An air traffic controller notices two signals from two planes on the radar monitor. One plane is at altitude 1162 m and a 10.1-km horizontal distance to the tower in a direction 34.2° south of west. The second plane is at altitude of 4162 m and its horizontal distance is 9.5 km directed 21.5° south of west. What is the distance between these planes in kilometers?

Answer:

Explanation:

Point charge is Placed at the center of a spherical Gaussian surface and according to Gauss law electric flux through Gaussian surface is given by [tex]\frac{1}{\epsilon _0}[/tex] times the charge enclosed

where [tex]\epsilon _0=[/tex]permittivity of free space

So unless we take charge outside of the surface flux will not change for the Gaussian surface

So Statement A,B,C and E are false

Option D is correct because we take the charge outside of the Gaussian surface                    

The electric flux depends on the charge enclosed by the Gaussian surface and the location of the charges. If the volume of the surface changes, but the charge remains inside, the flux remains the same. If the charge is moved off-center or outside the surface, or if a second charge is added outside the surface, the flux will change.

The electric flux depends on the charge enclosed by the Gaussian surface. So, if the sphere is replaced by a cube of the same volume with the charge at the center, the electric flux will not change. However, if the cube is one-tenth the volume of the original sphere, the electric flux will decrease by a factor of 10, since the charge is still at the center.

If the point charge is moved off center, but still inside the original sphere, the electric flux will not change, as long as the charge remains enclosed by the Gaussian surface.

If the point charge is moved to just outside the sphere, the electric flux will increase, as some electric field lines are now passing through the Gaussian surface.

If a second point charge is placed just outside the sphere, the electric flux through the Gaussian surface will increase even more, as the electric field lines from both charges are now passing through the surface.

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Answer:

Distance traveled by projectile in horizontal direction will be 4.95 m  

Explanation:

We have given horizontal velocity of the projectile v = 9 m/sec

We have to find the distance traveled by projectile in 0.550 sec in horizontal direction.

We know that distance is equal to multiplication of speed and time

So distance traveled in horizontal direction will be equal to [tex]d=v\times t=9\times 0.550=4.95m[/tex]

So distance traveled by projectile in horizontal direction will be 4.95 m  

Answer:

4.95 m

Explanation:

Horizontal velocity of projectile, u = 9 m/s

time, t = 0.550 s

The horizontal distance, d = horizontal velocity x time

d = 9 x 0.55

d = 4.95 m

Thus, the horizontal distance traveled is 4.95 m.

Answer: 3.35 m/s

Explanation:

This problem is related to the Conservation of Momentum principle, which states the following:

"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision will be the same as the total momentum of these same two objects after the collision".

Of course, the momentum of each object may change after the collision, however, the total momentum of the system does not change.  In addition, according to this law it is established that the initial momentum [tex]p_{o}[/tex] must be equal to the final momentum [tex]p_{f}[/tex]:

[tex]p_{o}=p_{f}[/tex] (1)

Where:

[tex]p_{o}=m_{1}V_{1}+m_{2}V_{2}[/tex] (2)

[tex]p_{f}=m_{1}U_{1}+m_{2}U_{2}[/tex] (3)

Being:

[tex]m_{1}=282 kg[/tex] the mass of the first car

[tex]V_{1}=3.70 m/s[/tex] the initial velocity of the first car

[tex]m_{2}=155 kg[/tex] the mass of the second car

[tex]V_{2}=-1.38 m/s[/tex] the initial velocity of the second car

[tex]U_{1}=1.10 m/s[/tex] the final velocity of the first car

[tex]U_{2}[/tex] the final velocity of the second car

Substituting (2) and (3) in (1):

[tex]m_{1}V_{1}+m_{2}V_{2}=m_{1}U_{1}+m_{2}U_{2}[/tex] (4)

Isolating [tex]U_{2}[/tex]:

[tex]U_{2}=\frac{m_{1}(V_{1}-U_{1})+m_{2}V_{2}}{m_{2}}[/tex] (5)

Solving:

[tex]U_{2}=\frac{282 kg(3.70 m/s-1.10 m/s)+(155 kg)(-1.38 m/s)}{155 kg}[/tex] (6)

Finally:

[tex]U_{2}=3.35 m/s[/tex]

Answer:

3.35

Explanation:

Answer:

(a) [tex]t=\frac{H}{v_0}[/tex]

(b) [tex]H=\frac{v_0^2}{g}[/tex]

Explanation:

Let the two balls collide at a height x from the ground. Therefore, ball 2 travels a distance of (H-x) before colliding with ball 1.

Using the following Newton's law of motion,

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

where,

[tex]S[/tex] = displacement

[tex]u[/tex] = initial velocity

[tex]a[/tex] = acceleration

[tex]t[/tex] = time

we can write the equations of motion of the two balls(ball 1 and ball 2 respectively):

[tex]x=v_0t-\frac{1}{2}gt^2[/tex]     ......(1)   ([tex]a=-g[/tex], ball is moving against gravity)

[tex]H-x=\frac{1}{2} gt^2[/tex]      .......(2)    (initial velocity is zero; [tex]a=+g[/tex])

Substituting [tex]x[/tex] from equation (1) in (2),

[tex]H-v_0t+\frac{1}{2}gt^2=\frac{1}{2}gt^2[/tex]

or, [tex]t=\frac{H}{v_0}[/tex]      ......(a)

(b) Now, it is said that the collision will occur when ball 1 is at it's highest point. That is, it's final velocity must be zero.

This time we shall have to use another equation of motion given by,

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

where, [tex]v[/tex] = final velocity

therefore, we get for ball 1,

[tex]0=v_0^2-2gx[/tex]       ([tex]u=v_0,v=0,a=-g[/tex])

or, [tex]x=\frac{v_0^2}{2g}[/tex]

Putting the value of [tex]x[/tex] in equation (2) and rearranging, we get,

[tex]\frac{g}{2v_0^2}H^2-H+\frac{v_0^2}{2g}=0[/tex]

which is a quadratic equation, whose solution is given by,

[tex]H=\frac{+1\pm\sqrt{(-1)^2-(4\times\frac{g}{2v_0^2} \times\frac{v_0^2}{2g}) } }{2\times\frac{g}{2v_0^2} }[/tex]

[tex]=\frac{v_0^2}{g}[/tex]

(a) The time at which the balls collide is H/[tex]v_{0}[/tex]

(b) The height H is equal to [tex]\frac{v_{0} ^{2} }{g}[/tex]

Let the balls collide at a height x above the ground.

Then the distance traveled by the ball thrown above is x.

And the distance traveled by the ball dropped from height H is (H-x).

(i) Both the balls will take the same time to travel respective distances in order to collide.

[tex]H-x=\frac{1}{2}gt^{2}[/tex]

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

We get:

[tex]x=v_{0}t-(H-x)[/tex]

[tex]t=\frac{H}{v_{0}}[/tex] , is the time after which the balls collide.

(ii) Let the ball thrown up attains its maximum height x at the time of thecollision

[tex]v^{2} = u^{2}-2gx[/tex] here v is the final velocity which is 0 when the ball attains maximum height

[tex]0=v_{0} ^{2}-2gx[/tex]

[tex]x=\frac{v_{0} ^{2} }{2g}[/tex] is the maximum height attained.

Now, the ball thrown downward travels distance (H-x) just before collision:

[tex]H-x=\frac{1}{2}gt^{2}[/tex]

[tex]H-\frac{v_{0} ^{2} }{2g}=\frac{1}{2}g\frac{H^{2} }{v_{0} ^{2} }[/tex]

Solving the quadratic equation we get:

[tex]H=\frac{v_{0} ^{2} }{g}[/tex]

Learn more about equations of motion:

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From the work theorem, this is defined as the amount of force applied on an object displaced on a longitudinal unit. Mathematically this is

[tex]W = \vec{F} \times \vec{d}[/tex]

Here,

F = Force vector

d = Displacement vector

Of all the options presented, only in the last one there is no change in distance, so the work done there is zero.

The correct option is 5.

Work in Physics is when a force causes displacement. Of the choices, the example of a runner stretching by pushing against a wall does not involve work, as there is no displacement.

In the context of Physics, 'work' is defined as a force causing displacement on an object. Essentially, work is done when a force acts upon an object to cause or prevent motion. Among the options provided, 5. A runner stretches by pushing against a wall does not involve 'work'. This is because, despite the runner exerting a force against the wall, there is no displacement of the wall in response to this force.

Work (W) is calculated by multiplying the force (F) that is applied to an object and the distance (d) that the object is moved, i.e., W = F * d. In this case, since the displacement (d) is zero, the work done is also zero.

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Explanation:

It is known that inside a sphere with uniform volume charge density the field will be radial and has a magnitude E that can be expressed as follows.

           E = [tex]\frac{q}{4 \p \epsilon_{o}R^{3}}r[/tex]

    V = [tex]-\int_{0}^{r}E. dr[/tex]

        = [tex]-\int_{0}^{r}\frac{q}{4 \p \epsilon_{o}R^{3}}r dr[/tex]

        = [tex](\frac{q}{4 \p \epsilon_{o}R^{3}})(\frac{r^{2}}{2})[/tex]

        = [tex]\frac{-qr^{2}}{8 \pi \epsilon_{o}R^{3}}[/tex]

At r = 1.45 cm = [tex]1.45 \times 10^{-2}[/tex]  (as 1 m = 100 cm)

     V = [tex]\frac{2.50 \times 10^{-15} \times 1.45 \times 10^{-2}}{8 \times 3.1416 \times 8.85 \times 10^{-12} \times 2.36 \times 10^{-2}}[/tex]

         = [tex]6.905 \times 10^{-6}[/tex] mV

Thus, we can conclude that value of V at radial distance r = 1.45 cm is  [tex]6.905 \times 10^{-6}[/tex] mV.

Answer:

a. 1.174 rad[/tex] or 67.3 degree

b. t = 49.28 s

Explanation:

Let [tex]v_v[/tex] be the vertical component of the boat velocity with respect the the river, pointing North. Let [tex]v_h[/tex] be the horizontal component of the boat velocity with respect to the river, pointing West, aka upstream. Since the total velocity of the boat is 4.4m/s

[tex]v_v^2 + v_h^2 = 4.4^2 = 19.36[/tex]

The time it takes for the boat to cross 200m-wide river at [tex]v_v[/tex] rate is

[tex]t = 200 / v_v[/tex] or [tex]v_v = 200 / t[/tex]

This is also the time it takes for the boat to travel 35m upstream, horizontally, at the rate of [tex] v_h - 0.99[/tex] m/s

[tex]t = \frac{35}{v_h - 0.99}[/tex]

[tex]v_h - 0.99 = 35/t[/tex]

[tex]v_h = 35/t + 0.99[/tex]

We can substitute [tex]v_v,v_h[/tex] into the total velocity equation to solve for t

[tex]\frac{200^2}{t^2} + (\frac{35}{t} + 0.99)^2 = 19.36[/tex]

[tex]\frac{40000}{t^2} + \frac{35^2}{t^2} + 2*0.99*\frac{35}{t} + 0.99^2 = 19.36[/tex]

From here we can multiply both sides by [tex]t^2[/tex]

[tex]40000 + 1225 + 69.3t + 0.9801t^2 = 19.36t^2[/tex]

[tex]18.38 t^2 - 69.3t - 41225 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{69.3\pm \sqrt{(-69.3)^2 - 4*(18.3799)*(-41225)}}{2*(18.38)}[/tex]

[tex]t= \frac{69.3\pm1742.31}{36.7598}[/tex]

t = 49.28 or t = -45.51

Since t can only be positive we will pick t = 49.28

[tex]v_h = 35 / t + 0.99 = 35 / 49.38 + 0.99 = 1.7 m/s[/tex]

The angle, relative to the flow of river direction is

[tex]cos(\alpha) = \frac{v_h}{v} = \frac{1.7}{4.4} = 0.3864[/tex]

[tex]\alpha = cos^{-1}(0.3864) = 1.174 rad[/tex] or 67.3 degree

Answer:

Power of the string wave will be equal to 5.464 watt

Explanation:

We have given mass per unit length is 0.050 kg/m

Tension in the string T = 60 N

Amplitude of the wave A = 5 cm = 0.05 m

Frequency f = 8 Hz

So angular frequency [tex]\omega =2\pi f=2\times 3.14\times 8=50.24rad/sec[/tex]

Velocity of the string wave is equal to [tex]v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{60}{0.050}}=34.641m/sec[/tex]

Power of wave propagation is equal to [tex]P=\frac{1}{2}\mu \omega ^2vA^2=\frac{1}{2}\times 0.050\times 50.24^2\times 34.641\times 0.05^2=5.464watt[/tex]

So power of the wave will be equal to 5.464 watt

Answer:

[tex]-70\hat{i} m+19\hat{j}m+21\hat{k} m[/tex]

Explanation:

We are given that

Displacement along x- axis =-[tex]70\hat{i}[/tex] m

Displacement along y-axis=[tex]19\hat{j}[/tex]m

Displacement along z-axis=[tex]21\hat{k} m[/tex]

Where [tex]\hat{i},\hat{j},\hat{k}[/tex] are unit vector along x, y and z-axis

We have to find the displacement from start to end in unit vector notation.

Total displacement from start to end=Displacement along x-axis+displacement along y-axis+displacement along z- axis

Total displacement from start to end=[tex]-70\hat{i} m+19\hat{j}m+21\hat{k} m[/tex]

Hence, the displacement of the sign from start to end=[tex]-70\hat{i} m+19\hat{j}m+21\hat{k} m[/tex]

The aqueous solution with the highest concentration of particles will have the lowest freezing point.

The freezing point of a solution is determined by the concentration of solute particles in the solution. The greater the concentration of particles, the lower the freezing point. In this case, the aqueous solution with the lowest freezing point will have the highest concentration of particles.

Out of the options given, the aqueous solution with the highest concentration of particles is 0.60 m aqueous sucrose. Therefore, aqueous sucrose (0.60 m) will have the lowest freezing point.

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Answer:

The change of the bar's length is [tex] 3.9\times10^{-4} m[/tex]

Explanation:

The bar length is a function of temperature T above room temperature:

[tex] L(T)=1.0000+2.4\times10^{-5} T[/tex]

So, if we evaluate at T= 16.1 C above room temperature

[tex]L(16.1)=1.0000+2.4\times10^{-5} (16.1)[/tex]

[tex]L=1.00039 m [/tex]

Now we can find the change of the bar length with the difference of L and Lo (the length at room temperature)

[tex]L- L_0=1.00039-1.0000 = 3.9\times10^{-4} m[/tex]

To solve this problem we will make a diagram in the Cartesian plane that will allow us to find and understand more accurately the displacement and the angle of rotation.

According to Pythagoras, the distance traveled would be equivalent to

[tex]d = \sqrt{(2.7)^2+(3.5)^2}[/tex]

[tex]d = 4.4 miles[/tex]

The individual had a displacement of 4.4 thousand from the starting point.

Now the angle [tex]\theta[/tex] plus the previously given angle will allow us to find the direction of travel.

[tex]tan\theta = \frac{\text{Opposite side}}{\text{Adjacent side}}[/tex]

[tex]tan\theta = \frac{3.5}{2.7}[/tex]

[tex]\theta = tan^{-1} (\frac{3.5}{2.7})[/tex]

[tex]\theta = 52.35\°[/tex]

[tex]\angle =[/tex] [tex]\theta + 35 = 52.35+35 = 87.35\°[/tex]

Therefore the net direction of the man is S 87.35° W

Answer: Height of building = 17.69m, velocity of brick = 18.6m/s

Explanation: From the question, the body has a zero initial speed, thus initial velocity (u) all through the motion is zero.

By ignoring air resistance makes it a free fall motion thus making it to accelerate constantly with a value of [tex]a = 9.8m/s^{2}[/tex].

Time taken to fall = 1.90s

a)

thus the height of the building is calculated using the formulae below

[tex]H = ut + \frac{1}{2} gt^{2}[/tex]

but u = 0 , hence

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

[tex]H = \frac{1}{2} *9.8* 1.9^{2} \\\\H = 17.69m[/tex]

b)

to get the value of velocity (v) as the brick hits the ground, we use the formulae below

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

but u= 0, hence

[tex]v^{2} = 2gH\\[/tex]

[tex]v^{2} = 2 * 9.8 * 17.69\\v = \sqrt{2 *9.8* 17.69} \\v = 18.62m/s[/tex]

find the attachment in this answer for the accleration- time graph, velocity- time graph and distance time graph

a. The height of building, in meters, is equal to 17.69 meters.

b. The magnitude of the brick’s velocity just before it reaches the ground is equal to 18.72 m/s.

Given the following data:

We know that acceleration due to gravity (a) for an object in free fall is equal to 9.8 meter per seconds square.

a. To determine the height of building, in meters, we would use the second equation of motion:

Mathematically, the second equation of motion is given by the formula;

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

Where:

Substituting the given parameters into the formula, we have;

[tex]S = 0(1.90) + \frac{1}{2} \times 9.8 \times 1.90^2\\\\S = 0 + 4.9 \times 3.61[/tex]

Distance, S = 17.69 meters.

b. To determine the magnitude of the brick’s velocity just before it reaches the ground, we would use the third equation of motion;

[tex]V^2 = U^2 + 2aS[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]V^2 = 0^2 + 2 \times 9.8 \times 17.69\\\\V^2 = 350.26\\\\V = \sqrt{350.26}[/tex]

V = 18.72 m/s

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Final answer:

The distance from the train station where the car catches up with the train is 250 miles.

Explanation:

To find the distance from the train station where the car catches up with the train, we need to determine the time it takes for the car to catch up.

Let's assume the car catches up with the train after t hours. Since the train leaves 2 hours before the car, the train has already been traveling for t + 2 hours.

Distance traveled by the train = speed of the train × time = 50t + 100 miles

Distance traveled by the car = speed of the car × time = 70t miles

Since the car catches up with the train, the distances traveled by both must be equal. Equating the distances, we get:

50t + 100 = 70t

20t = 100

t = 5

Now, substitute the value of t back into one of the distance formulas to find the distance from the train station:

Distance from the train station = speed of the train × time = 50 × 5 = 250 miles

Final answer:

The car catches up with the train in 5 hours, and they are 350 miles from the station when the car catches up.

Explanation:

The problem you presented is related to finding when two objects, moving at different constant speeds, meet. To solve this, we use the concept of relative speed.

The train travels for 2 hours before the car starts, so by that time, the train would have covered a distance of (50 mi/hr × 2 hr) = 100 miles. When the car starts, it has to cover not only the distance to the train but also the additional distance the train covers as the car approaches.

Let's consider the time it takes for the car to catch up with the train to be 't' hours. In that time, the car would travel 70t miles, and the train would travel an additional 50t miles.

Since the car needs to cover the initial 100 miles plus whatever distance the train covers in 't' hours, we can set up the equation:

70t = 100 + 50t

Subtracting 50t from both sides gives us:

20t = 100

Dividing both sides by 20:

t = 5

So, the car catches up with the train in 5 hours. To find out how far from the station they are when the car catches up with the train:

x = 70 mi/hr × 5 hr = 350 miles

Answer:

[tex]\theta=119.8138^{\circ}[/tex]

Explanation:

Given:

a)

From the vector addition rule we know:

[tex]R^2={A^2+B^2+2A.B\cos\theta}[/tex]

where

[tex]\theta=[/tex] angle between the two vectors with their tail at common point.

b)

Putting respective values:

[tex]7.6^2=7.1^2+5^2+7.1\times 5\times \cos \theta[/tex]

[tex]\theta=119.8138^{\circ}[/tex]

To add the two vectors, draw them graphically and connect the tip of one vector to the tail of the other. The angle between the original two vectors can be determined using the adjacent and hypotenuse sides of a right triangle.

To add the two vectors, draw the first vector as a line segment with a length of 7.1 units and draw the second vector as a line segment with a length of 5.0 units. Place the tail of the second vector at the tip of the first vector. The sum of the two vectors is the line segment connecting the tail of the first vector to the tip of the second vector. This sum vector should have a length of 7.6 units, as mentioned in the question.

Using the sketch from Part (a), draw a horizontal line from the tail of the first vector to the tip of the second vector. Next, draw a vertical line from the tip of the first vector to the tip of the second vector. Connect the tail of the first vector to the tip of the second vector to form a right triangle. The angle between the original two vectors is the angle opposite to the horizontal line. To determine this angle, we can use the trigonometric inverse function. Let's call the angle 'theta'. We have the adjacent side as 5.0 units and the hypotenuse as 7.6 units. Using the arccosine function (inverse cosine), we can calculate theta using the equation: theta = arccos(5.0/7.6).

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To solve this problem we will apply the concept related to the Poisson ratio for which the longitudinal strains are related, versus the transversal strains.  First we need to calculate the longitudinal strain as follows

[tex]\epsilon_x = \frac{l_f-l_i}{l_i}[/tex]

[tex]\epsilon_x = \frac{(9.80554)-(9.8)}{9.8}[/tex]

[tex]\epsilon_x = 0.0005653[/tex]

Second we will calculate the lateral strain as follows

[tex]\epsilon_y = \frac{a_f-a_i}{a_i}[/tex]

[tex]\epsilon_y = \frac{2.59952-2.6}{2.6}[/tex]

[tex]\epsilon_y = -0.0001846153[/tex]

The Poisson's ratio is the relation between the two previous strain, then,

[tex]\upsilon = -\frac{\epsilon_y}{\epsilon_x}[/tex]

[tex]\upsilon = -\frac{(-0.0001846153)}{0.0005653}[/tex]

[tex]\upsilon = 0.3265[/tex]

Therefore the Poisson's ratio for the material is 0.3265

Answer:

0.43 s

Explanation:

We have the following parameters:

Initial velocity, u = 7.4 m/s

Acceleration of gravity, g = 9.8 [tex]m/s^2[/tex]

Distance, s = 43 in + 10 ft = 1.092 m + 3.048 m = 4.14 m

Time, t = ?

Using the equation of motion [tex]s=ut +\frac{1}{2}gt^2[/tex], we have

[tex]4.14 = 7.4t + 0.5\times9.8t^2[/tex]

[tex]4.9t^2 + 7.4t - 4.14 =0[/tex]

Using the quadratic formula [tex]\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex] where a = 4.9, b = 7.4 and c = - 4.14, and solving for the positive value of t only, we have

[tex]t = 0.43[/tex] s

Answer:

12J

Explanation:

Kinetic Energy before collision = 1/2mv1^2 = 1/2×2×4^2 = 16J

Velocity after collision (v2) = m1v1/m1+m2 = 2×4/2+2 = 8/4 = 2m/s

Kinetic Energy after collision = 1/2mv2^2 = 1/2×2×2^2 = 4J

Kinetic Energy lost = 16J - 4J = 12J

Answer:

Lost in kinetic energy = 12 J

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')..................................... Equation 1

Where m = mass of first body, u = initial of the first body, m' = mass of the second body, u' = initial velocity of the second body, V = common velocity.

Making V the subject of the equation,

V = mu+m'u'/(m+m')........................... Equation 2

Where m = 2.0 kg, m' = 2.0 kg, u = 4.0 m/s, u' = 0 m/s ( stationary).

Substitute into equation 2

V = (2×4 + 2×0)/(2+2)

V = 8/4

V = 2 m/s.

Total kinetic energy before collision = 1/2mu² = 1/2(2)(2)² = 16 J.

Total Kinetic energy after collision = 1/2V²(m+m') = 1/2(2²)(4) = 4 J.

Thus

Lost in kinetic energy = 16 - 4

Lost in kinetic energy = 12 J

Answer:

The distinction can be understood by their individual definitions given below.

Explanation:

A meteoroid is a small rocky/metallic body that can be found in outer space (space beyond the Earth's atmosphere). Their sizes are much smaller than asteroids (often called planetoids) and even more smaller than that of any planets or their moons. They generally originate from comets, asteroids (fragments of them) and even from planets or moons when there occurs heavy collisions.

A meteor is basically what we know to be "shooting stars". When a meteoroid, asteroid, etc. passes through the Earth's atmosphere, they heat up and begin to glow because of the frictional force experienced due to gas molecules in the atmosphere. But the important thing is that they do not reach the surface of the Earth as they completely burn out long before coming close. If some object does manage to reach the Earth's surface, we then call it a meteorite.

(These definitions are not restricted to the Earth but applies to all panets and moons.)

(Also check the gif provided here: https://en.wikipedia.org/wiki/Meteoroid)

Answer:

The beat frequency is 2 Hz.

Explanation:

Given that,

Frequency of A = 440 Hz

Frequency of E = 659 Hz

Suppose, piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note A should have a frequency of 440 Hz and the note E should be at 659 Hz.

We need to calculate the third harmonic of A

Using formula of harmonic

[tex]f_{3}=n\times f[/tex]

Put the value into the formula

[tex]f_{3} =3\times440[/tex]

[tex]f_{3}=1320\ Hz[/tex]

We need to calculate the second harmonic of E

Using formula of harmonic

[tex]f_{2}=n\times f[/tex]

Put the value into the formula

[tex]f_{2} =2\times659[/tex]

[tex]f_{2}=1318\ Hz[/tex]

We need to calculate the beat frequency

Using formula of beat frequency

[tex]f_{b}=f_{3}-f_{2}[/tex]

Put the value into the formula

[tex]f_{b}=1320-1318[/tex]

[tex]f_{b}=2\ Hz[/tex]

Hence, The beat frequency is 2 Hz.

The beat frequency indicating proper tuning of the E string relative to the A string tuned to 440 Hz should be zero. This corresponds to a situation where the harmonics of both strings match exactly, with the second harmonic of the E string (659.26 Hz) aligning with the third harmonic of the A string (660 Hz), reducing the beat frequency to zero.

To determine the beat frequency that indicates proper tuning for the E string of a guitar, we must consider the fundamental frequencies of the A and E strings. The A string is tuned to 440 Hz. The E string above it should be tuned to a fundamental frequency of 329.63 Hz.

When the A and E strings are played together, harmonics of these frequencies are heard. The second harmonic of the E string would be 2 * 329.63 Hz = 659.26 Hz. This is very close to the 660 Hz, which is the third harmonic of the A string (3 * 440 Hz = 1320 Hz). For the strings to resonate without beats, the harmonics should match as closely as possible. Any difference in these harmonic frequencies would cause beats corresponding to the frequency difference.

Therefore, if the second harmonic of the E string (659.26 Hz) matches the third harmonic of the A string (660 Hz) closely, there will be no beats. In this ideal case, the beat frequency would be 660 Hz - 659.26 Hz = 0.74 Hz. However, for the E string to be perfectly in tune, we would want the beat frequency to be zero, indicating no difference between the expected harmonic frequency and the actual harmonic frequency.

Explanation:

A.

Displacement,S; Θ = S/r

= 180/360 * 2π * 3500

= 10995.6 m

Θ = 10995.6/3500

= 3.142

B.

Angular speed,w = Θ/t

= 3.142/1.503102

= 2.09 rad/s

Velocity = w * r

= 2.09 × 3500

= 7315 m/s.

C.

The same as B.

The airplane's Motion in a Circle displacement is 7.0km, its average velocity is 0.0467 km/s, and its average speed is 0.0737 km/s.

The airplane is flying half a circle, which is essentially semi-circular motion. Under this condition, we can calculate the values asked in the question as follows:

(a) Displacement: Displacement is a vector quantity and represents the shortest distance between the initial and the final points of an object's path. But since the airplane is rounding half the circle, the displacement is essentially the diameter of the circular path. The diameter will be two times the radius, therefore 2*3.50km = 7.00 km.

(b) Average velocity: Average velocity is the total displacement divided by the total time. So, the average velocity would be 7.00km / (1.50*10^2) s = 0.0467 km/s.

(c) Average speed: Average speed is defined as the total distance traveled by the object divided by the total time taken. Here, the airplane travels half the circumference of the circle in the given time. The formula for the circumference of a circle is 2*pi*r, so half the circle's circumference will be pi*3.50 km. Then, Average speed = (pi*3.50 km) / (1.50*10^2) s = 0.0737 km/s.

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To find the distance from the line where the electric field strength is 2000 N/C, we can use the formula for the electric field of an infinitely long charged wire. The distance from the line where the electric field strength is 2000N/C is d = lambda / 2 x 10^6 m.

To find the distance from the line where the electric field strength is 2000 N/C, we can use the formula for the electric field of an infinitely long charged wire:

E = k * (lambda / d)

Where E is the electric field strength, k is the Coulomb's constant, lambda is the charge density of the wire, and d is the distance from the wire. In this case, since the line of charge is infinite, the charge density is simply the charge per unit length.

To solve for the distance d, we can rearrange the formula: d = k * (lambda / E)

Plugging in the given values, the distance d is:

d = (9.0 x 10^9 Nm^2/C^2) * (lambda / 2000 N/C)

So, the distance from the line where the electric field strength is 2000 N/C is d = lambda / 2 x 10^6 m.

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Answer:

[tex] v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s[/tex]

Explanation:

For this case we know that the initial velocity is given by:

[tex] v_i = 7 \frac{m}{s}[/tex]

The final velocity on this case is given by:

[tex] v_f = 13 \frac{m}{s}[/tex]

And we know that it takes 8 seconds to go from 7m/s to 13m/s. We can use the following kinematic formula in order to find the acceleration during the first interval:

[tex] v_f = v_i +at[/tex]

If we solve for the acceleration we got:

[tex] a = \frac{v_f -v_i}{t} = \frac{13 m/s -7 m/s}{8 s}= 0.75 \frac{m}{s^2}[/tex]

So for the other traject we assume that the acceleration is constant and the train travels for 16 s. The initial velocity on this case would be 13m/s from the first interval and we can find the final velocity with the following formula:

[tex] v_f = v_ i +a t[/tex]

And if we replace we got:

[tex] v_f = 13m/s + 0.75 \frac{m}{s^2} * 16 s= 13 m/s +12m/s = 25 m/s[/tex]

Answer

given,

length of tube = 120 cm

a) Open at both ends.

distance between two successive nodes or anti nodes = λ/2

for first possibility =  λ/2 = 120

                          λ  =2 x 120 = 240 cm

for second Possibility, distance between two node or antinode acting symmetrically

                       λ = 120 cm

for three nodes in the tube, distance between them is equal to

[tex]\dfrac{3}{2}\lambda = 120[/tex]

[tex]\lambda = 80 cm[/tex]

b) Open at one end

 First possibility, The distance between one node or anti node

  [tex]\dfrac{\lambda}{4} = 120[/tex]

             λ = 480 cm

 Second Possibility, distance between two node or anti node is equal to

[tex]\dfrac{3}{4}\lambda = 120[/tex]

    λ = 160 cm

Third possibility, distance between three nodes and three anti nodes is equal to

[tex]\dfrac{5}{4}\lambda = 120[/tex]

   λ = 100 cm

Answer:

130.165636364°C

Explanation:

P = Pressure

V = Volume

n = Number of moles

R = Gas constant = 0.082 L atm/mol K

From ideal gas law we have

[tex]PV=nRT\\\Rightarrow n=\dfrac{PV}{RT}\\\Rightarrow n=\dfrac{22\times 25}{0.082\times (25+273.15)}\\\Rightarrow n=22.496451696\ moles[/tex]

[tex]PV=nRT\\\Rightarrow T=\dfrac{PV}{nR}\\\Rightarrow T=\dfrac{48\times 15.5}{22.496451696\times 0.082}\\\Rightarrow T=403.315636364\ K[/tex]

The initial temperature is [tex]403.315636364-273.15=130.165636364\ ^{\circ}C[/tex]

Answer: 130 degrees Celsius

Explanation:

P1V1 / T1 = P2V2 / T2

Let the subscript 1 represent the initial 15.5L of gas and the subscript 2 represent the gas at the final volume of 25L. Rewrite the temperature in Kelvin by adding 273.

P1=48atm, V1=15.5L, P2=22atm, V2=25L, T2=298K, and T1 is unknown.

Substitute the known values into the combined gas law equation.

P1V1 / T1=P2V2 / T2 → (48atm)(15.5L) / T1 = (22atm)(25L) / 298K

Solve the equation for T1, simplify, and round to the nearest degree.

T1 = P1V1T2 / P2V2

T1 = (48atm)(15.5L)(298K)(22atm)(25L)

T1 = 403K

To get the temperature in Celsius, subtract 273.

T1 = 130∘C

Answer:

The rate at which radar must rotate is 0.335 rad/s.

Explanation:

Given that,

Velocity = 65 m/h = 29.0576 m/s

Angle = 15°

Suppose, the radius given by

[tex]r=(100\cos2\theta)\ m[/tex]

We need to calculate the rate at which radar must rotate

Using formula of linear velocity

[tex]v=r\omega[/tex]

[tex]\omega=\dfrac{v}{r}[/tex]

Where, v = velocity

r = radius

Put the value into the formula

[tex]\omega=\dfrac{29.0576}{100\cos30}[/tex]

[tex]\omega=0.335\ rad/s[/tex]

Hence, The rate at which radar must rotate is 0.335 rad/s.

Answer:

a) [tex]\Delta s=5\ m[/tex] is the distance between deer and the vehicle

b) [tex]u'=22.36\ m.s^{-1}[/tex] is the maximum speed the driver can be at and still not hit the deer.

Explanation:

Given:

a)

Now the distance travelled after application of the brakes till the vehicle stops:

[tex]v^2=u^2+2a_m.s[/tex]

(assuming that the brakes are applied with maximum acceleration)

where:

[tex]s=[/tex] displacement of the vehicle after braking till it stops

[tex]v=[/tex] final velocity of the vehicle = 0 (stops)

putting the values:

[tex]0^2=20^2-2\times 10\times s[/tex]

[tex]s=20\ m[/tex]

Now before the application of the brakes 0.5 second is taken to react and the vehicle travels during this time as well.

So, distance covered before applying the brakes:

[tex]s'=u.t'[/tex]

[tex]s'=20\times 0.5[/tex]

[tex]s'=10\ m[/tex]

The distance between the deer and the vehicle:

[tex]\Delta s=x-(s+s')[/tex]

[tex]\Delta s=35-(20+10)[/tex]

[tex]\Delta s=5\ m[/tex]

b)

The maximum speed the driver can have with the vehicle and still not hit the deer is given as:

[tex]v^2=u'^2+2. a_m.(x-s')[/tex]

because s' is the distance covered before braking during the reaction time.

[tex]0^2=u'^2-2\times 10\times (35-10)[/tex]

[tex]u'=22.36\ m.s^{-1}[/tex] is the maximum speed the driver can be at and still not hit the deer.

Using the equations of motion under constant acceleration, a) the distance between the driver and the deer when the car comes to a stop is 5 m and b) the maximum speed the driver could have and still not hit the deer is approximately 23.45 m/s.

The subject of the question, Problem-Solving Strategy 2.1 Motion with constant acceleration, involves using the equations of motion under constant acceleration. Let's break down the problem into two parts:

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Answer:

The net work produced is 30.37 KJ

The efficiency of cycle is 72.3%

Explanation:

For the net work produced, we have the formula:

Work = (Th - Tc)(Sh - Sc)(M)

Where,

Th = higher temperature = 800° C + 273 = 1073 k

Tc = lower temperature = 25° C + 273 = 298 k

Sh = specific entropy at higher temperature

Sc = specific entropy at lower temperature

M = molar mass of air

Using, ideal gas table to find entropy. The table is attached.

therefore,

Work = (1073 k - 298 k)(3.0485235 KJ/kg.k - 1.69528 KJ/kg.k)(0.02896 kg)

Work = (775 k)(1.3532435 KJ/kg.k)(0.02896 kg)

Work = 30.37 KJ

Now, for the efficiency (n), we have a formula:

n = 1 - Tc/Th

n = 1 - (298 k)/(1073 k)

n = 0.723 = 72.3 %

Answer:

Explanation:

Single turn implies that N=1

The rectangle is 6cm (0.06m) wide and 10 cm (0.1m) long.

Current in loop is 3A

Magnetic Field B= 4×10^-3T

To take the maximum torque then, sinθ=1

Then

τ=NiABsinθ

Where N is number of turns, and in this case N is 1

i is current in coil and it this case it is 3Amps.

A is area of of coil

And in this case it is a rectangle

Area of rectangle is Lenght × breadth

A= 0.06×0.1= 0.006m^2

And B is magnetic field given as B=4×10^-3T

And θ is angle between the field and the normal to the coil.

a. Torque parallel to the short side of the loop,

The length of the short side is 6cm=0.06m.

This is actually the maximum possible torque, when the field is in the plane of the loop.

τ=NiABsinθ

τ=1×3×0.006×4×10^-3

τ=0.072 ×10^-3

τ=7.2 ×10^-5Nm

b. Torque parallel to the long side of the loop,

The length of the short side is 10cm=0.1m.

This is actually the maximum possible torque, when the field is in the plane of the loop.

τ=NiABsinθ

τ=3×0.006×4×10^-3

τ=0.072 ×10^-3

τ=7.2 ×10^-5Nm.

c. Torque perpendicular to the plane. When the field is perpendicular to the loop the torque is zero.

If the torque is perpendicular to the plane, we said theta is the angle between the normal and the magnetic field

Then if the torque is perpendicular to the normal, then the angle between the torque and the normal is 0, the sinθ = 0

τ=NiABsinθ

Since sinθ =0

Then,

τ=0Nm

Answer:

a) 8.83*10⁵ m/s  b) 2.80*10⁶ m/s

Explanation:

a) Assuming no other forces acting on the proton, the acceleration on it is produced by the electric field.

By definition, the  force due to the electric field is as follows:

F = q*E = e*E (1)

where e is the elementary charge, the charge carried by only one proton, and is e = 1.6*10⁻¹⁹ C.

According to Newton's 2nd law, this force is at the same time, the product of the mass of the proton, times the acceleration a:

F = mp*a (2)

From (1) and(2), being left sides equal, right sides must be equal too:

[tex]F = e*E = mp*a[/tex]

Solving for a:

[tex]a = \frac{e*E}{mp} =\frac{1.6e-19C*1.36e5N/C}{1.67e-27kg} =1.3e13 m/s2[/tex]

⇒ a = 1.3*10¹³ m/s²

As we have the value of a (which is constant due to the field is uniform), the displacement x, and we know that the initial velocity is 0, in order to get the value of the speed, we can use the following kinematic equation:

[tex]vf^{2} -vo^{2} = 2*a*x[/tex]

Replacing by v₀ = 0, a= 1.3*10¹³ m/s² and  x = 0.03 m, we can find vf as follows:

[tex]vf =\sqrt{2*(1.3e13 m/s2)*0.03m} = 8.83e5 m/s[/tex]

⇒ vf = 8.83*10⁵ m/s

b) We can just repeat the equation from above, replacing x=0.03 m by x=0.3 m, as follows:

[tex]vf =\sqrt{2*(1.3e13 m/s2)*0.3m} = 2.80e6 m/s[/tex]

⇒ vf = 2.80*10⁶ m/s

To find the speed of the proton after traveling a certain distance in a uniform electric field, we can use the equations of motion and the equation for force. By plugging in the known values and solving for acceleration, we can then use the equation for speed to find the desired velocity.

To find the speed of the proton, we can use the equation:

F = ma

where F is the force, m is the mass, and a is the acceleration.

The force on the proton is given by:

F = qE

where q is the charge of the proton and E is the electric field strength.

Plugging in the values, we get:

qE = ma

Solving for acceleration, we have:

a = qE/m

For the first part of the question, with a distance of 3.00 cm, we can use the equation:

v^2 = u^2 + 2as

where v is the final velocity, u is the initial velocity (which is 0), a is the acceleration, and s is the distance.

Plugging in the known values, we have:

v^2 = 0 + 2 * (qE/m) * s

Simplifying:

v = sqrt(2 * (qE/m) * s)

For the second part of the question, with a distance of 30.0 cm, we can use the same equation:

v^2 = u^2 + 2as

Plugging in the known values:

v^2 = 0 + 2 * (qE/m) * s

Simplifying:

v = sqrt(2 * (qE/m) * s)

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