You drop a ball from a window on an upper floor of a building and it is caught by a friend on the ground when the ball is moving with speed vf. You now repeat the drop, but you have a friend on the street below throw another ball upward at speed vf exactly at the same time that you drop your ball from the window. The two balls are initially separated by 41.1 m. (a) At what time do they pass each other? s (b) At what location do they pass each other relative to the window?

Answer:

The gravitational force between them is [tex]1.079\times10^{-7}\ N[/tex].

Explanation:

Given that,

Distance = 1.50 m

Mass of one student = 70.0 kg

Mass of other student = 52.0 kg

We need to calculate the gravitational force

Using formula of gravitational force

[tex]F=\dfrac{Gm_{1}m_{2}}{r^2}[/tex]

Where, m₁ = mass of one student

m₂ = mass of other studen

r = distance between them

Put the value into the formula

[tex]F=\dfrac{6.67\times10^{-11}\times70.0\times52.0}{1.50^2}[/tex]

[tex]F=1.079\times10^{-7}\ N[/tex]

Hence, The gravitational force between them is [tex]1.079\times10^{-7}\ N[/tex].

To calculate the gravitational force between the two students, we use Newton's law of universal gravitation, substituting the given values for mass and distance into the formula. The result suggests that the gravitational force would be incredibly small, aligning with our daily experiences.

The subject of this question is Physics, specifically gravitational force. From Newton's law of universal gravitation, we know that the gravitational force between two masses is given by the equation F = G(M₁M₂)/r², where F is the gravitational force, G is the gravitational constant, M₁ and M₂ are the two masses, and r is the distance between them.

Given that one student has a mass of 70 kg (M₁), the other a mass of 52 kg (M₂) and the distance between them is 1.5 m (r), we can substitute these values into the formula. Using a gravitational constant (G) of approximately 6.67 × 10-¹¹ Nm²/kg², the gravitational force (F) becomes:

F = (6.67 × 10-¹¹ Nm²/kg²)(70 kg)(52 kg)/(1.50 m)²

Note, though, that the gravitational force between two people sitting 1.50 m apart would be incredibly small due to the immense smallness of the gravitational constant. This is inline with our daily experiences where we don't feel any noticeable gravitational pull from an ordinary object.

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Answer:

(a) - 42700 m/s

(b) - 6.8 x 10^-4 m/s^2

Explanation:

initial velocity of star, u = 20.7 km/s

Final velocity of star, v = - 22 km/s

time, t = 1.99 years

Convert velocities into m/s and time into second

So, u = 20700 m / s

v = - 22000 m/s

t = 1.99 x 365.25 x 24 x 3600 = 62799624 second

(a) Change in planet's velocity = final velocity - initial velocity

  = - 22000 - 20700 = - 42700 m/s

(b) Accelerate is defined as the rate of change of velocity.

Acceleration = change in velocity / time

                     = ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2

Answer:

4.427 m

Explanation:

We shall consider east as + x- axes and north as + ve y- axes. .

We shall represent every displacement in vector form as follows

D₁ = 5.2 km due west = - 5.2 i

D₂ = 2.1 km due north = 2.1 j

D₃ = 3.7 km towards north east at 30 degree from east

= 3.7 cos30 i + 3.7 sin 30 j = 3.2 i + 1.85 j

Total displacement D = D₁ + D₂ +D₃ +D₄.

- 5.2 i + 2.1 j + 3.2 i + 1.85 j

= - 2 i + 3.95 j

Magnitude of D

D² = (2)² + (3.95)²

= 4 + 15.6025

D = 4.427 m

Answer:

a)[tex]E=2.88*10^{11}N/C[/tex]

b)[tex]E=1.44*10^{11}N/C[/tex]

c)[tex]F=4.61*10^{-8}N[/tex]

Explanation:

We use the definition of a electric field produced by a point charge:

[tex]E=k*q/r^2[/tex]

a)Electric Field  due to the alpha particle:

[tex]E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(10^{-10})^2=2.88*10^{11}N/C[/tex]

b)Electric Field  due to electron:

[tex]E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(10^{-10})^2=1.44*10^{11}N/C[/tex]

c)Electric Force on the alpha particle, on the electron:

The alpha particle and electron feel the same force but with opposite direction:

[tex]F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(10^{-10})^2=4.61*10^{-8}N[/tex]

The magnitude of your average velocity for the whole trip is 102.19 miles per hour.

First, let's calculate the time taken for each segment of the trip:

1. For the first 60 miles at 55 mi/h:

  Time = Distance / Speed

            = 60 miles / 55 mi/h

            = 1.0909 hours

2. During the traffic jam, you're not moving, so the time is 20 minutes, which needs to be converted to hours:

  Time = 20 minutes / 60 minutes/hour

            = 1/3 hours

3. For the last 40 miles at 75 mi/h:

  Time = 40 miles / 75 mi/h

           = 0.5333 hours

Now, calculate the total time for the trip:

Total Time = 1.0909 hours  + 1/3 hours + 0.5333 hours

                  = 1.9572 hours

Since you traveled 100 miles to your aunt's house and then returned,

the total displacement = 2 * 100 miles = 200 miles.

Now, calculate the average velocity:

Average Velocity = Total Displacement / Total Time

                            = 200 miles / 1.9572 hours

                             = 102.19 mi/h

So, the average velocity is 102.19 miles per hour.

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If you and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The magnitude of your average velocity for the whole trip is 51.04 miles per hour.

To determine the magnitude of your average velocity for the whole trip, we can use the formula for average velocity:

Average Velocity = Total Displacement / Total Time

Time for the first 60 miles at 55 mi/h:

Time = Distance / Speed

= 60 miles / 55 mi/h

= 1.0909 hours

Time spent in the traffic jam:

20 minutes = 20 / 60

= 1/3 hours

Time for the remaining 40 miles at 75 mi/h:

Time = Distance / Speed

= 40 miles / 75 mi/h

= 0.5333 hours

Total Time = 1.0909 + 1/3 + 0.5333

= 1.9572 hours

Now we can calculate the average velocity:

Average Velocity = Total Displacement / Total Time

Average Velocity = 100 miles / 1.9572 hours

≈ 51.04 mi/h

Therefore  the magnitude of your average velocity for the whole trip is 51.04 miles per hour.

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Answer:

5953.42 J

Explanation:

Given:

Initial volume, [tex]V_i[/tex]= 100 in³

Final Volume, [tex]V_f[/tex] = 10 in³

Initial pressure = 50 psia

Temperature = 100° F = 310.93 K

For isothermal reversible process, work done is given as:

Work done = [tex]-2.303RTlog_{10}\frac{V_f}{V_i}[/tex]

Where,

R is the ideal gas constant = 8.314 J/mol.K

or

Work done = [tex]-2.303\times8.314\times310.93log_{10}\frac{10}{100}[/tex]

or

Work done = 5953.42 J

Answer:

The greatest height reached by the ball is 0.8 m.

Explanation:

Given that,

Initial velocity = 4 m/s

We need to calculate the greatest height reached by the ball

Using equation of motion

[tex]v^2=u^2+2gh[/tex]

Where, v = final velocity

u = initial velocity

g = acceleration due to gravity

Put the value in the equation

[tex]0=4^2+2\times(-10)\times h[/tex]

[tex]h =\dfrac{16}{20}[/tex]

[tex]h =0.8\ m[/tex]

Hence, The greatest height reached by the ball is 0.8 m.

The ratio of the orbital periods (Ta/Tb) of two satellites, where the orbital speed of satellite A is twice that of satellite B, can be found using Kepler's third law, which relates the orbital period squared to the radius of the orbit cubed. By understanding the relationship between orbital speed and radius, the ratio of the orbital periods can be calculated.

When comparing two satellites, A and B, with orbital speeds such that the orbital speed of satellite A is twice that of satellite B, we are tasked with finding the ratio of their orbital periods (Ta/Tb). This problem is grounded in the principles of classical mechanics and specifically relates to Kepler's laws of planetary motion.

According to Kepler's third law, the square of the orbital period (T) is proportional to the cube of the radius of the orbit (r). Mathematically, this is expressed as T² ≈ r³ for a satellite orbiting a much larger body, such as the Earth. Since the gravitational attraction provides the necessary centripetal force for the satellite's circular motion, the gravitational force is also centrally involved in this relationship.

To compare the periods of two satellites, we use this proportionality. If the orbital speed of satellite A is twice that of satellite B, the radius of the orbit is also related to the speed. Specifically, speed is directly related to the square root of the radius of the orbit based on centripetal force considerations. Since the speed of satellite A is twice that of satellite B (VA = 2*VB), it follows that the radius of A's orbit would be four times that of B's orbity (rA = 4*rB). Applying Kepler's law then allows us to find the period ratio as (Ta/Tb)² = (rA/rB)³, and after substituting the relation for the radii, we can solve for (Ta/Tb).

Answer:

v = 8.96 m/s

Explanation:

Initial speed of the ball, u = 10 m/s

It caught 1 meter above its initial position.

Acceleration due to gravity, [tex]g=-9.8\ m/s^2[/tex]

We need to find the final speed of the ball when it is caught. Let is equal to v. To find the value of v, use third equation of motion as :

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

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

[tex]v^2=2(-9.8)\times 1+(10)^2[/tex]

v = 8.96 m/s

So, the speed of the ball when it is caught is 8.96 m/s. Hence, this is the required solution.

Answer:

8.96 m/s, upward direction

Explanation:

Given that, the initial velocity of the ball is,

[tex]u=10m/s[/tex]

And the acceleration in the downward direction is positive but in this situation the acceleration will be negative so,

[tex]a=9.8\frac{m}{s^{2} }[/tex]

And according to question vertical displacement is,

[tex]s=1m[/tex]

Now suppose v be the final velocity of the ball.

Applying third equation of motion,

[tex]v^{2}=u^{2}+2as[/tex]

Here, u is the initial velocity, a is the acceleration, s is the displacement.

Substitute all the variables.

[tex]v=\sqrt{10^{2}+2(-9.8)\times 1 } \\v=\sqrt{80.4}\\ v=8.96\frac{m}{s}[/tex]

Therefore, the speed of ball when it is caught is 8.96 m/s in the upward direction.

Answer:

In order to use the parallelogram method, we have to select first two vectors.  We move the vectors until their initial points coincide. Then we draw lines to form a complete parallelogram, as is shown in the figure annexed. The diagonal from the initial point to the opposite vertex of the parallelogram is the resultant.

We use the last vector resultant with a third vector, and we use again the parallelogram method to add them. We use the new resultant and we add it with a 4th vector. We repeat this task until all the vectors are used.

Answer:Due to the pressure difference created by rotating fans.

Explanation:

In most of the  vacuum cleaners, there is an area which is of disc shape and it is in right next to the motor. There are several fans within the disc that spin at a very high velocity.

The blades will push the air outside of the disk.There is no air in inside of the disc and  air pressure creates which  pushes air inside the disk to replace the missing air.

So motor is pushing the air outside and to maintain this pressure the air is pushing toward inside with dirt.

Answer:

The force exerted by the charge q  on the rod is [tex]5\times 10^{10}\ \rm N[/tex]

Explanation:

Given:

We have to find the force exerted by the charge on the rod. this will be equal to the force exerted by the rod on the charge according to coulombs law.

The Electric field due the the rod at the location of the charge is given by

[tex]E=\dfrac{kQL}{d(d+L}\\E=\dfrac{9\times10^9\times2\times2.5}{2\times4.5}\\\\E=5\times 10^9\ \rm N/C\\[/tex]

Force  between them is given by F

[tex]F=qE\\\\=10\times5\times10^9\\=5\times10^{10}\ \rm N[/tex]

Answer:

[tex]\lambda=1.23[/tex]Å

Explanation:

Bragg's Law refers to the simple equation:

[tex]n\lambda = 2d sin(\theta)[/tex]

In this case:

n=1

θ = 12.062°

d=2.9344 Å

[tex]\lambda = 2*2.9344sin(12.062)=1.23[/tex]Å

Answer:

[tex]\Delta s\ =\ 21.33\ J/K[/tex]

Explanation:

Given,

Let T be the final temperature of the mixture,

Therefore From the law of mixing, heat loss by the water is equal to the heat gained by the ice,

[tex]m_iL_f\ +\ m_is_w(T_f\ -\ 0)\ =\ m_ws_w(T_w\ -\ T_f)\\\Rightarrow 333000\times 0.0138\ +\ 0.0138\times 4190T_f\ =\ 0.134\times 4190\times(70.7\ -\ T_f)\\\Rightarrow 4595.4\ +\ 57.96T_f\ =\ 39789.96\ -\ 562.8T_f\\\Rightarrow 620.76T_f\ =\ 35194.56\\\Rightarrow T_f\ =\ \dfrac{35194.56}{620.76}\\\Rightarrow T_f\ =\ 56.69^o\ C[/tex]

Now, We know that,

Change in the entropy,

[tex]\Delta s\ =\ s_f\ -\ s_i\ =\ \dfrac{Q}{T}\\\Rightarrow \displaystyle\int_{s_i}^{s_f} ds\ =\ \displaystyle\int_{T_i}^{T_f}\dfrac{msdT}{T}\\\Rightarrow \Delta s =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)[/tex]

Now, change in entropy for the ice at 0^o\ C to convert into 0^o\ C water.

[tex]\Delta s_1\ =\ \dfrac{Q}{T}\\\Rightarrow \Delta s_1\ =\ \dfrac{m_iL_f}{T}\ =\ \dfrac{0.0138\times 333000}{273.15}\ =\ 16.82\ J/K.[/tex]

Change in entropy of the water converted from ice  from [tex]273.15\ K[/tex] to water 330.11 K water.

From the equation (1),

[tex]\therefore \Delta s_2\ =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\\\Rightarrow \Delta s_2\ =\ 0.0138\times 4190\times \ln \left (\dfrac{273.15}{330.11}\ \right )\\\Rightarrow \Delta s_2\ =\ 6.88\ J/K[/tex]

Change in entropy of the original water from the temperature 342.85 K to 330.11 K

From the equation (1),

[tex]\therefore \Delta s_3\ =\ ms \ln \left (\dfrac{T_i}{T_f}\ \right )\\\Rightarrow \Delta s_3\ =\ 0.134\times 4190\times \ln \left (\dfrac{330.11}{343.85} \right )\\\Rightarrow \Delta s_3\ =\ -2.36\ J/K[/tex]

Total entropy change = [tex]\Delta s\ =\ \Delta s_1\ +\ \Delta s_2\ +\ \Delta s_3\\\Rightarrow \Delta s\ =\ (16.82\ +\ 6.88\ -\ 2.36)\ J/K\\\Rightarrow \Delta s\ =\ 21.33\ J/K.[/tex]

Hence, the change in entropy of the system form then untill the system reaches the final temperature is 21.33 J/K

Answer:

D. Small group, whole group

Explanation:

Small group activities give the teacher the opportinity to deliver instruction on a more personal level than whole group activities because in a smaller group there is more time to work with individuals as opposed to have to dedicate all time to generalized lessons for the whole group.

"The correct answer is D. Small group, whole group. Small group activities give the teacher the opportunity to deliver instruction on a more personal level than whole group activities

When comparing the types of activities listed, it is clear that small group activities allow for a more personal level of instruction compared to whole group activities. Small group activities involve fewer students, which means that the teacher can focus more on the individual needs of each student, provide more targeted feedback, and facilitate more in-depth discussions. This setting is conducive to addressing specific questions, adapting the pace of instruction to the group's needs, and fostering a sense of community among the students.

On the other hand, whole group activities involve the entire class and are less personal by nature. In this setting, the teacher must address a larger audience, which can make it challenging to meet every student's individual needs. Instruction is typically more general and may not be as tailored to each student's learning style or pace.

To contrast, independent activities do not involve direct instruction from the teacher, so they do not provide the opportunity for personal-level instruction. Kinesthetic activities involve movement and can be done individually or in groups, but they are not inherently more personal than small group activities. Therefore, the comparison between independent and holistic activities (A), or whole group and independent activities (C), does not accurately reflect the level of personalization in instruction.

Thus, the correct pair that illustrates the difference in the level of personalization in instruction is small group activities, which are more personal, compared to whole group activities, which are less personal."

Answer:

We can calculate the atmospheric pressure using the given formula:

[tex]P = P_{o} - \rho\times g\times h[/tex]

where;

[tex]P_{o} = 1.013 \times 10^{5} pa[/tex]

[tex]\rho = \frac{1}{V}[/tex]

[tex]\rho = \frac{1}{1.334} = 0.75 kg/m^{3}[/tex]

h = 10,700 m

equating the following variables in above equation, we get;

[tex]P = 1.013\times 10^{5} - 0.75 \times 9.8 \times 10700[/tex]

P = 0.227 bar

Answer:

 507599.78 J

Explanation:

Charge input = current x time

=15 x 53 x 60

= 47000 coulomb

increase in voltage

= 12.6 - 9 = 3.6

capacity of the battery C = Charge input / increase in voltage

= 47000 / 3.6 = 13055.55

energy of the capacitor = 1/2 CV²

Initial energy of car battery =  .5 x 13055.55 x 9 x 9

= 528749.77 J

Final energy of car battery

= .5 x 13055.55 x 12.6 x 12.6

= 1036349.55

Increase in energy = 507599.78 J

Final answer:

The total energy delivered to the car battery during its charging process from 9 V to 12.6 V over 53 minutes with a charging current of 15 A is 515,160 Joules.

Explanation:

To calculate the energy delivered to the car battery while it is being charged, we need to consider the change in voltage, the charge current, and the time it was charged. Since the voltage changed linearly from 9 V to 12.6 V over 53 minutes with a constant charging current of 15 A, we first convert the charging time to seconds (53 min  imes 60 s/min = 3180 s) and then use the formula Energy (E) = Power (P)  imes Time (t), where Power (P) is the product of voltage (V) and current (I).

Assuming a linear voltage increase, we take the average voltage (9 V + 12.6 V)/2 during the charge time. The average voltage is then 10.8 V. The energy delivered is given by:

E = P  imes t = V  imes I  imes t = 10.8 V  imes 15 A  imes 3180 s

E = 515160 Joules

Therefore, the energy delivered to the car battery during the charging process is 515,160 Joules.

Answer:

A) The speed at the first point is 6.91 m/s.

B) The acceleration is 1.12 m/s²

Explanation:

The equations of velocity and position of the antelope will be as follows:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x = position of the antelope at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

v = velocity at time t

A) If we place the origin of the frame of reference at the first point, we can say that at t = 6.60 the position of the antelope is 70.0 m and its velocity is 14.3 m/s. In this way, we will have 2 equations with 2 unknowns, the initial velocity (the velocity at the first point) and the acceleration.

Let´s start finding the speed at the first point:

v = v0 + a · t       (solving for "a")

(v - v0)/t = a

Replacing a = (v - v0)/t  in the equation of the position:

x = x0 + v0 · t + 1/2 · (v - v0)/t · t²             (x0 = 0)

x = v0 · t  + 1/2 · v · t - 1/2 · v0 · t

x - 1/2 · v · t = 1/2 · v0 · t

2/t · (x - 1/2 · v · t) = v0

2/6.60 s · (70.0 m - 1/2 · 14.3 m/s · 6.60s) = v0

v0 = 6.91 m/s

The speed at the first point is 6.91 m/s.

B) Using the equation of velocity

a = (v - v0)/t

a = (14,3 m/s - 6.91 m/s) / 6.60 s

a = 1.12 m/s²

The acceleration is 1.12 m/s²

Answer:28 m

Explanation:

Given

Direction is [tex]58^{\circ}[/tex] North of east i.e. [tex]58 ^{\circ}[/tex] with x axis

Also ball moved by 33 m

therefore its east component is 33cos58=17.48 m

Northward component [tex]=33sin58=27.98 m\approx 28 m[/tex]

Answer:

The answer is 535 nanometers.

Explanation:

[tex]1\ micrometer = 1\ \mu m = 1.0\times 10^{-6}\ m.[/tex]

and

[tex]1\ nanometer = 1\ nm = 1.0\times 10^{-9}\ m.[/tex],

so

[tex]1\ \mu m = 1.0 \times 10^{3}\ nm[/tex]

which means that

[tex]\lambda = 0.535\ \mu m = 535\ nm[/tex].

In fact we can say that the light is green, because its wavelength is in the range of 500 nm to 565 nm.

Answer:

E. all of these

Explanation:

The designation of a point in space all the points that necessary

- reference point

- a direction

- fundamental units

- a direction

- motion

all are necessary to designate a point in space. Hence option E is correct.

For example in simple harmonic motion we need to specify all the above factors of the object in order to designate the position of the object.  

To designate a position, the necessary elements are A. A reference point, B. A direction, C. Fundamental units, D. Motion. Therefore option E is correct.

A. A reference point: A reference point is needed to establish a starting point or a fixed location from which the position is measured. It serves as a point of comparison and allows for consistent measurements.

B. A direction: The direction specifies the orientation or path followed from the reference point to the object or location being described. It provides information on the relative position or displacement of the object.

C. Fundamental units: Fundamental units, such as meters or feet, are used to quantify and measure the distance or displacement between the reference point and the object. These units provide a standardized way to express the position.

D. Motion: While motion itself is not necessary to designate a position, it can be relevant in certain cases when describing the position of a moving object. In such situations, the position is defined with respect to both the reference point and the object's movement.

Therefore, the correct answer is E. All of these elements - a reference point, a direction, fundamental units, and in some cases, motion - are necessary to designate a position accurately.

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Answer:

option D

Explanation:

given,

angle of the snow-covering hill (θ) = 40°

coefficient of kinetic friction = 0.12

acceleration of the shed = ?

we know,

F = m a...................(1)

now,

[tex]F = m g sin\theta -\mu_k N sin\theta[/tex].......(2)

comparing  equation (1) with (2)

[tex]m a =m g sin\theta -\mu_k m g sin\theta[/tex]

[tex]a = g sin \theta - \mu_k sin\theta[/tex]

[tex]a = 9.8\times sin 40^0 - 0.12\times 9.8\times  sin40^0[/tex]

a = 5.54 m/s²

Hence, the correct answer is option D

Answer:

0.8c and -0.14c

Explanation:

The first fragment will have a speed of +0.5c respect of a frame of reference moving at +0.5c

Lest name v the velocity of the frame of reference, and u' the velocity of the object respect of this moving frame of reference.

The Lorentz transform for velocity is:

u = (u' + v) / (1 + (u' * v) / c^2)

u = (0.5c + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = 0.8c

The other fragment has a velocity of u' = -0.6c respect of the moving frame of reference.

u = (-0.6v + 0.5c) / (1 + (0.5c * 0.5c) / c^2) = -0.14c

Answer:

astronauts age is 32 years

correct option is e 32 years

Explanation:

given data

travels = 20 light year

stay = 2 year

return = 52 years

to find out

astronauts aged

solution

we know here they stay 2 year so time taken in traveling is

time in traveling = ( 52 -2 )  = 50 year

so it mean 25 year in going and 25 years in return

and distance is given 20 light year

so speed will be

speed = distance / time

speed = 20 / 25 = 0.8 light year

so time is

time = [tex]\frac{t}{\sqrt{1-v^2} }[/tex]

time =  [tex]\frac{25}{\sqrt{1-0.8^2} }[/tex]

time = 15 year

so age is 15 + 2 + 15

so astronauts age is 32 years

so correct option is e 32 years

Answer:84.405m,[tex]\theta =48.876^{\circ}[/tex]

Explanation:

Given

Person walk 42 miles due to north  so its position vector is

[tex]r_1=42\hat{j}[/tex]

Now he deviates [tex]78^{\circ}[/tex] to east and walk 65 miles

so its new position vector

[tex]r_2=42\hat{j}+65cos78\hat{j}+65sin78\hat{i}[/tex]

[tex]r_2=65sin78\hat{i}+\left ( 42+65cos78\right )\hat{j}[/tex]

So magnitude of acceleration is

[tex]|r_2|=\sqrt{\left ( 65sin78\right )^2+\left ( 42+65cos78\right )^2}[/tex]

[tex]|r_2|=\sqrt{63.58^2+55.514^2}[/tex]

[tex]|r_2|=84.405 m[/tex]

for direction

[tex]tan\theta =\frac{42+65cos78}{65sin78}[/tex]

[tex]tan\theta =0.8731 [/tex]

[tex]\theta =41.124^{\circ}with\ respect\ to\ east[/tex]

[tex]\theta =48.876^{\circ}with\ respect\ to\ North[/tex]

No, the momenta of two particles with equal kinetic energies are not necessarily equal.

No, the momenta of two particles with equal kinetic energies are not necessarily equal. Momentum is given by the equation:

p = mv

Where p is momentum, m is mass, and v is velocity. Kinetic energy is given by the equation:

K = mv²/2

In order for two particles to have equal kinetic energies, their masses and velocities must satisfy the equation:

mv₁²/2 = mv₂²/2

But this does not necessarily mean that their momenta are equal, as the masses and velocities can still differ.

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Final answer:

To find the electric field strength between the disks, we can use Coulomb's law and the formula for electric field. The launch speed of the proton can be found using the conservation of energy.

Explanation:

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

Electric Field = Force / Charge

The force between the disks is given by Coulomb's law:

Force = (k * q1 * q2) / r^2

Where k is the electrostatic constant (9 * 10^9 Nm^2/C^2), q1 and q2 are the charges on the disks, and r is the distance between them.

Substituting the given values, we have:

Force = (9 * 10^9 Nm^2/C^2) * (16 * 10^-9 C)^2 / (2.3 * 10^-2 m)^2

Calculating this expression, we get the force between the disks. Then, we can divide this force by the charge on either disk to find the electric field strength.

For part B, we can use the conservation of energy to find the launch speed of the proton. The potential energy difference between the disks can be calculated as:

Potential Energy = charge * voltage

Given the charge of the proton and the distance between the disks, we can find the voltage. Since the proton starts from rest, all of its initial potential energy will be converted into kinetic energy:

Kinetic Energy = (1/2) * mass * velocity^2

Where the mass of the proton is known.

Solving for velocity, we can find the launch speed of the proton.

Answer:

(a): 37.94 m.

(b): [tex]323.16^\circ.[/tex]

(c): 126.957 m.

(d): [tex]0.93^\circ.[/tex]

(e): 49.92 m.

(f): [tex]130.08^\circ.[/tex]

Explanation:

Given:

Any vector [tex]\vec A[/tex], making an angle [tex]\theta[/tex] with respect to the positive x-axis, can be written in terms of its x and y components as follows:

[tex]\vec A = A\cos\theta\ \hat i+A\sin\theta \ \hat j.[/tex]

where, [tex]\hat i,\ \hat j[/tex] are the unit vectors along the x and y axes respectively.

Therefore, the given vectors can be written as

[tex]\vec a = 50\cos30^\circ \ \hat i+50\sin 30^\circ\ \hat j = 43.30\ \hat i +25\ \hat j\\\vec b = 50\cos195^\circ \ \hat i+50\sin 195^\circ\ \hat j = -48.29\ \hat i +-12.41\ \hat j\\\vec c = 50\cos 315^\circ \ \hat i+50\sin 315^\circ\ \hat j = 35.35\ \hat i +-35.35\ \hat j\\[/tex]

(a):

[tex]\vec a +\vec b + \vec c=  (43.30\ \hat i +25\ \hat j)+(-48.29\ \hat i +-12.41\ \hat j)+(35.35\ \hat i +-35.35\ \hat j)\\=(43.30-48.29+35.35)\hat i+(25-12.41-35.35)\hat j\\=30.36\hat i-22.75\hat j.\\\\\text{Magnitude }=\sqrt{30.36^2+(-22.75)^2}=37.94\ m.[/tex]

(b):

Direction [tex]\theta[/tex] can be found as follows:

[tex]\tan\theta = \dfrac{\text{x component of }(\vec a + \vec b +\vec c)}{\text{y component of }(\vec a + \vec b +\vec c)}=\dfrac{-22.75}{30.36}=-0.749\\\Rightarrow \theta = \tan^{-1}(-0.749)=-36.84^\circ.[/tex]

The negative sign indicates that the sum of the vectors is [tex]36.84^\circ.[/tex] below the positive x axis.

Therefore, direction of this vector sum counterclockwise with respect to positive x-axis = [tex]360^\circ-36.84^\circ=323.16^\circ.[/tex]

(c):

[tex]\vec a -\vec b + \vec c=  (43.30\ \hat i +25\ \hat j)-(-48.29\ \hat i +-12.41\ \hat j)+(35.35\ \hat i +-35.35\ \hat j)\\=(43.30+48.29+35.35)\hat i+(25+12.41-35.35)\hat j\\=126.94\hat i+2.06\hat j.\\\\\text{Magnitude }=\sqrt{126.94^2+2.06^2}=126.957\ m.[/tex]

(d):

Direction [tex]\theta[/tex] can be found as follows:

[tex]\tan\theta = \dfrac{\text{x component of }(\vec a - \vec b +\vec c)}{\text{y component of }(\vec a - \vec b +\vec c)}=\dfrac{2.06}{126.94}=0.01623\\\Rightarrow \theta = \tan^{-1}(0.01623)=0.93^\circ.[/tex]

(e):

[tex](\vec a + \vec b)-(\vec c + \vec d)=0\\(\vec a + \vec b)=(\vec c + \vec d)\\\vec d = \vec a + \vec b -\vec c.[/tex]

[tex]\vec d = \vec a +\vec b - \vec c=  (43.30\ \hat i +25\ \hat j)+(-48.29\ \hat i +-12.41\ \hat j)-(35.35\ \hat i +-35.35\ \hat j)\\=(43.30-48.29-35.35)\hat i+(25-12.41+35.35)\hat j\\=-40.34\hat i+47.94\hat j.\\\\\text{Magnitude }=\sqrt{(-40.34)^2+47.94^2}=62.65\ m.[/tex]

(f):

Direction [tex]\theta[/tex] can be found as follows:

[tex]\tan\theta = \dfrac{\text{x component of }\vec d}{\text{y component of }\vec d}=\dfrac{47.94}{-40.34}=-1.188\\\Rightarrow \theta = \tan^{-1}(-1.188)=-49.92^\circ.[/tex]

The x component of this vector is negative and y component is positive therefore the vector lie in second quadrant, which means, the direction of this vector, counterclockwise with respect to positive x axis = [tex]180^\cir.

c-49.92^\circ=130.08^\circ.[/tex]

The electric field measurements indicate a point charge located to the right of both measurement positions on the x-axis, being more than 5.00 m away. Using the electric field values (3.000 N/C at 2.00 m and 0.750 N/C at 5.00 m), and applying the equation E = kQ/r^2, we can solve for both the point charge's magnitude and location.

The student's question revolves around the electric field produced by a point charge and involves finding both the location of the point charge and its magnitude. To solve this, we need to apply Coulomb's Law and the principle that the intensity of the electric field (E) from a point charge can be described by the equation E = kQ/r^2, where k is Coulomb's constant (8.988  imes 10^9 N m^2/C^2), Q is the charge, and r is the distance from the charge to the point where the electric field is measured.

Given the electric field measurements at two different points both pointing to the left, we can deduce that the point charge is located to the right of both points on the x-axis. The point charge must thus be more than 5.00 m to the right of the origin.

By using the measured electric field values and the distances to set up two equations, we can determine the charge's value:

E1 = kQ/r1^2 (at 2.00 m)
E2 = kQ/r2^2 (at 5.00 m)

Where E1 = 3.000 N/C, E2 = 0.750 N/C, r1 = 2.00 m, and r2 = 5.00 m. By solving these equations simultaneously, we can find the value of Q and the exact location along the x-axis.

Answer:

[tex]\theta=28.07^{\circ}[/tex]

Explanation:

Speed of van, v = 28 m/s

Radius of unbanked curve, r = 150 m

Let [tex]\theta[/tex] is the angle with the vertical. In case of banking of road,

[tex]T\ cos\theta=mg[/tex].............(1)

And

[tex]T\ sin\theta=\dfrac{mv^2}{r}[/tex]..........(2)

From equation (1) and (2) :

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

[tex]tan\theta=\dfrac{(28)^2}{150\times 9.8}[/tex]

[tex]\theta=28.07^{\circ}[/tex]

So, the string makes an angle of 28.07 degrees with the vertical. Hence, this is the required solution.