons in the ball. The total number of electrons in the ball can be calculated using the formula:

Total number of electrons = Number of copper atoms x Number of electrons per copper atom

The number of copper atoms can be calculated using the formula:

Number of copper atoms = Mass of the ball / Atomic mass of copper

The mass of the ball can be calculated using the formula:

Mass of the ball = Volume of the ball x Density of copper

Given that the diameter of the ball is 2.0 mm, the volume of the ball can be calculated using the formula for the volume of a sphere:

Volume of the ball = (4/3) x pi x (radius)^3

As the ball is a sphere, the radius is half the diameter, so the radius is 1.0 mm or 1 x 10^-3 m.

Using the given density of copper (8900 kg/m^3) and atomic mass of copper (63.5 g/mol), we can now calculate the fraction of electrons removed:

Fraction of electrons removed = Number of electrons removed / Total number of electrons = (2.5 x 10^10 electrons) / (Number of copper atoms x Number of electrons per copper atom)

Answer:

The final velocity of the car is 30 m/s.

Explanation:

Given that,

Initial speed of the car, u = 0

Acceleration of the car, [tex]a=5\ m/s^2[/tex]

Time taken, t = 6 s

Let v is the final velocity of the car. It can be calculated using first equation of kinematics as :

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

[tex]v=at[/tex]

[tex]v=5\ m/s^2\times 6\ s[/tex]

v = 30 m/s

So, the final velocity of the car is 30 m/s. Hence, this is the required solution.

Answer:

a) 38.27      b) 322.5°

c) 126.99    d) 1.17°

e) 62.27     e) 139.6°

Explanation:

First of all we have to convert the coordinates into rectangular coordinates, so:

a=( 43.3 , 25)

b=( -48.3 , -12.94)

c=( 35.36 , -35.36)

Now we can do the math easier (x coordinate with x coordinate, and y coordinate with y coordinate):

1.)  a+b+c=( 30.36 , -23.3) = 38.27 < 322.5°

2.)  a-b+c=( 126.96 , 2.6) = 126.99 < 1.17°

3.)  (a+b) - (c+d)=0   Solving for d:

     d=(a+b) - c = ( -40.36 , 47.42) = 62.27 < 139.6°

Answer:

b. the air

Explanation:

David Scott's experiment was performed on the moon, where there is gravity but there is no air. This experiment consisted of letting a hammer and a feather fall at the same time. The result was that the two objects touch the ground simultaneously.

Since these two objects obviously have a different mass, the experiment shows that in a vacuum, objects fall with the same acceleration regardless of their mass.

Answer:

The spring constant is [tex]1.09\times10^{9}\ N/m[/tex]

Explanation:

Given that,

length = 500 mm

Diameter = 2 cm

Young's modulus = 17.4 GPa

We need to calculate the young's modulus

Using formula of young's modulus

[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}[/tex]....(I)

[tex]Y=\dfrac{Fl}{\Delta l A}[/tex]

From hook's law

[tex]F=kx[/tex]

[tex]k=\dfrac{F}{x}[/tex]

[tex]F=k\times\Delta l[/tex]....(II)

Put the value of F in equation

[tex]Y=\dfrac{k\times\Delta l\times l}{\Delta l A}[/tex]

[tex]Y=\dfrac{kl}{A}[/tex]

We need to calculate the spring constant

[tex]k = \dfrac{YA}{l}[/tex]....(II)

We need to calculate the area of cylinder

Using formula of area of cylinder

[tex]A=2\pi\times r\times l[/tex]

Put the value into the formula

[tex]A=2\pi\times 1\times10^{-2}\times500\times10^{-3}[/tex]

[tex]A=0.0314\ m^2[/tex]

Put the value of A in (II)

[tex]k=\dfrac{1.74\times10^{10}\times0.0314}{500\times10^{-3}}[/tex]

[tex]k=1.09\times10^{9}\ N/m[/tex]

Hence, The spring constant is [tex]1.09\times10^{9}\ N/m[/tex]

Answer:

The shell hit at a distance of 1.9 x 10² km

The time of flight of the shell was 5.3 x 10² s

Explanation:

The position of the shell is given by the vector "r":

r  = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)

where:

x0 = initial horizontal position

v0 = magnitude of the initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration of gravity

When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.

Then:

ry = 0 =  y0 + v0 * t * sin α + 1/2 g t²

Since the gun is at the center of our system of reference, y0 and x0 = 0

0 = t (v0 sin α + 1/2 g t)

t= 0 is discarded as solution

v0 sin α + 1/2 g t = 0

t = -2v0 sin α / g

t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.

Then, the distance at which the shell hit is:

Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α  

Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km

Answer:

The current through [tex]9 \Omega[/tex] is 0.297 A

Solution:

As per the question:

[tex]R_{5} = 5.0 \Omega[/tex]

[tex]R_{9} = 9.0 \Omega[/tex]

[tex]R_{4} = 5.0 \Omega[/tex]

V = 6.0 V

Now, from the given circuit:

[tex]R_{5}[/tex] and [tex]R_{9}[/tex] are in parallel

Thus

[tex]\frac{1}{R_{eq}} = \frac{1}{R_{5}} + \frac{1}{R_{9}}[/tex]

[tex]R_{eq} = \frac{R_{5}R_{9}}{R_{5} + R_{9}}[/tex]

[tex]R_{eq} = \frac{5.0\times 9.0}{5.0 + 9.0} = 3.2143 \Omega[/tex]

Now, the [tex]R_{eq}[/tex] is in series with [tex]R_{4}[/tex]:

[tex]R'_{eq} = R_{eq} + R_{4} = 3.2143 + 4.0 = 7.2413 \Omega[/tex]

Now, to calculate the current through [tex]R_{9}[/tex]:

[tex]V = I\times R'_{eq}[/tex]

[tex]I = {6}{7.2143} = 0.8317 A[/tex]

where

I = circuit current

Now,

Voltage across [tex]R_{eq}[/tex], V':

[tex]V' = I\times R_{eq}[/tex]

[tex]V' = 0.8317\times 3.2143 = 2.6734 V[/tex]

Now, current through [tex]R_{9}[/tex], I' :

[tex]I' = \frac{V'}{R_{9}}[/tex]

[tex]I' = \frac{2.6734}{9.0} = 0.297 A[/tex]

Final answer:

To find the current through the 9.0 ohm resistor in a series-parallel circuit, we can calculate the equivalent resistance of the parallel combination.

Explanation:

To find the current through the 9.0 ohm resistor, we need to first determine the equivalent resistance of the circuit. The two resistors of 5.0 and 9.0 ohms that are connected in parallel have an equivalent resistance given by the formula:

1/Req = 1/R1 + 1/R2

1/Req = 1/5.0 + 1/9.0

1/Req = (9.0 + 5.0)/(5.0 * 9.0)

1/Req = 14.0/45.0

Req = 45.0/14.0 ≈ 3.21 ohms

The equivalent resistance of the parallel combination is approximately 3.21 ohms.

Answer:

magnitude is 2.52 m  with 66.15° north east

Explanation:

given data

displace 1 = 3.45 m north = 3.45 j

displace 2 = 1.53 m southeast = 1.53 cos(315) i + sin(315) j

displace 3 = 0.877 m southwest  = 0.877 cos(225) i- sin(225) j

to find out

displacement and direction

solution

we consider here direction i as east and direction j as north

so here

displacement = displace 1 + displace 2 + displace 3

displacement =  3.45 j + 1.53 cos(315) i + sin(315) j + 0.877 cos(225) i- sin(225) j

displacement =  3.45 j + 1.081 i -1.0818 j - 0.062 i -0.062 j

displacement = 1.019 i + 2.306 j

so magnitude

magnitude = [tex]\sqrt{1.019^{2} + 2.306^{2}}[/tex]

magnitude = 2.52

and

angle will be = arctan(2.306/1.019)

angle = 1.15470651 rad

angle is 66.15 degree

so magnitude is 2.52 m  with 66.15° north east

Answer:

a) 0 m/s

b) - 24 m/s

c)  - 68 m

Explanation:

Given:

Initial distance = - 4 m

Initial velocity, u = 4 m/s

1) acceleration, a = - 2 m/s² for time, t = 2 seconds

thus,

velocity after 2 seconds will be

from Newton's equation of motion

v = u + at

v = 4 + (-2) × 2

v = 0 m/s

2) Velocity after 2 second is the initial velocity for this case

given acceleration = - 6 m/s² for 4 seconds

thus,

final velocity, v = 0 + ( - 6 ) × 4 = - 24 m/s

here the negative sign depicts the velocity in opposite direction to the initial direction of motion

thus, velocity after 6 seconds = - 24 m/s

3) Now,

Total displacement in 6 seconds

= Displacement in 2 seconds + Displacement in 4 seconds

From Newton's equation of motion

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

where,  

s is the distance

u is the initial speed  

a is the acceleration

t is the time

thus,

= [tex]0\times2+\frac{1}{2}\times(-2)\times2^2[/tex]  + [tex]0\times4+\frac{1}{2}\times-6\times4^2[/tex]

= - 16 - 48

= - 64 m

Hence, the final displacement = - 64 - 4 = - 68 m

Answer:

Zero

Explanation:

The overall work done by gravitational force on completion of a complete turn is zero.

Since the work done by gravitaional force is conservative and depends only on the initial and end position here height and the path followed does not matter.

Since, in a complete turn the wheel return to its final position as a result of which displacement is zero and as work is the dot product of Force exerted and displacement, the work done is zero.

Also the work done in half cycle by gravity is counter balance by the work which is done against the gravity in the other half cycle.

The net work done by the gravitational force when a person is riding on a Ferris Wheel and it makes one complete turn is zero.

To understand why the net work done by the gravitational force is zero, we need to consider the definition of work and the nature of the motion on a Ferris Wheel. Work done by a force is defined as the product of the force and the displacement in the direction of the force. Mathematically, this is expressed as:

[tex]\[ W = F \cdot d \cdot \cos(\theta) \][/tex]

where ( W ) is the work, ( F ) is the force, ( d ) is the displacement, and [tex]\( \theta \)[/tex] is the angle between the force and the displacement.

In the case of the Ferris Wheel, the gravitational force acts vertically downward towards the center of the Earth, while the displacement of the person on the Ferris Wheel is along the circumference of the wheel, which is horizontal at any given point. Since the force and displacement are perpendicular to each other at every point in the circle, the angle \( \theta \) between them is always 90 degrees. Therefore, the cosine of 90 degrees is zero, which means that the work done by the gravitational force at each point is zero:

[tex]\[ W = F \cdot d \cdot \cos(90^\circ) = F \cdot d \cdot 0 = 0 \][/tex]

Moreover, over one complete turn of the Ferris Wheel, the initial and final positions of the person are the same. This means that the total displacement over one complete cycle is zero. Even if we consider the components of the gravitational force along the direction of displacement during different parts of the cycle, the net effect is zero because the person is raised to a certain height and then lowered back to the starting point. The work done against gravity to raise the person is equal in magnitude and opposite in sign to the work done by gravity as the person descends, resulting in a net work of zero for the entire cycle.

Therefore, the gravitational force does no net work on the person over one complete turn of the Ferris Wheel.

Answer:

[tex]F_1+F_2= (28.26, 9.05) N[/tex]

[tex]\alpha = 17.7\º[/tex]

[tex]F = 29.67 N[/tex]

Explanation:

Hi!

In a (x, y) coordinate representation, the two forces are:

[tex]F_1=(12.7N, 0)\\F_2=(18.1N\cos(30\º), 18.1N \sin(30\º) )\\\cos(\º30)=0.86\\\sin(\º30)= 0.5[/tex]

The sum of the two forces is:

[tex]F_1 + F_2 = ( 12.7 + 0.86*18.1, 18.1*0.5) N[/tex]

[tex]F_1+F_2= (28.26, 9.05) N[/tex]

The angle to x-axis is calculated using arctan:

[tex]\alpha = \arctan(\frac{F_y}{F_x}) = \arctan(\frac{9.05}{28.26} = 17.7\º[/tex]

The magnitude is:

[tex]F = \sqrt {F_x^2 + F_y^2}= \sqrt{798.6 + 81.9} = 29.67 N[/tex]

Answer:

d = 3.44 *10^{-7} m  

Explanation:

given data:

length of metal plates = 16.50 cm

capacitor charge = 18.5 nC

potential difference = 37.8 V

capacitance of parallel plate capacitor

[tex]C = \frac{A\epsilon _{0}}{d}[/tex]

area of the individual plate

A=[tex] a^2 = (16.5*10^{-2})^2 = 272.25 *10^{-4}[/tex] m2

capacitance

[tex]C = QV = 18.5 *10^{-9} *37.8 = 699.3 * 10^{-9} C[/tex]

separation between plates d  is given as[tex] = \frac{A\epsilon _{0}}{C }[/tex]

[tex]d =  \frac{272.25 *10^{-4} *8.85*10^{-12}}{699.3 *10^-9}[/tex]

d = 3.44 *10^{-7} m  

Answer:

a) [tex]y=16.53m[/tex]

b) [tex]t_{up}=1.83s[/tex]

c)[tex]t_{down}=1.84s[/tex]

d) [tex]v=-18m/s[/tex]

Explanation:

a) To find the highest point of the ball we need to know that at that point the ball stops going up and its velocity become 0

[tex]v^{2} =v^{2} _{o} +2g(y-y_{o})[/tex]

[tex]0=(18)^{2} -2(9.8)(y-0)[/tex]

Solving for y

[tex]y=\frac{(18)^{2} }{2(9.8)}=16.53m[/tex]

b) To find how long does it take to reach that point:

[tex]v=v_{o}+at[/tex]

[tex]0=18-9.8t[/tex]

Solving for t

[tex]t_{up} =\frac{18m/s}{9.8m/s^{2} }= 1.83s[/tex]

c) To find how long does it take to hit the ground after it reaches its highest point we need to find how long does it take to do the whole motion and then subtract the time that takes to go up

[tex]y=y_{o}+v_{o}t+\frac{1}{2}gt^{2}[/tex]

[tex]0=0+18t-\frac{1}{2}(9.8)t^{2}[/tex]

Solving for t

[tex]t=0 s[/tex] or [tex]t=3.67s[/tex]

Since time can not be negative, we choose the second option

[tex]t_{down}=t-t_{up}=3.67s-1.83s=1.84s[/tex]

d) To find the velocity when it returns to the level from which it started we need to use the following formula:

[tex]v=v_{o}+at[/tex]

[tex]v=18m/s-(9.8m/s^{2} )(3.67s)=-18m/s[/tex]

The sign means the ball is going down