An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the positive x -direction. Another extremely long wire laying parallel to the x -axis and passing through the y -axis at ry= 1.8m carries a current of 50A running in the negative x -direction. What is the magnitude of the net magnetic field that the wires generate on the y -axis at ry= −3.510m ?

Answer:

μ= 0.408 : coefficient of kinetic friction

Explanation:

Kinematic equation for the box:

[tex]a=\frac{v_{f} -v_{i} }{t}[/tex] Formula( 1)

a= acceleration

v_i= initial speed =0

v_f= final speed= 20 m/s

t= time= 5 s

We replace data in the formula (1):

[tex]a=\frac{0-20}{5}[/tex]

[tex]a= -\frac{20}{5}[/tex]

a= - 4m/s²

Box kinetics: We apply Newton's laws in x-y:

∑Fx=ma : second law of Newton

-Ff= ma Equation (1)

Ff is the friction force

Ff=μ*N Equation (2)

μ is the coefficient of kinetic friction

N is the normal force

Normal force calculation

∑Fy=0  : Newton's first law

N-W=0   W is the weight of the box

N=W= m*g  : m is the mass of the box and g is the acceleration due to gravity

N=9.8*m

We replace N=9.8m in the equation (2)

Ff=μ*9.8*m

Coefficient of kinetic friction ( μ) calculation

We replace Ff=μ*9.8*m and a= -4m/s² in the equation (1):

-μ*9.8*m: -m*4 : We divide by -m on both sides of the equation

9.8*μ=4

μ=4 ÷ 9.8

μ= 0.408

Answer:

Ff= 8.9N:Force of friction acting on the block.

Explanation

Box kinetics:  in x₁-y₁ :We apply  the second law of Newton

∑Fx₁=ma : second law of Newton  Formula (1)

Where:

∑Fx₁:algebraic sum of forces in the direction of x1, positive in the direction of movement of the block, which is down and negative in the direction opposite to the movement of the block

m: is the mass of the block

The x₁ axis coincides with the plane of sliding of the block, that is, the x₁ axis forms 45 degrees with the horizontal

The total weight (W) of the block is in the vertical direction and the tip of the vector down and its magnitude is calculated as follows:

W=m*g

Where:

m: is the mass of the block

g: is the acceleration due to gravity

Calculation of the weight in the direction x₁

Wx₁=Wcos45°= m*g*cos 45 Equation( 1)

Data

m=1.8kg

g=9.8 m/s²

a=2m/s

Wx₁=m*g*cos 45°= 1.8*9.8 *[tex]\frac{\sqrt{2} }{2}[/tex]= 12.47 N

Friction force (Ff ) calculation

We apply formula (1)

∑Fx₁=m*a

Wx₁ - Ff= m*a

Ff=Wx₁ - m*a

We replace data

Ff= 12.47 - 1.8*2 =8.87

Ff= 8.9N

Answer:

Gravitational potential energy

Explanation:

Hydroelectric dams are the power plants which generates electricity by using the energy of falling water from a great height.

A the water is stored in big reservoirs and at a great height, it contain lot of potential energy due to the height. As it falls downwards, the potential energy is converted into kinetic energy of the water. this kinetic energy of the falling water is used to run the turbine, and then the electric energy is generated.

So, in hydroelectric power stations, the potential energy of water is converted into the electric energy.

Answer:

33 seconds.

Explanation:

The equation for speed with constant acceleration at time t its:

[tex]V(t) \ = \ V_0 \ + \ a \ t[/tex]

where [tex]V_0[/tex] is the initial speed, and  a its the acceleration.

Starting at rest, the initial speed will be zero, so

[tex]V_0 = 0[/tex]

the final speed is

[tex]V(t_{f1}) = 23 \frac{m}{s}[/tex]

and the acceleration is

[tex]a = 2.3 \frac{m}{s^2}[/tex].

Taking all this together, we got

[tex]V(t_{f1}) = 23 \frac{m}{s} = 0 + 2.3 \ \frac{m}{s^2} t_{f1}[/tex]

[tex]23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}[/tex]

[tex]\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} =  t_{f1}[/tex]

[tex]10 s =  t_{f1}[/tex]

So, for the first half of the problem we got a time of 10 seconds.

Now, the initial speed will be

[tex]V_0 = 23 \frac{m}{s}[/tex],

the acceleration

[tex]a=-1.0 \frac{m}{s^2}[/tex],

with a minus sign cause its slowing down, the final speed will be

[tex]V(t_{f2}) = 0[/tex]

Taking all together:

[tex]V(t_{f2}) = 0 = 23 \frac{m}{s} -  1.0 \frac{m}{s^2} t_{f2}[/tex]

[tex] 23 \frac{m}{s} =  1.0 \frac{m}{s^2} t_{f2}[/tex]

[tex] \frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}[/tex]

[tex] 23 s = t_{f2}[/tex]

So, for the first half of the problem we got a time of 23 seconds.

[tex]t_total = t_{f1} + t_{f2} = 33  \ s[/tex]

Answer:

option (d) 7.1 kN

Explanation:

Given:

Mass of the car, m = 1600 kg

Acceleration of the car, a = 1.5 m/s²

Coefficient of kinetic friction = 0.3

let the tension be 'T'

Now,

ma = T - f .................(1)

where f is the frictional force

also,

f = 0.3 × mg

where g is the acceleration due to the gravity

thus,

f = 0.3 × 1600 × 9.81 =

therefore,

equation 1 becomes

1600 × 1.5 = T - 4708.8

or

T = 2400 + 4708.8

or

T = 7108.8 N

or

T = 7.108 kN

Hence,

The correct answer is option (d) 7.1 kN

The rate at which the area of the triangle formed by the wall, the ground, and the ladder is changing, at the instant the bottom of the ladder is 36 feet from the wall, is -216 square feet per second.

This question revolves around the concept of related rates in calculus. The area of the triangle formed by the ground, the wall, and the ladder is given by the formula ½ * base * height. In this situation, the base is x, which represents the distance from the wall to the base of the ladder, and the height is the vertical reach of the ladder onto the wall, represented by y (which can be found using the Pythagorean theorem). Differentiating this equation with respect to time gives us dA/dt = ½ (x dy/dt + y dx/dt).

Given that dx/dt is 8 feet per second, and x=36 feet, to find dy/dt, we use the Pythagorean relationship (39² = 36² + y²), and solve for y to get y = 15 feet. Therefore, substituting all these values into the equation we find:** dA/dt = -216 sq.ft/sec**.

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The area of the triangle is changing at a rate of ( -285.6 ) square feet per second at the instant the bottom of the ladder is 36 feet from the wall. The negative sign indicates that the area is decreasing.

Let ( x ) be the distance of the bottom of the ladder from the wall, and ( y) be the height of the ladder on the wall. We are given that [tex]\( \frac{dx}{dt} = 8 \)[/tex] feet per second, and we want to find [tex]\( \frac{dA}{dt} \) when \( x = 36 \) feet.[/tex]

Using the Pythagorean theorem, we have [tex]\( x^2 + y^2 = 39^2 \).[/tex]Differentiating both sides with respect to time ( t ), we get:

[tex]\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \][/tex]

We can solve for [tex]\( \frac{dy}{dt} \):[/tex]

[tex]\[ 2y\frac{dy}{dt} = -2x\frac{dx}{dt} \][/tex]

[tex]\[ \frac{dy}{dt} = -\frac{x}{y}\frac{dx}{dt} \][/tex]

Now, we can express the area ( A ) in terms of ( x ) and ( y ):

[tex]\[ A = \frac{1}{2}xy \][/tex]

Differentiating ( A ) with respect to time ( t ), we get:

[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(x\frac{dy}{dt} + y\frac{dx}{dt}\right) \][/tex]

Substituting [tex]\( \frac{dy}{dt} \)[/tex] from above, we have:

[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(x\left(-\frac{x}{y}\frac{dx}{dt}\right) + y\frac{dx}{dt}\right) \][/tex]

[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(-\frac{x^2}{y}\frac{dx}{dt} + y\frac{dx}{dt}\right) \][/tex]

[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(y - \frac{x^2}{y}\right)\frac{dx}{dt} \][/tex]

We know [tex]\( \frac{dx}{dt} = 8 \)[/tex] feet per second, and we need to find [tex]\( \frac{dA}{dt} \)[/tex]when ( x = 36 ) feet. To find ( y ) at this instant, we use the Pythagorean theorem:

[tex]\[ 36^2 + y^2 = 39^2 \][/tex]

[tex]\[ y^2 = 39^2 - 36^2 \][/tex]

[tex]\[ y^2 = 1521 - 1296 \][/tex]

[tex]\[ y^2 = 225 \][/tex]

[tex]\[ y = 15 \][/tex]

Now we can find [tex]\( \frac{dA}{dt} \):[/tex]

[tex]\[ \frac{dA}{dt} = \frac{1}{2}\left(15 - \frac{36^2}{15}\right) \times 8 \][/tex]

[tex]\[ \frac{dA}{dt} = 4\left(15 - \frac{1296}{15}\right) \][/tex]

[tex]\[ \frac{dA}{dt} = 4\left(15 - 86.4\right) \][/tex]

[tex]\[ \frac{dA}{dt} = 4\left(-71.4\right) \][/tex]

[tex]\[ \frac{dA}{dt} = -285.6 \][/tex]

So, the area of the triangle is changing at a rate of ( -285.6 ) square feet per second at the instant the bottom of the ladder is 36 feet from the wall. The negative sign indicates that the area is decreasing.

Answer:

The work done on the jet is [tex]W_{jet} = 2.49\times 10^{7} J[/tex]

Given:

Force, F = [tex]2.3\times 10^{5} N[/tex]

Distance moved by the jet, x = 87 m

Kinetic Energy, KE = [tex]4.5\times 10^{7} J[/tex]

Solution:

Now, to calculate the work done on the fire fighter jet by the catapult, we find the difference between the KE of the jet at lift off and the work done by the engines.

This difference provides the amount of work done on the jet and is given by:

[tex]W_{jet} = KE - W_{engines}[/tex]              (1)

Now, the work done by the engines is given by:

[tex]W_{engines} = Fx = 2.3\times 10^{5}\times 87 = 2.001\times 10^{7} J[/tex]

Now, using eqn (1):

[tex]W_{jet} = 4.5\times 10^{7} - 2.001\times 10^{7} = 2.49\times 10^{7} J[/tex]

Answer: The two places altitudes are: 16.17 m and 40.67 m

Explanation:

Hi!

Lets call z to the vertical direction (z= is ground) . Then the positions of the balloon and the pellet, using the values of the velocities we are given, are:

[tex]z_b =\text{balloon position}\\z_p=\text{pellet position}\\z_b=(7\frac{m}{s})t\\z_p=30\frac{m}{s}(t-t_0)-\frac{g}{2}(t-t_0)^2\\g=9.8\frac{m}{s^2}[/tex]

How do we know the value of t₀? This is the time when the pellet is fired. At this time the pellet position is zero: its initial position. To calculate it we know that the pellet is fired when the ballon is in z = 12m. Then:

[tex]t_0=\frac{12}{7}s[/tex]

We need to know the when the z values of balloon and pellet is the same:

[tex]z_b=z_p\\(7\frac{m}{s})t =30\frac{m}{s}(t-\frac{12}{7}s)-\frac{g}{2}(t-\frac{12}{7}s)^2[/tex]

We need to find the roots of the quadratic equation. They are:

[tex]t_1=2.31s\\t_2=5.81[/tex]

To know the altitude where the to objects meet, we replace the time values:

[tex]z_1=16,17m\\z_2=40,67m[/tex]

Answer:

The horizontal distance is 478.38 m

Solution:

As per the question:

Initial Speed of the bullet in horizontal direction, [tex]v_{x} = 670 m/s[/tex]

Initial vertical velocity of the bullet, [tex]v_{y} = 0 m/s[/tex]

Vertical distance, y = 0.025 m

Now, for the horizontal distance, 'x':

We first calculate time, t:

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

(since, motion is vertically downwards under the action of 'g')

[tex]0.025 = 0 - \frac{1}{2}\times 9.8t^{2}[/tex]

[tex]t = \sqrt{0.05}{9.8} = 0.0714 s[/tex]

Now, the horizontal distance, x:

[tex]x = v_{x}t + \frac{1}{2}a_{x}t^{2}[/tex]

[tex]x = v_{x}t + \frac{1}{2}0.t^{2}[/tex]

(since, the horizontal acceleration will always be 0)

[tex]x = 670\times 0.714 = 478.38 m[/tex]

Answer:

b. precise

Explanation:

Accuracy means how close a measurement is to the true value (in our example the true value is score a goal )

Precision refers how close two or more measurements to each other.

So if the playing soccer hit always the left goal post, he is not accurate (e doesn't score any goals ) but he is very precise

Answer:1.902 m

Explanation:

Given

height of apartment=1.5 m

It takes 0.21 sec to reach the bottom from apartment

So

[tex]s=u_1t+\frac{gt^2}{2}[/tex]

[tex]1.5=u_1\times 0.21+\frac{9.81\times 0.21^2}{2}[/tex]

[tex]u_1=6.11 m/s[/tex]

i.e. if ball is dropped from top its velocity at window is 6.11 m/s

So height of upper floor above window

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

where s= height of upper floor above window

here u=0

[tex]6.11^2=2\times 9.81\times s[/tex]

[tex]s=\frac{6.11^2}{2\times 9.81}[/tex]

s=1.902 m

Answer:

Explanation:

Shoveling gravel into the water will increase the buoyancy of the barge, which will make it float higher. Without data it is hard to tell if it will raise the barge enough to be counterproductive, but in any case throwing away the payload is not a good idea. Adding some weights would make the barge float lower, and then maybe the plie can make it under the bridge.

Answer:

The initial horizontal velocity was 21 m/s

Explanation:

Please, see the figure for a better understanding of the problem.

The equation for the position of an object moving in a parabolic trajectory is as follows:

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

Where:

r = vector position at time t

x0 = initial horizontal position

v0 = initial velocity

t = time

α = launching angle

y0 = initial vertical position

g = acceleration due to gravity

Notice that at time t = 2.4 s the vector "r" is the one in the figure. We know that the x-component of that vector is 50 m. Then using the equation for the x-component of the vector "r", we can calculate the initial velocity:

x = x0 + v0 · t · cos α

Let´s place the center of the frame of reference at the point of the kick so that x0 = 0.

x =  v0 · t · cos α

x/t = v0 · cos α

Notice in the figure that v0 · cos 30° = v0x which is the initial horizontal velocity. Remember trigonometry of right triangles:

cos α = adjacent / hypotenuse = v0x / v0

Then:

50 m/ 2.4 s = v0 · cos 30° = v0x

v0x = 21 m/s

Answer:

the radius of bigger loop = 6 cm

Explanation:

given,

two concentric current loops

smaller loop radius = 3.6 cm

]current in smaller loop = 12 A

current in the bigger loop = 20 A

magnetic field at the center of loop = 0

Radius of the bigger loop = ?

[tex]B_t = B_1 + B_2[/tex]

[tex]0 = \dfrac{\mu_0I_1}{2R_1} +\dfrac{\mu_0I_2}{2R_2}[/tex]

now, on solving

[tex]\dfrac{I_1}{R_1} = \dfrac{I_2}{R_2}[/tex]

[tex]R_2 = I_2\dfrac{R_1}{I_1}[/tex]

       = [tex]20\times \dfrac{3.6}{12}[/tex]

       = 6 cm

hence, the radius of bigger loop = 6 cm

The forces acting on you are:

That is. There is not need for any other force, cause the car its going at constant speed, so the acceleration its zero, as the net force its mass multiplied by acceleration, the net force is zero.

The forces acting on you are:

That is. Again. The car its slowing down. But, it cant apply a force in you against the direction of movement to slow you down.

As you can't be slowed down, you will go forward at the same speed you had when the car went steady, and will crash against whatever its in front of you.

D)

The friction force, will pull you against the direction of movement, and slow you down will the car slows down.

When a car moves at a steady pace, the forces acting on a passenger include gravity and the normal force from the bench. When the car begins to slow down, the passenger experiences inertia, which might give the effect of being 'pushed' forward. If the bench is not slippery, friction will be the force that aids in decelerating the passenger along with the car.

(a) If you are sitting on a frictionless bench in a car that is moving at a perfectly steady speed, the forces acting on you are: the gravitational force directed down and the normal force from the bench acting upward. Since the car moves at a constant speed, there is no horizontal force.

(b) If the car starts slowing down, then in addition to the forces mentioned before, there is one more force in play, which is inertia. It's not a force like gravity or the normal force, but it is a consequence of Newton's first law of motion. You experience this as a force pushing you forward, even though the car is slowing down or stopping.

(c) Because of inertia, as the car slows down, you tend to keep moving. Without friction or anything to hold onto, you would slide off the bench and continue moving forward at the original speed of the car.

(d) If the bench is not slippery and you don't slide off as the car slows down, the force responsible for your deceleration is friction. Even though you can't feel it, friction between you and the bench affects you when the car decelerates.

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Answer:

Along North,  1.816 km

Along east , 3.564 km

Explanation:

d = 4 km 27° North of east

As we need to find the distance traveled due north and the distance traveled due east, they are the components along north and along east. The component along north is the vertical component and the distance traveled along east is the horizontal component.

Distance traveled due North

y = d Sin 27 = 4 Sin 27 = 1.816 km

Distance traveled due east

x = d cos 27 = 4 cos 27 = 3.564 km

Answer:

mean = 54

standard deviation = 6.24

Explanation:

given data

mean a = 72

standard deviation σx = 12

average b = 95

standard deviation σy = 4

final grade Z = 70%  = 0.70

project y = 30%  = 0.30

to find out

mean and standard deviation

solution

final grade equation will be here

final grade = 0.70x + 0.30 y   .....a

here x is statics grade and y is class project

so

mean will be 0.70x + 0.30 y

mean = 0.70 (a) + 0.30 (b)

mean = 0.70 (12) + 0.30 (95)

mean = 54

and

standard deviation = 0.70x + 0.30 y

standard deviation = 0.70(σx) + 0.30 (σy)

standard deviation = 0.70(12) + 0.30 (4)

standard deviation = 6.24

Answer:

[tex]r=\frac{1}{75}[/tex] Ω≈0.01333Ω

Explanation:

You can check the internal resistance model in the image I attached you.

Now using Kirchhoff's voltage law:

ε=[tex]Ir+IR[/tex]  (1)

Where:

[tex]IR=V_R=LoadVoltage=10V[/tex]

ε=12V

[tex]I=150A[/tex]

Evaluating the data provide in (1)

[tex]r=\frac{12-10}{150} =\frac{1}{75}[/tex]≈0.01333Ω

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

[tex]H=ut+\dfrac{1}{2}at^2[/tex]

Here, a = g

[tex]H=\dfrac{1}{2}gt^2[/tex]            

[tex]H=\dfrac{1}{2}\times 9.8\times (2.261)^2[/tex]

H = 25.04 meters

So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

Final answer:

Using the equation of motion under gravity, it was calculated that a ball dropped and taking 2.261 s to hit the ground was released from a height of approximately 25.07 meters.

Explanation:

To find out from what height H a ball was released if it takes 2.261 s to fall to the ground, we use the equation of motion under gravity, which is H = 0.5 * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2 on the surface of the Earth) and t is the time in seconds.

Substituting the given values in the equation, we get: H = 0.5 * 9.8 * (2.261)^2. After calculating, we find that the height H approximately equals 25.07 meters.

Therefore, the ball was released from a height of approximately 25.07 meters.

Answer:

2.82 s

Explanation:

The ball will be subject to the acceleration of gravity which can be considered constant. Therefore we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

Y0 is the starting position, 2.3 m in this case.

Vy0 is the starting speed, 13 m/s.

a will be the acceleration of gravity, -9.81 m/s^2, negative because it points down.

Y(t) = 2.3 + 13 * t - 1/2 * 9.81 * t^2

It will reach the ground when Y(t) = 0

0 = 2.3 + 13 * t - 1/2 * 9.81 * t^2

-4.9 * t^2 + 13 * t + 2.3 = 0

Solving this equation electronically gives two results:

t1 = 2.82 s

t2 = -0.17 s

We disregard the negative solution. The ball spends 2.82 seconds in the air.

Answer:

longest length is 13304.05 m

Explanation:

given data

diameter = 25 mm

so radius r = 12.5 × [tex]10^{-3}[/tex] m

density = 5000 kg/m³

tension = 320000 N

to find out

longest length of of this rope

solution

we know that here tension is express as

T = m × g

and m is mass and g is  acceleration due to gravity

here mass = density × volume

and volume = π×r²×l

here r is radius and l is length

so T = density c volume  × g

put here value

320000 = 5000 × π×12.5² × [tex]10^{-3}[/tex] × l × 9.8

solve it we get length l

l = 13304.05

so longest length is 13304.05 m

Answer:

The the decibel level be when three sleep deprived students are typing furiously in the same room is 64.77 dB.

Explanation:

Given that,

Decibel level = 60.0 dB

Using formula of decibel level

[tex]\beta= 10 log\dfrac{I}{I_{0}}[/tex]

For one student,

[tex]60.0=10 log\dfrac{I}{I_{0}}[/tex]

For 3 students,

[tex]\beta'=10 log\dfrac{3I}{I_{0}}[/tex]

[tex]\beta'=10log 3+10 log\dfrac{I}{I_{0}}[/tex]

[tex]\beta'=4.77+60.0[/tex]

[tex]beta'=64.77\ dB[/tex]

Hence, The the decibel level be when three sleep deprived students are typing furiously in the same room is 64.77 dB.

Planets like Mars exhibit direct and retrograde motion, resulting in complex paths across the night sky, which differ from the Sun's steady eastward movement along the ecliptic through the zodiac signs. Earth's faster orbit leads to the phenomenon of retrograde motion when observing other planets.

The motion of planets like Mars among the stars of the zodiac is different from the motion of the Sun because planets exhibit both direct motion and retrograde motion in their paths across the sky, as observed from Earth. Planets periodically appear to move in the opposite direction in the sky, known as retrograde motion, while the Sun moves steadily along the ecliptic through the zodiac signs. This distinct planetary motion is due to the orbital speed and path of Earth relative to other planets, leading to the phenomenon where Earth, moving faster along its orbit, occasionally overtakes other planets like Mars, changing how these planets appear to move against the backdrop of stars.

Additionally, planets can have stationary points where they appear to pause before they switch from direct to retrograde motion or vice versa. These motions are perceived from Earth because of the combination of the planets' own orbital movement around the Sun and Earth's simultaneous revolution. By contrast, the Sun follows a consistent eastward path among the fixed stars, circling the zodiac once per year, spending about a month in each zodiac sign.

Answer:

(a): [tex]247.77^\circ.[/tex]

(b): [tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \dfrac{\hat i-12\hat j}{\sqrt{145}}.[/tex]

(c): Magnitude = [tex]0.604\ C.[/tex]

     Sign = negative.

Explanation:

(a):

Given that first charge is located at (3,-5) and second charge is located at (2,7).

The electrostatic force between two charges acts long the line joining the two charges, therefore, the direction of electrostatic force of attraction between these two charges is along the position vector of charge at (2,7) with respect to the position of charge at (3,-5).

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

Position vector of charge at (3,-5) with respect to origin is

[tex]\vec r_1 = 3\hat i+(-5)\hat j[/tex]

Position vector of charge at (2,7) with respect to origin is

[tex]\vec r_2 = 2\hat i+7\hat j[/tex]

The position vector of at (2,7) with respect to the position of charge at (3,-5) is

[tex]\vec r = \vec r_2-\vec r_1\\=(3\hat i-5\hat j)-(2\hat i+7\hat j)\\=1\hat i-12\hat j.[/tex]

If [tex]\theta[/tex] is the angle this position vector is making with the positive x axis then,

[tex]\rm \tan\theta =\dfrac{y\ component\ of\ (\hat i-12\hat j)}{x\ component\ of\ (\hat i-12\hat j}=\dfrac{-12}{1} = -12.\\\theta =\tan^{-1}(-12)=-85.23^\circ.[/tex]

The negative sign indicates that this position vector is [tex]85.23^\circ.[/tex] below the positive x axis, therefore, in counterclockwise direction from positive x axis, its direction = [tex]360^\circ-85.23^\circ = 247.77^\circ.[/tex]

The force is also along the same direction.

(b):

According to Coulomb's law, the expression of this force is given by

[tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \hat r[/tex]

where,

We have,

[tex]\vec r = \hat i-12\hat j\\\therefore |\vec r| = \sqrt{1^2+(-12)^2}=\sqrt{145}.\\\Rightarrow \hat r = \dfrac{\vec r}{|\vec r|} = \dfrac{\hat i-12\hat j}{\sqrt{145}}[/tex]

Putting this value, we get,

[tex]\vec F = k\ \dfrac{q_1q_2}{|\vec r|^2}\ \dfrac{\hat i-12\hat j}{\sqrt{145}}[/tex]

(c):

If charge at (3,-5) is [tex]+2\ pC[/tex], then the magnitude of the force between the two charges is given by

[tex]F=k\ \dfrac{q_1q_2}{|\vec r|^2}[/tex]

where, we have,

[tex]F=0.75\ N\\|\vec r| = \sqrt{145}\ cm = \sqrt{145}\times 10^{-2}\ m.\\q_1 = +2\ pC = +2\times 10^{-12}\ C.[/tex]

Putting all these values,

[tex]0.75=9\times 10^9\times \dfrac{(+2\times 10^{-12})\times q_2}{(\sqrt{145}\times 10^{-2})^2}\\\Rightarrow q_2 = \dfrac{0.75\times(\sqrt{145}\times 10^{-2})^2}{9\times 10^9\times 2\times 10^{-12}} =0.604\ C.[/tex]

Since, the electric force between the two spheres is given to be attractive therefore this charge must be negative as the other charge is positive.

(a) Initial velocity (v₀) = 0 m/s,  Final velocity (v) = 15.0 m/s.

(b) The dolphin rises approximately 11.5 meters above the water.

(c) The dolphin is in the air for approximately 1.5 seconds.

(a)

Initial velocity (v₀) = 0 m/s (since the dolphin jumps straight up from the water, its initial velocity is zero)

Final velocity (v) = 15.0 m/s (given)

(b)

To solve for the height, we can use the kinematic equation that relates the final velocity (v), initial velocity (v₀), acceleration (a), and displacement (Δy):

v² = v₀² + 2aΔy

v² = v₀² + 2gh

v² = 2gh

Solving for h:

h = (v²) / (2g)

h = (15.0)² / (2 × 9.8)

h ≈ 11.5 m

Therefore, the dolphin rises approximately 11.5 meters above the water.

(c) To determine the time the dolphin is in the air, we can use the equation for time (t) derived from the kinematic equation:

Δy = v₀t + (1/2)at²

Δy = (1/2)at²

t = √((2Δy) / a)

t = √((2 × 11.5) / -9.8)

t ≈ 1.5 s

Therefore, the dolphin is in the air for approximately 1.5 seconds.

To know more about the velocity:

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The dolphin rises 8.64 meters above the water and is in the air for 1.33 seconds.

(a) The knowns in this problem are:

(b) To find the height above the water, we need to find the time taken by the dolphin to reach that height and use the equation:

Final velocity = Initial velocity + Acceleration × Time

Solving for time, we have:

Time = (Final velocity - Initial velocity) / Acceleration

Substituting the known values into the equation, we get:

Time = (0 m/s - 13.0 m/s) / -9.8 m/s² = 1.33 s

Next, we can use the formula for height:

Height = Initial velocity × Time + 0.5 × Acceleration × Time²

Substituting the known values, we have:

Height = 13.0 m/s × 1.33 s + 0.5 × -9.8 m/s² × (1.33 s)² = 8.64 m

(c) The time the dolphin is in the air is equal to the time calculated above, which is 1.33 seconds.

Answer:

8717 meters.

Explanation:

We need to know the conversion factors. We know that:

1 mile = 1609.34 meters

1 yard = 0.9144 meters

This means that:

[tex]\frac{1609.34 meters}{1 mile}=1[/tex]

[tex]\frac{0.9144 meters}{1 yard}=1[/tex]

It is convenient to leave the units we want at the end in the numerator so the ones in the denominator cancel out with the ones we want to remove, as will be seen in the next step.

We will convert first the miles, then the yards, and add them up.

[tex]5miles=5miles\frac{1609.34 meters}{1 mile}=8046.7meters[/tex]

[tex]733yards=733yards\frac{0.9144 meters}{1 yard}=670.2552meters[/tex]

So total distance is the sum of these,  8717 meters.

The straight-line distance from city A to city C is approximately 458.80 km. The direction from city A to city C is 14.9° north of west. The approximation is due to ignoring Earth's curvature.

To solve this problem, we will use the laws of vector addition as airplanes movements are vector quantities as it involves both magnitude (distance) and direction. Initially, the airplane flies 200 km west. Then, it changes its direction 26.0° north of west and flies an additional 275 km.

When we decompose the second segment of the flight into its westward and northward components, we can use trigonometric principles. The horizontal (westward) distance is 275 km * cos(26.0°) = 245.56 km. Adding that to the 200 km the airplane flew initially gives us a total westward distance of 445.56 km. The vertical (northward) distance is 275 km * sin(26.0°) = 118.40 km.

To find the straight-line distance between City A and City C, we apply the Pythagorean theorem: √[(445.56 km)² + (118.4 km)²] = 458.80 km. To find the direction, we take the arctan of the northward distance over the westward distance: arctan(118.4 km / 445.56 km) = 14.9° north of west.

It's only approximately correct because we ignore the curvature of the Earth, which would slightly modify the distances and angles involved.

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The straight-line distance from city A to city C is approximately 460 km and the direction is 15° north of west. The answer is only an approximation because we've considered a flat plane, ignoring the spherical nature of the Earth.

The question is asking for the linear distance and direction from city A to city C, which can be solved through vector addition and trigonometric calculations in physics. The displacement from A to B is 200 km west. The displacement from B to C is 275 km, 26.0° north of west. These can be seen as components of a right triangle, and by applying the Pythagorean theorem, the straight-line distance, or the hypotenuse, can be calculated.

First, we derive the north and west components of the 275 km using cosine and sine respectively as the motion is angled. The westward motion is 275cos(26) = 244 km. Add this to the 200 km westward motion to get a total westward motion of 444 km. The northward motion is 275sin(26) = 121 km.

Then, we apply Pythagorean theorem. √(444^2 + 121^2) ≈ 460 km. This is the straight-line distance from city A to city C.

For the direction, we find the angle, using tan^(-1)(121/444) ≈ 15° north of west.

The answer is only approximately correct because in real situation we have to take into account the spherical nature of the Earth, but here we've considered a flat plane for simplicity.

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An additional force of 33.73 lb must be applied to the left to cancel out the initial force and create a resultant force of zero, according to Newton's first law of motion.

To ensure that the resultant force on the object is zero, another force of equal magnitude but in the opposite direction must be applied. In this case, since a force of 33.73 lb is directed toward the right, an additional force must be applied to the left with the same magnitude, i.e., 33.73 lb to the left. This balances out the forces acting upon the object, creating a state of equilibrium where the sum of the forces equals zero, thus satisfying Newton's first law of motion.

Now, considering the provided example where a man applies a force of +50 N, it's clear that the force of friction opposing the motion must also be 50 N but in the negative direction (-50 N) for the forces to cancel each other and fulfill the condition of Newton’s first law where Fnet = 0.

In the context of three forces acting on an object, discussing the necessary condition for an object to move straight down, the forces in the horizontal plane (to the right and left) must be balanced. This means that force A and force B, which act in opposite directions, must be equal in magnitude.

Answer:

Explanation:

First, we will find the Cartesian Representation of the [tex]\vec{X}[/tex] and [tex]\vec{Y}[/tex] vectors. We can do this, using the formula

[tex]\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]

where [tex]| \vec{A} |[/tex] its the magnitude of the vector and θ the angle. For  [tex]\vec{X}[/tex] we have:

[tex]\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )[/tex]

[tex]\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )[/tex]

where the unit vector [tex]\hat{i}[/tex] points east, and [tex]\hat{j}[/tex] points north. Now, the [tex]\vec{Y}[/tex] will be:

[tex]\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )[/tex]

Now, taking the sum:

[tex]\vec{X} + \vec{Y} + \vec{Z} = 0[/tex]

This is

[tex]\vec{Z} = - \vec{X} - \vec{Y}[/tex]

[tex](Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )[/tex]

[tex](Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 2200 m \ - \ 956.86 m \ )[/tex]

[tex](Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 1243.14 m\ )[/tex]

Now, for the magnitude, we just have to take its length:

[tex]|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}[/tex]

[tex]|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}[/tex]

[tex]|\vec{Z}| = 1635.43 m[/tex]

For its angle, as the vector lays in the second quadrant, we can use:

[tex]\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m})  [/tex]

[tex]\theta = 180\° - arctan( -1.1720)  [/tex]

[tex]\theta = 180\° - 45\°31'40'' [/tex]

[tex]\theta = 130\°28'20''  [/tex]

Final answer:

To estimate the electric field near a uniformly charged wire, one must determine the linear charge density and integrate the contributions from infinitesimal charge elements along the wire, considering the symmetry that leads to the cancellation of horizontal components.

Explanation:

To estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire, we use the principle of superposition and integrate the contributions of the infinitesimally small charge elements along the wire.

First, we determine the linear charge density λ by dividing the total charge Q by the length L of the wire:

λ = Q / L

Then, we apply the formula for the electric field due to an infinitesimal charge element dQ at a distance r:

dE = (k * dQ) / r²

Where k is the Coulomb's constant (approximately 8.99 × 10⁹ Nm²/C²). Since we need to integrate over the length of the wire to find the total electric field, we use the linear charge density λ and a differential element of the wire's length dl:

dQ = λ * dl

The resulting integral, when evaluated, will give us the magnitude of the electric field E at the specified distance from the wire.

It is important to note that due to the symmetry of the problem, the horizontal components of the electric field due to charge elements on the wire will cancel out, leaving only the vertical components to contribute to the total electric field at the point of interest