Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observer on the distant star is 950 nm. What is the velocity of the star relative to you (in units of the speed of light c), and is it moving towards or away from you?

Answer:

f=171.43Hz

Explanation:

Wave frequency is the number of waves that pass a fixed point in a given amount of time.

The frequency formula is: f=v÷λ, where v is the velocity and λ is the wavelength.

Then replacing with the data of the problem,

f=[tex]\frac{120\frac{m}{s} }{0.7m}[/tex]

f=171.43[tex]\frac{1}{s}[/tex]

f=171.43 Hz (because [tex]\frac{1}{s} = Hz[/tex], 1 hertz equals 1 wave passing a fixed point in 1 second).

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle [tex]\angle \mathrm{B\hat{C}D}[/tex] which corresponds to this minor arc.

This angle comes can be split into two parts:

[tex]\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}[/tex].

Also,

[tex]\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}[/tex].

The radius of this circle is:

[tex]\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}[/tex].

The lengths of segment DC, AC, BC can all be found:

In the two right triangles [tex]\triangle\mathrm{DAC}[/tex] and [tex]\triangle \rm BAC[/tex], the value of [tex]\angle \mathrm{B\hat{C}A}[/tex] and [tex]\angle \mathrm{A\hat{C}D}[/tex] can be found using the inverse cosine function:

[tex]\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}} [/tex]

[tex]\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}} [/tex]

[tex]\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}[/tex].

The length of the minor arc will be:

[tex]\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km[/tex].

Answer:

(A) Yes, the driver hits the deer with a speed of 14.66 m/s.

(B) 22.21 m/s

Explanation:

Assume:

Part (A):

Before the brakes were applied, the driver moves with a constant velocity of 25 m/s for 1.5 s.

Let us find out distance traveled travel by the driver in this time interval.

[tex]x = 25\times 1.5 = 37.5\ m[/tex]

After the brakes were applied, we have

[tex]u = 25\ m/s\\v = 0\ m/s\\a = -10\ m/s^2\\[/tex]

Using the formula for constant acceleration motion, we have

[tex]v^2=u^2+2as\\\Rightarrow s = \dfrac{v^2-u^2}{2a}\\\Rightarrow s = \dfrac{(0)^2-(25)^2}{2\times (-10)}\\\Rightarrow s = 31.25\ m[/tex]

This means the driver moves a distance of 31.25 m to stop from the time he applies brakes.

So, the total distance traveled by the driver in the entire journey = 37.5 m + 31.25 m = 68.75 m.

Since, the distance traveled by the driver (68.75 m) is greater than the distance of the deer from the driver (58.0 m). So, the driver hits the deer before stopping.

For calculating the speed of the driver with which the driver hits the deer, we have to calculate the actual distance to be traveled by the driver from the instant he applies brake so that the deer does not get hit. This distance is given by:

s = 58 m - 37.5 m = 20.5 m

Now, again using the equation for constant acceleration, we have

[tex]v^2=u^2+2as\\\Rightarrow v^2=(25)^2+2(-10)(20.5)\\\Rightarrow v^2=625-410\\\Rightarrow v^2=215\\\Rightarrow v=\pm \sqrt{215}\\\Rightarrow v=\pm 14.66\ m/s\\[/tex]

Since the driver hits the deer, this means the speed must be positive.

[tex]\therefore v = 14.66 m/s[/tex]

Hence, the driver hits the deer at a speed of 14.66 m/s.

Part (B):

Let the maximum velocity of the driver so that it does not hit the deer be u.

Then,

[tex]x = u\times 1.5 = 1.5u[/tex]

After travelling this distance, the driver applies brakes till it just reaches the position of the deer and does not hit her.

This much distance is S (let).

Using the equation of constant acceleration, we have

[tex]v^2=u^2+2aS\\\Rightarrow (0)^2=u^2+2(-10)(58-x)\\\Rightarrow 0=u^2-1160+30u\\\Rightarrow u^2+30u-1160=0\\[/tex]

On solving the above quadratic equation, we have

[tex]u = 22.21\ m/s\,\,\, or u = -52.21\ m/s[/tex]

But, this speed can not be negative.

So, u = 22.21 m/s.

Hence, the driver could have traveled with a maximum speed of 22.21 m/s so that he does not hit the deer.

Answer:

40.5 W/m²

Explanation:

Intensity of unpolarized light = 432 W/m² = I₀

Intensity of light as it passes through first polarizer

I = I₀×0.5

⇒I = 432×0.5

⇒I = 216 W/m²

Intensity of light as it passes through second polarizer. So, Intensity after it passes through first polarizer is the input for the second polarizer

I = I₀×0.5 cos²30

⇒I = 216×0.75

⇒I = 162 W/m²

Intensity of light as it passes through third polarizer.

I = I₀×0.5 cos²30×cos²60

⇒I = 162×0.25

⇒I = 40.5 W/m²

Intensity of light at the end is 40.5 W/m²

The electric field strength between the disks can be found using the formula E = (k × q₁ × q₂) / (q × r²), where k is the Coulomb constant, q₁ and q₂ are the charges on the disks, and r is the distance between them. To calculate the launch speed of the proton, we can use the conservation of energy principle and equate the potential energy gained by the proton with its kinetic energy.

Part A:

To find the electric field strength between the disks, we can use the formula:

Electric field strength (E) = Electric force (F) / Charge (q)

The electric force between the disks can be calculated using the formula:

Electric force (F) = (k × q₁ × q₂) / r₂

where k is the Coulomb constant, q₁ and q₂ are the charges on the disks, and r is the distance between them.

Substituting the given values, we get:

E = (k × q₁ × q₂) / (q × r₂)

Now, we can calculate the electric field strength.

Part B:

To calculate the launch speed of the proton, we can use the conservation of energy principle. The potential energy gained by the proton when moved from negative disk to positive disk can be equated to its kinetic energy.

Potential energy (PE) = Kinetic energy (KE)

PE = qV (where V is the potential difference between the disks)

KE = (1/2)mv2 (where m is the mass of the proton, v is the launch speed)

Equating the two equations, we get:

qV = (1/2)mv2

Now, we can solve for the launch speed.

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The electric field strength between the disks is around 79402.08N/C. The proton must be launched at a speed of roughly 2.86*10^6 m/s to just barely reach the positive disk.

This problem involves concepts from electrodynamics and quantum mechanics. To answer Part A, we can use the formula E=σ/ε0 where σ is the surface charge density and ε0 is the permittivity of free space. The surface charge density σ=Q/A where Q is charge and A is the area of the disk, A = π*(d/2)^2. Plugging this in, we get E=~79402.08 N/C.

For Part B, the energy that a proton must have to reach the other plate is equal to the work done by the electric field in bringing the proton from the negative to the positive plate, which is W=qE*d, where q is the charge of the proton and d is the distance between the plates. Equating this to kinetic energy,1/2mv^2, we get v = sqrt((2*q*E*d)/m). Substituting appropriate values gives us v=~2.86*10^6 m/s.

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Answer:

Work done, W = 0.0219 J

Explanation:

Given that,

Force constant of the spring, k = 290 N/m

Compression in the spring, x = 12.3 mm = 0.0123 m

We need to find the work done to compress a spring. The work done in this way is given by :

[tex]W=\dfrac{1}{2}kx^2[/tex]

[tex]W=\dfrac{1}{2}\times 290\times (0.0123)^2[/tex]

W = 0.0219 J

So, the work done by the spring is 0.0219 joules. Hence, this is the required solution.

Final answer:

The correct answer is b) 1.78 J, which is calculated using the formula W = 1/2kx² for the work done on a spring.

Explanation:

The work done to compress a spring is given by the formula W = 1/2kx², where k is the spring constant and x is the displacement in meters. For a spring with a force constant of 290.0 N/m compressed by 12.3 mm (which should be converted to meters by dividing by 1000, thus 0.0123 m), the work done can be calculated as follows:

W = 1/2 × 290.0 N/m × (0.0123 m)² = 1/2 × 290.0 × 0.00015129 = 1.78 J.

Therefore, the correct answer is b) 1.78 J.

Answer:

Maximum speed, v = 36 m/s                                                          

Explanation:

Given that,

The radius of the curved road, r = 120 m

Road is at an angle of 48 degrees. We need to find the maximum speed of stay on the curve in the absence of friction. On a banked curve, the angle at which it is cant is given by :

[tex]tan\theta=\dfrac{v^2}{rg}[/tex]

g is the acceleration due to gravity  

[tex]v=\sqrt{rg\ tan\theta}[/tex]

[tex]v=\sqrt{120\times 9.8\times \ tan(48)}[/tex]

v = 36.13 m/s

or

v = 36 m/s

So, the maximum speed to stay on the curve in the absence of friction is 36 m/s. Hence, this is the required solution.

Answer:

3. v(t) = (0.05 m/s 3 )t 2

Explanation:

Instantaneous acceleration  = 0.1 t

dv /dt = 0.1 t

dv = 0.1 t dt

Integrating on both sides

v(t)  = ( 0.1 /2) t²

= .05 t²

Answer:

Acceleration, [tex]a=-31.29\ m/s^2[/tex]

Explanation:

It is given that,

Initial speed of the aircraft, u = 140 mi/h = 62.58 m/s

Finally, it stops, v = 0

Time taken, t = 2 s

Let a is the acceleration of the aircraft. We know that the rate of change of velocity is called acceleration of the object. It is given by :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{0-62.58}{2}[/tex]

[tex]a=-31.29\ m/s^2[/tex]

So, the acceleration of the aircraft is [tex]31.29\ m/s^2[/tex] and the car is decelerating. Hence, this is the required solution.

Answer:

The speed of ocean currents approximately is 0.4667 m/s

Solution:

As per the question:

The ocean currents water moves at 30 W near Africa and 70 W near Puerto Rico

Time taken, t = 1 year = [tex]365\times 24\times 3600 = 3.15\times 10^{7} s[/tex]

The distance between Puerto Rico and Africa, x = [tex]1.46\times 10^{7} m[/tex]

Thus

The speed of the ocean currents there:

speed, [tex]v = \frac{d}{t} = \frac{1.47\times 10^{7}}{3.15\times 10^{7}}[/tex]

[tex]s = 0.4667 m/s[/tex]

Answer:

A) 3.11 seconds

B) 82.33 m

C)  5.32 seconds

D) 25.2 m/s

Explanation:

t = Time taken

u = Initial velocity = 12 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

[tex]v=u+at\\\Rightarrow 0=12-9.8\times t\\\Rightarrow \frac{-12}{-9.81}=t\\\Rightarrow t=1.22 \s[/tex]

Time taken to reach maximum height is 1.22 seconds

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=12\times 1.22+\frac{1}{2}\times -9.81\times 1.22^2\\\Rightarrow s=7.33\ m[/tex]

So, the stone would travel 7.33 m up

B) So, total height stone would fall is 7.33+75 = 82.33 m

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 82.33=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{82.33\times 2}{9.81}}\\\Rightarrow t=4.1\ s[/tex]

A) Total time taken by the stone to reach the ground from the time it left the person's hand is 4.1+1.22 = 5.32 seconds.

C) When s = 82.33-50 = 33.33 m

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 32.33=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{32.33\times 2}{9.81}}\\\Rightarrow t=2.57\ s[/tex]

The time it takes from the person's hand throwing the ball to the distance of 50 m from the ground is 2.57+1.22 = 3.777 seconds

D)

[tex]v=u+at\\\Rightarrow v=0+9.81\times 2.57\\\Rightarrow v=25.2\ m/s[/tex]

The stone is moving at 25.2 m/s when it is 50m from the bottom of the building

The iron vat will be 0.01056 meters (or 10.56 millimeters) longer when it contains boiling water at 1 atm pressure.

Given data:

To calculate the change in length of the iron vat when it is heated from room temperature (20°C) to boiling water temperature, we can use the linear thermal expansion formula:

ΔL = α * L * ΔT

Where:

ΔL = Change in length

α = Coefficient of linear expansion of iron (given)

L = Original length of the vat

ΔT = Change in temperature

Given:

Original length, L = 11 m

Coefficient of linear expansion of iron, α = 12 x 10^-6 /°C (approximate value for iron)

Change in temperature, ΔT = Boiling point of water (100°C) - Room temperature (20°C) = 80°C

Now, plug in the values and calculate:

[tex]\Delta L = (12 * 10^{-6} ) * (11 m) * (80 ^\circ C)[/tex]

ΔL ≈ 0.01056 m

Hence, the iron vat will be 0.01056 meters longer.

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The iron vat will be approximately 1.056 cm longer when it contains boiling water at 1 atm pressure compared to at room temperature due to linear thermal expansion. This is calculated using the relevant formula and the coefficients for iron.

The subject of this question is linear thermal expansion, related to physics. When the iron vat is heated by boiling water, it expands. The length of the expansion can be calculated using the formula ΔL = αLΔT, where ΔL is the change in length, α is the coefficient of linear expansion for iron (12 ×10^-6/°C), L is the original length, and ΔT is the change in temperature. Considering the original length L as 11 m and ΔT as the difference between 100°C (boiling point of water at 1 atm pressure) and 20°C (room temperature), which is 80°C. So, plug in these values we have ΔL = 12 ×10^-6/°C * 11m * 80°C = 0.01056 m or about 1.056 cm. Therefore, the iron vat is about 1.056 cm longer when it contains boiling water at 1 atm pressure compared to at room temperature.

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The electric potential at the center (C) of the circle formed by the rod is zero due to the symmetry and cancellations of charges. Calculating the potential at point P requires a detailed approach considering the distance D and the charge distribution.

The question involves calculating the electric potential at two different points with reference to a specially configured plastic rod. For part (a), the total charge along the circumference is given by the sum of Q1 and Q2, where Q2 = -6Q1. However, due to the uniform distribution and the symmetry of the setup, the electric potential at the center of the circle (point C) would effectively be zero because the contributions from the positively and negatively charged segments cancel each other out.

For part (b), calculating the electric potential at point P, which lies on the axis and is a distance D from the center, would require the application of principles of superposition and the integration of electric potential contributions from each infinitesimal segment of the charged parts of the rod. Considering the arrangement and the charges' nature, this part of the solution involves more intricate calculations and requires knowing the specific distribution of the charges along the rod.

Answer:

Magnitude of velocity V = 6.015 m/sec

Y component of velocity = 3.61 m/sec

Explanation:

We have given magnitude of velocity in x direction, that is horizontal component [tex]V_X=3.8m/sec[/tex]

Angle with x axis = 37°

We know that horizontal component [tex]V_X=Vcos\Theta[/tex]

So [tex]4.8=Vcos37^{\circ}[/tex]

[tex]V=6.015m/sec[/tex]

Y component of velocity is given by [tex]V_Y=Vsin\Theta =6.015sin37^{\circ}=3.61m/sec[/tex]

Answer:

500 m

Explanation:

t = Time taken

u = Initial velocity = 50 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -2.5 m/s²

Equation of motion

[tex]v=u+at\\\Rightarrow 0=50-2.5\times t\\\Rightarrow \frac{-50}{-2.5}=t\\\Rightarrow t=20\ s[/tex]

Time taken by the train to stop is 20 seconds

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=50\times 20+\frac{1}{2}\times -2.5\times 20^2\\\Rightarrow s=500\ m[/tex]

∴ The engineer applied the brakes 500 m from the station

Answer:

[tex]E_{polarization} = 2.21 \times 10^3(-0.99\hat i -0.132\hat j)[/tex]

Explanation:

As we know that electric field inside any conducting shell is always zero

so we can say that

[tex]E_{charge} + E_{polarization} = 0[/tex]

here we know that

[tex]E_{charge} = \frac{kq}{r^2} \hat r[/tex]

here we know that

[tex]\hat r = \frac{(0 - -0.6)\hat i + (0.08 - 0)\hat j}{\sqrt{0.6^2 + 0.08^2}}[/tex]

[tex]\hat r = 0.99\hat i + 0.132\hat j[/tex]

now we will have

[tex]E_{polarization} = - E_{charge}[/tex]

[tex]E_{polarization} = - \frac{(9\times 10^9)(9 \times 10^{-8})}{0.6^2 + 0.08^2}(0.99\hat i + 0.132\hat j)[/tex]

[tex]E_{polarization} = 2.21 \times 10^3(-0.99\hat i -0.132\hat j)[/tex]

We have that the electric field contributed by the polarization charges on the surface of the metal sphere is

From the question we are told

Generally the equation for the electric field inside any conducting shell    is mathematically given as

E charge+E polarization=0

Therefore

[tex]E charge=\frac{kq}{r^2}r'\\\\E charge=\frac{0--0.6i+0.08-0j}{\sqrt{0.6^2+0.08^2}}\\\\r'=0.99i+0.132j[/tex]

Therefore

E polarization=-E charge

Therefore

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Answer:

4.56 seconds

63.25 m

Explanation:

t = Time taken for the car to stop

u = Initial velocity = 60 km/h = 100 km/h = 100000/3600 = 27.78 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = -6.1 m/s²

Time taken by the car to stop

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-27.78}{-6.1}\\\Rightarrow t=4.56\ s[/tex]

Time taken by the car to stop is 4.56 seconds

[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow s=27.78\times 4.56+\frac{1}{2}\times -6.1\times 4.56^2\\\Rightarrow s=63.25\ m[/tex]

The total stopping distance would be 63.25 m

Answer:

35 mph

Explanation:

The key of this problem lies in understanding the way that projectile motion works as we are told to neglect the height of the javelin thrower and wind resistance.

When the javelin is thown, its velocity will have two components: a x component and a y component. The only acceleration that will interact with the javelin after it was thown will be the gravety, which has a -y direction. This means that the x component of the velocity will remain constant, and only the y component will be affected, and can be described with the constant acceleration motion properties.

When an object that moves in constant acceleration motion, the time neccesary for it to desaccelerate from a velocity v to 0, will be the same to accelerate the object from 0 to v. And the distance that the object will travel in both desaceleration and acceleration will be exactly the same.

So, when the javelin its thrown, it willgo up until its velocity in the y component reaches 0. Then it will go down, and it will reach reach the ground in the same amount of time it took to go up and, therefore, with the same velocity.

Answer:

Explanation:

For an object to move with constant velocity, the acceleration of the object must be zero:

[tex]\vec{a} = \vec{0}[/tex].

As the net force equals acceleration multiplied by mass , this must mean:

[tex]\vec{F}_{net} = m \vec{a} = m * \vec{0} = \vec{0}[/tex].

So, the sum of the three forces must be zero:

[tex]\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{0}[/tex],

this implies:

[tex]\vec{F}_3  = - \vec{F}_1 - \vec{F}_2[/tex].

To obtain this sum, its easier to work in Cartesian representation.

First we need to define an Frame of reference. Lets put the x axis unit vector [tex]\hat{i}[/tex] pointing east,  with the y axis unit vector [tex]\hat{j}[/tex] pointing south, so the positive angle is south of east. For this, we got for the first force:

[tex]\vec{F}_1 = 83.7 \ N \ (-\hat{j})[/tex],

as is pointing north, and for the second force:

[tex]\vec{F}_2 = 59.9 \ N \ (-\hat{i})[/tex],

as is pointing west.

Now, our third force will be:

[tex]\vec{F}_3  = - 83.7 \ N \ (-\hat{j}) - 59.9 \ N \ (-\hat{i})[/tex]

[tex]\vec{F}_3  =  83.7 \ N \ \hat{j}  + 59.9 \ N \ \hat{i}[/tex]

[tex]\vec{F}_3  =  (59.9 \ N , 83.7 \ N ) [/tex]

But, we need the magnitude and the direction.

To find the magnitude, we can use the Pythagorean theorem.

[tex]|\vec{R}| = \sqrt{R_x^2 + R_y^2}[/tex]

[tex]|\vec{F}_3| = \sqrt{(59.9 \ N)^2 + (83.7 \ N)^2}[/tex]

[tex]|\vec{F}_3| = 102.92 \ N[/tex]

this is the magnitude.

To find the direction, we can use:

[tex]\theta = arctan(\frac{F_{3_y}}{F_{3_x}})[/tex]

[tex]\theta = arctan(\frac{83.7 \ N }{ 59.9 \ N })[/tex]

[tex]\theta = 57 \° 24 ' 48''[/tex]

and this is the angle south of east.

Answer:

Explanation:

mass of block, m = 20 kg

let teh angle of inclination of the ramp with the horizontal is θ.

Weight of the block = mg

There are two components of the weight

The component of weight of block along the plane = mg Sinθ

The component of weight of block normal to the plane = mg Cosθ

Friction force is acting opposite to the motion, i.e., along the plane and upwards, f = μN = μ mg Cosθ

Look at the diagram to find the directions of force.

Answer:

A.) t₁= 4.37 s

B.) d₁= 19.09 m

C.) [tex]v_{f1} = 8,74 \frac{m}{s}[/tex]

Explanation:

Bicyclist kinematics :

Bicyclist moves with uniformly accelerated movement:

d₁ = v₀₁*t₁ + (1/2)a₁*t₁²  equation (1)

d₁ : distance traveled by the cyclist

v₀₁: initial speed (m/s)

a₁: acceleration  (m/s²)

t₁:  time it takes the cyclist to catch his friend (s)

Friend  kinematics:

Friend moves with constant speed:

d ₂= v₂*t₂ Equation (2)

d₂ : distance traveled by the friend (m)

v₂: friend speed (m/s)

t₂ : time that has elapsed since the cyclist meets the friend until the cyclist catches him.

Data

v₀₁=0

a₁ = 2.0 m/s²

v₂ = 3.0 m/s

Problem development

A.) When the bicyclist catches his friend, d₁ = d₂=d  and t₂=t₁+2

in the equation (1) :

d = 0 + (1/2)2*t₁²  = t₁²

d = t₁²  equation (3)

in the equation (2) :

d = 3*(t₁+2) ₂ = 3*t₁+6

d =  3t₁+6 equation (4)

Equation (3) = Equation (4)

                t₁²  =  3t₁+6  

t₁² - 3t₁ - 6  = 0 ,we solve the quadratic equation

t₁= 4.37 s : time it takes the cyclist to catch his friend

B.) d₁ : distance traveled by the cyclist

In the Equation (2) d₁= (1/2)2* 4.37² = 19.09 m

C.)  speed of the cyclist when he catches his friend

[tex]v_{f1} = v_{o1} + a*t[/tex]

[tex]v_{f1} =0 + 2* 4.37 [/tex]

[tex]v_{f1} = 8.74 \frac{m}{s}[/tex]

Answer:

(A). The location of the charge is 26.45 cm.

(B). The magnitude of the charge is 51.1 pC.

Explanation:

Given that,

Distance in x axis = 5.00 cm

Electric field = 10.0 N/C

Distance in x axis = 10.0 cm

Electric field = 17.0 N/C

Since, q is the same charge, two formulas can be set equal using the two different electric fields.

(a). We need to calculate the location of the charge

Using formula of force

[tex]F = qE[/tex]....(I)

Using formula of electric force

[tex]F =\dfrac{kq^2}{d^2}[/tex]....(II)

From equation (I) and (II)

[tex]qE=\dfrac{kq^2}{d^2}[/tex]

[tex]E=\dfrac{kq}{d^2}[/tex]

[tex]q=\dfrac{E(x-r)^2}{k}[/tex]...(III)

For both points,

[tex]\dfrac{E(x-r)^2}{k}=\dfrac{E(x-r)^2}{k}[/tex]

Put the value into the formula

[tex]\dfrac{10.0\times(x-5.00)^2}{k}=\dfrac{17.0\times(x-10.0)^2}{k}[/tex]

[tex]10.0\times(x-5.0)^2=17.0\times(x-10.0)^2[/tex]

Take the square root of both sides

[tex]3.162(x-5.0)=4.123(x-10.0)[/tex]

[tex]3.162x-3.162\times5.0=4.123x-4.123\times10.0[/tex]

[tex]3.162x-4.123x=-4.123\times10.0+3.162\times5.0[/tex]

[tex]0.961x=25.42[/tex]

[tex]x=\dfrac{25.42}{0.961}[/tex]

[tex]x=26.45\ cm[/tex]

(B). We need to calculate the charge

Using equation (III)

[tex]q=\dfrac{E(x-r)^2}{k}[/tex]

Put the value into the formula

[tex]q=\dfrac{10.0(26.45\times10^{-2}-5.00\times10^{-2})^2}{9\times10^{9}}[/tex]

[tex]q=5.11\times10^{-11}\ C[/tex]

[tex]q=51.1\ pC[/tex]

Hence, (A). The location of the charge is 26.45 cm.

(B). The magnitude of the charge is 51.1 pC.

The charge is located at x = 2.25 cm along the x-axis and has a magnitude of approximately 8.41 x 10⁻¹² C. These were determined using the relationships of electric field magnitude and distance from the point charge. The analysis involved solving the electric field equations at different points based on Coulomb's law.

The electric field due to a point charge follows the equation:

E = k * |q| / 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 charge.

Part A: Finding the Charge's Location

At x = 5.00 cm, E1 = 10.0 N/C and at x = 10.0 cm, E2 = 17.0 N/C.

Assume the charge q is located at x = x0.

For x = 5.00 cm, the distance r1 is |x0 - 5.00 cm|.

For x = 10.0 cm, the distance r2 is |x0 - 10.0 cm|.

Using the electric field equations:
10.0 = k * |q| / (|x0 - 5.00|²)
17.0 = k * |q| / (|x0 - 10.0|²)

Divide the second equation by the first to eliminate q:

(17.0 / 10.0) = (|x0 - 5.00|² / |x0 - 10.0|²)

Solving this, we get |x0 - 5.00|² = 1.7 * |x0 - 10.0|².

Let u = x0 - 5.00 and v = x0 - 10.0, hence v = u - 5.00.

Substituting and solving gives the location x0 = 2.25 cm.

Part B: Finding the Magnitude of the Charge

Use the equation E = k * |q| / r² with one of the electric field values from Part A.

Let's use E1 = 10.0 N/C and r1 = |2.25 cm - 5.00 cm| = 2.75 cm = 0.0275 m:

10.0 = (8.99 x 10⁹ N*m²/C²) * |q| / (0.0275 m)².

Solving for q, we get q ≈ 8.41 x 10⁻¹² C.

Thus, the charge is located at 2.25 cm along the x-axis with a magnitude of 8.41 x 10⁻¹² C.

Answer:

They all hit at the same time

Explanation:

Let the time of flight is T.

The maximum height is H and the horizontal range is R.

The formula for the time of flight is

[tex]T=\frac{2uSin\theta }{g}[/tex] ..... (1)

Te formula for the maximum height is

[tex]H=\frac{u^{2}Sin^{2}\theta }{2g}[/tex]    .... (2)

From equation (1) and (2), we get

[tex]\frac{T^{2}}{H}=\frac{\frac{4u^{2}Sin^{2}\theta }{g^{2}}}{\frac{u^{2}Sin^{2}\theta }{2g}}[/tex]    

[tex]\frac{T^{2}}{H}=\frac{8}{g}[/tex]

[tex]T=\sqrt{\frac{8H}{g}}[/tex]

here, we observe that the time of flight depends on the maximum height and according to the question, the maximum height for all the three balls is same so the time of flight of all the three balls is also same.

Answer:

The electric force on the proton is 8.2x10^-10 N

Explanation:

We use the formula to calculate the distance between two points, as follows:

r = ((x2-x1)^2 + (y2-y1)^2)^1/2, where x1 and x2 are the x coordinate, y2, y1 are the y coordinate. replacing values:

r = ((0.36-0)^2 + (0.39-0)^2)^1/2 = 0.53 nm = 5.3x10^-10 m

We will use the following expression to calculate the electrostatic force:

F = (q1*q2)/(4*pi*eo*r^2)

Here we have:

q1 = q2 = 1.6x10^-19 C, 1/4*pi*eo = 9x10^9 Nm^2C^-2

Replacing values:

F = (1.6x10^-19*1.6x10^-9*9x10^9)/((5.3x10^-10)^2) = 8.2x10^-10 N

The electric force on a proton can be calculated using Coulomb's Law and the charges and distance between the proton and electron. The charge of both particles is ±1.602 × 10^{-19} C, and the distance is calculated from their coordinates on the x and y-axis.

To find the electric force on the proton, we can use Coulomb's Law. Coulomb's Law states that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is:
F = k * |q1 * q2| / r2,

where F is the force, k is Coulomb's constant (approximately 8.9875 × 109 N m2/C2), q1 and q2 are the charges of the particles, and r is the distance between them.

The charge of a proton (q1) and an electron (q2) is approximately ±1.602 × 10−16 C.

The distance r can be found using the Pythagorean theorem, since we have the x and y coordinates:

r = √(x2 + y2) = √(0.362 + 0.392).

After calculating the distance r, we can plug all the values into the Coulomb's Law formula to get the magnitude of the force.

Answer:

Answer:

77.58°

Explanation:

Let the projectile is projected at an angle θ and the velocity of projection is u.

At the maximum height, the projectile has only horizontal component of velocity.

Let the velocity at maximum height is v.

According to the question, the velocity at maximum height is 21.5% of initial velocity.

v = 21.5 % of u

u Cos θ = 21.5 % of u

u Cos θ = 0.215 u

Cos θ = 0.215

θ = 77.58°

Thus, the angle of projection is  77.58°.

Explanation:

Answer:

minimum amount of energy required is 8.673 MJ

Explanation:

given data

mass m1 = 80 kg

additional mass m2 = 20 kg

height = 8850 m

to find out

minimum amount of energy required

solution

we know here total mass is m = m1 + m2

m = 80 + 20

m = 100 kg

so now apply energy equation for climb high that is

energy = m×g×h   ....................1

here m is mass and g is 9.8 and h is height

put here all value in equation 1

energy = m×g×h

energy = 100 ×9.8×8850

energy = 8673000 J

so minimum amount of energy required is 8.673 MJ

Final answer:

To climb Mt. Everest at 8850 m high, an 80-kg climber carrying a 20-kg pack requires a minimum of 867,300 Joules (867.3 kJ) of energy, based on the gravitational potential energy calculation.

Explanation:

The question relates to calculating the minimum amount of energy required for an 80-kg climber carrying a 20-kg pack to climb Mt. Everest, 8850 m high. To calculate the energy, we use the formula for gravitational potential energy, which is Potential Energy (PE) = mass (m) × gravity (g) × height (h). The mass of the climber and the pack combined is 100 kg (80 + 20), the acceleration due to gravity (g) is approximately 9.8 m/s2, and the height (h) is 8850 m. Therefore, the required energy can be calculated as PE = 100 kg × 9.8 m/s2 × 8850 m.

The calculation gives us PE = 867,300 Joules or 867.3 kJ. This is the minimum amount of energy required for the climber and his pack to reach the summit of Mt. Everest, assuming no energy is lost to friction or other forces.

Answer: a) 278 * 10^-12 F and 19.4 * 10^-9 C

b) 17.44 * 10^3 N/C and c) C=k*C0 and V=70/k

Explanation: In order to solve this problem we have to use the expression of the capacitor of parallel plates as:

C=A*ε0/d where A is the area of the plates and d the distance between them

C=Π r^2*ε0/d

C=π*0.2^2*8.85*10^-12/0.004=278 * 10^-12F= 278 pF

then

ΔV= Q/C

so Q= ΔV*C=70V*278 pF=19.4 nC

The electric field between the plates is given by:

E= Q/(A*ε0)=19.4 nC/(π*0.2^2*8.85*10^-12)=17.44 *10^3 N/C

If it is introduced a dielectric between the plates, then the new C is increased a factor k while the potential between the plates decreases a factor 1/k.

Answer:

4.555 x 10^-11 m

Explanation:

Potential difference, V = 27.3 kV

Let the wavelength is λ.

The energy associated with the potential difference, E = 27.3 keV

E = 27.3 x 1000 x 1.6 x 10^-19 J = 4.368 x 10^-15 J

The energy associated with the wavelength is given by

[tex]E=\frac{hc}{\lambda }[/tex]

Where, h is Plank's constant = 6.63 x 10^-34 Js

c is velocity of light  3 x 10^8 m/ s

By substituting the values, we get

[tex]4.368\times10^{-15}=\frac{6.634\times10^{-34}\times 3\times 10^{8}}{\lambda }[/tex]

λ = 4.555 x 10^-11 m

Answer:

a) 3.65 seconds

b) 35.87 m/s

Explanation:

s = Displacement = 65.6 m

u = Initial velocity

v = Final velocity

t = Time taken

a = Acceleration due to gravity = 9.81 m/s² (downward direction is taken as positive and upward is taken as negative)

b) Equation of motion

[tex]v^2-u^2=2as\\\Rightarrow 0^2-u^2=2\times -9.81\times 65.6\\\Rightarrow u=\sqrt{2\times 9.81\times 65.6}\\\Rightarrow u=35.87\ m/s[/tex]

Initial pop up velocity is 35.87 m/s

a)

[tex]v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-35.87}{-9.81}\\\Rightarrow t=3.65\ s[/tex]

It took 3.65 seconds to reach this height

Answer:

Electric force, [tex]F=1.29\times 10^6\ N[/tex]

Explanation:

Given that,

Charge 1, [tex]q_1=24\ C[/tex]

Charge 2, [tex]q_2=-24\ C[/tex]

Distance between charges, d = 2 km

The electric force is given by :

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

k is the electrostatic constant

[tex]F=8.98755\times 10^9\times \dfrac{(24)^2}{(2\times 10^3)^2}[/tex]

F = 1294207.2 N

or

[tex]F=1.29\times 10^6\ N[/tex]

Hence, this is the required solution.