Answer:
Electric field at radius r inside the solid sphere is
[tex]E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C\\[/tex]
Electric field at radius r between inner radius and outer radius of the shell is
E=0 N/C
Explanation:
Given
The radius of the solid sphere is [tex]r_{i}=0.105 \ m\\[/tex]
The charge on the solid sphere is [tex]q_{1}=+4.4\ \mu C[/tex]
The inner radius of the shell is [tex]r_{3}=0.15\ m[/tex]
The outer radius of the shell is [tex]r_{3}=0.242\ m[/tex]
The total charge on the shell is [tex]q_{2}=-1.1\ \mu C[/tex]
PART(A)
The magnitude of electric field at radius r where [tex] r<r_{1}[/tex] \\The volumetric charge density of the solid sphere will be
[tex]\rho=\dfrac{q_{1}}{V}\\ \rho=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}\\[/tex]
The charge enclosed by the radius r inside the solid sphere is
rho=[tex]q_{enc}=\rho\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}}{\dfrac{4}{3}\pi r_{1}^3}\times \dfrac{4}{3}\pi r^3\\q_{enc}=\dfrac{q_{1}r^3}{r_{1}^3}\\[/tex]
According to gauss law
[tex]EA=\dfrac{q_{enc}}{\epsilon_{o}}\\E=\dfrac{\dfrac{q_{1}r^3}{r_{1}^3}}{\epsilon_{o}\times 4\pi r^2}\\E=\dfrac{q_{1}r}{4\pi \epsilon_{o} r_{1}^3}\ N/C\\[/tex]
PART(B)
The electric field at radius r where [tex]r_{2}<r<r_{3}[/tex]
The shell is conducting so due to induction of charge
The charge induced on the inner surface of the shell is equal in magnitude of the total charge on the solid sphere but polarity will be changed because a conducting shell has no net electric field inside the shell
So,
The charge on the inner surface of the shell is
[tex]q_{i}=-q_{1}\\q_{i}=-4.4\ \mu C\\[/tex]
Due to conservation of the charge on the shell
The charge accumulated on the outer surface of the shell is
[tex]q_{o}=q_{2}-q_{i}\\q_{o}=-1.1\ \mu C-(-4.4\ \mu C)\\q_{o}=-1.1\ \mu C+4.4\ \mu C\\q_{o}=3.3\ \mu C\\[/tex]
The charge enclosed by the radius r where [tex]r_{1}<r<r_{2}[/tex]
[tex]q_{enc}=q_{1}+q_{i}\\q_{enc}=4.4\times 10^{-6}-4.4\times 10^{-6}\\ q_{enc}=0\\[/tex]
According to gauss law
[tex]EA=\dfrac{q_{enc}}{\epsilon_{o}}\\ E=0\ N/C\\[/tex]
(a) Inside the aluminum sphere, the electric field is 0 N/C due to electrostatic equilibrium.
(b) Inside the copper shell, for any radius between r2 and r3, the electric field is given by (3.95 x 10⁴) / r² N/C.
To solve this problem, we need to understand the behavior of the electric field in different regions around charged spherical conductors.
(a) Electric Field Inside the Aluminum Sphere
Inside a conductor in electrostatic equilibrium, the electric field is zero. Since the aluminum sphere is a conductor, the electric field inside it (for any radius r < r1) is zero.
Result: The magnitude of the electric field inside the aluminum sphere is 0 N/C.
(b) Electric Field Inside the Copper Shell
For the region inside the copper shell but outside the aluminum sphere (r2 < r < r3), we can apply Gauss's Law. According to Gauss's Law, the electric field at a distance r from the center due to a symmetric charge distribution can be calculated as follows:
Define the Gaussian surface: Here, it will be a spherical surface of radius r where r2 < r < r3.
Calculate the enclosed charge: The total charge enclosed within the Gaussian surface is only the charge of the aluminum sphere, q1 = +4.4 μC, because the copper shell's net charge is enclosed separately.
Apply Gauss's Law: Gauss's law states that the electric flux through a surface is proportional to the enclosed charge:
Equation: E * 4πr² = q1/ε₀
Solve for E:
E = q1 / (4πε₀ r²)
Substitute the values: q1 = 4.4 x 10⁻⁶ C, ε₀ = 8.854 x 10⁻¹² C²/(N·m²), and r is the distance within the copper shell.
[tex]E = \frac{4.4 \times 10^{-6} \, \text{C}}{4\pi \cdot (8.854 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2) \cdot r^2}[/tex]
Simplifying gives:
E = (4.4 x 10⁻⁶) / (1.112 x 10⁻¹⁰ * r²)
E = (3.95 x 10⁴) / r² N/C
Hence, the magnitude of the electric field inside the copper shell, at any point where r2 < r < r3, is (3.95 x 10⁴) / r² N/C.
Metal sphere A has a charge of -Q. An identical metal sphere B has a charge of 2Q. The magnitude of the electric force on B due to A is F. The magnitude of the electric force on A due to B must be(A) 2F(B) F/4(C) F/2(D) F(E) 4F
Answer:
option D ) is correct.
Explanation:
When two charges Q₁ and Q₂ which are unequal in magnitude as well as in nature are at a distance of R then a force F is created between them which is given by the expression
F = k ( Q₁ X Q₂) / R²
This force acts equally on both the charges Q₁ and Q₂ irrespective of the fact that they are unequal charges . In other words both Q₁ and Q₂ experience same force F whose value is given above.
In the given case too charges are unequal in magnitude and in nature . They will attract each other with equal force . So if A experiences force F , B too will experience the same force,
Hence option D ) is correct.
An echo is heard from a building 0.485 s after you shout "hello." How many feet away is the building if the air temperature is 30.6°C?
Answer: 84.63 m
Explanation:
The speed of sound in the air, at [tex]0 \°C[/tex] is [tex]331.5 m/s[/tex], and for each degree Celsius the temperature rises, the speed of the sound increases by [tex]0.6 m/s[/tex]. So, if we estimate the speed of sound at [tex]30 \°C[/tex] it will be [tex]349 m/s[/tex].
On the other hand, the speed of sound [tex]V[/tex] is defined as the distance traveled [tex]d[/tex] in a especific time [tex]t[/tex]:
[tex]V=\frac{d}{t}[/tex]
Where:
[tex]V=349 m/s[/tex] is the speed of sound
[tex]t=\frac{0.485 s}{2}=0.2425 s[/tex] is half the time the sound wave travels since you say "hello", the sound wave hits the building and then returns to you
[tex]d[/tex] is the distance between you and the building
[tex]349 m/s=\frac{d}{0.2425 s}[/tex]
Finding [tex]d[/tex]:
[tex]d=(349 m/s)(0.2425 s)[/tex]
Finally:
[tex]d=84.63 m[/tex]
To find the distance to the building, calculate the speed of sound at 30.6°C, multiply it by the echo time of 0.485 seconds, and divide by 2 to account for the round trip of the sound. Then, convert the distance from meters to feet to get approximately 280.95 feet.
Explanation:To calculate the distance of the building when an echo is heard 0.485 seconds after shouting, we need to consider the speed of sound at the given air temperature of 30.6°C. Since the speed of sound varies with temperature, we use the formula v = v_0 + 0.6t, where v_0 is the speed of sound at 0°C (approximately 331.5 m/s), t is the temperature in Celsius, and v is the speed of sound at t degrees Celsius. At 30.6°C, the speed of sound is v = 331.5 m/s + 0.6(30.6°C), which calculates to roughly 349.86 m/s. Since the echo travels the distance to the building and back, the actual distance to the building is half the total distance the sound wave traveled.
Therefore, the one-way distance to the building is 0.485 s × 349.86 m/s ÷ 2. Converting meters to feet (1 meter = 3.28084 feet), we find that the building is approximately 280.95 feet away.
A boat moves ar 10.0 m/s relative to
thewater. If the boat is in a river where the current is 2.00 m/s,
howlong does it take the boat to make a complete round trip of 1000
mupsteam followed by a 1000- m trip downstream?
Answer: 208.3 s
Explanation:
Hi!
You need to calculate the velocity of the boat relative to the shore, in each trip.
Relative velocities are transformed according to:
[tex]V_{b,s} = V_{b,w} + V_{w,s}\\ V_{b,s} = \text{velocity of boat relative to shore}\\V_{b,w} = \text{velocity of boat relative to water} \\V_{b,w} = \text{velocity of water relative to shore}\\[/tex]
Let's take the upstram direction as positive. Water flows downstream, so it's velocity relative to shore is negative , -2 m/s
In the upstream trip, velocity of boat relative to water is positive: 10m/s. But in the downstream trip it is negatoive: -10m/s
In upstream trip we have:
[tex]V_{b,s} = (10 - 2)\frac{m}{s} = 8 ms[/tex]
In dowstream we have:
[tex]V_{b,s} = (-10 - 2)\frac{m}{s} = -12 ms[/tex]
In both cases the distance travelled is 1000m. Then the time it takes the round trip is:
[tex]T = T_{up} + T_{down} = \frac{1000}{8}s + \frac{1000}{12}s = 208.3 s[/tex]
If a truck is heading south at a velocity of 90m/s for 10 seconds how far did it travel?
Answer:
Truck will be at a distance of 720 m
Explanation:
Speed of the truck is given as 90 m/sec
Time for which truck travel t = 10 seconds
We have to find the distance that how much truck travel in 8 sec with speed of 90 m/sec
We know that distance is given by [tex]distance=velocity\times time[/tex]
So distance traveled by truck in 8 sec will be equal to [tex]=90\times 8 =720m[/tex]
So truck will be at a distance of 720 m
A hydraulic press must exert a force of 20,000N to crush a wrecked car into a convienent block. If the output piston is eight times the area of the input piston, what input force is required?
Answer:
Explanation:
We shall apply Pascal's law here to find the solution .According to this law if we apply a pressure on a particular area over an in-compressible liquid it gets transferred to other point inside or over the surface of the liquid without getting diminished.
in the present case force is applied on a car to crush it which is enormous in magnitude. This force is 20000 N, which is also called output force Let the surface on which car is placed be A₂. We apply input force F₁ at other point over a small area be it A₁.
According to Pascal law ,
20000/ A₂ = F₁ / A₁
F₁ = 20000 X (A₁ / A₂)
= 20000 X 1/8
= 2500 N.
When point charges q = +8.4 uC and q2 = +5.6 uC are brought near each other, each experiences a repulsive force of magnitude 0.66 N. Determine the distance between the charges.ns
Answer:
it is separated by 80 cm distance
Explanation:
As per Coulombs law we know that force between two point charges is given by
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
here we know that
[tex]q_1 = +8.4\mu C[/tex]
[tex]q_2 = +5.6 \mu C[/tex]
force between two charges is given as
[tex]F = 0.66 N[/tex]
now we have
[tex]F = \frac{kq_1q_2}{r^2}[/tex]
[tex]0.66 = \frac{(9\times 10^9)(8.4 \mu C)(5.6 \mu C)}{r^2}[/tex]
[tex]r = 0.8 m[/tex]
so it is separated by 80 cm distance
Final answer:
The distance between the charges is approximately 20.73 cm.
Explanation:
To determine the distance between the charges, we can use Coulomb's Law. Coulomb's Law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is: F = k × (q1 × q2) / r², where F is the force, k is the Coulomb's constant (8.99 x 10⁹ Nm²/C²), q1 and q2 are the charges of the particles, and r is the distance between them.
In this case, we have the force (F) as 0.66 N, q1 as 8.4 uC, q2 as 5.6 uC, and we need to find the distance (r). Plugging these values into the formula, we get:
0.66 = (8.99 x 10⁹) × ((8.4 x 10⁻⁶) × (5.6 x 10⁻⁶)) / r²
Solving for r, we can rearrange the equation to get:
r² = ((8.99 x 10⁹) × ((8.4 x 10⁻⁶) × (5.6 x 10⁻⁶))) / 0.66
r² = 8.99 x 10⁹ x 8.4 x 10⁻⁶ x 5.6 x 10⁻⁶ / 0.66
r² = 8.99 x 8.4 x 5.6 / 0.66
r² = 429.643
r = √429.643
r ≈ 20.73 cm
A conveyor belt is used to move sand from one place to another in a factory. The conveyor is tilted at an angle of 18° above the horizontal and the sand is moved without slipping at the rate of 2 m/s. The sand is collected in a big drum 5 m below the end of the conveyor belt. Determine the horizontal distance between the end of the conveyor belt and the middle of the collecting drum.
Answer:
x = 2.044 m
Explanation:
given data
initial vertical component of velocity = Vy = 2sin18
initial horizontal component of velocity = Vx = 2cos18
distance from the ground yo = 5m
ground distance y = 0
from equation of motion
[tex]y = yo+ V_y t +\frac{1}{2}gt^2[/tex]
[tex]0 = 5 + 2sin18+ \frac{1}{2}*9.8t^2[/tex]
solving for t
t = 1.075 sec
for horizontal motion
[tex]x = V_x t[/tex]
x = 2cos18*1.075
x = 2.044 m
The motion of sand is due to the movement of conveyor belt. The horizontal distance between the end of the conveyor belt and the middle of the collecting drum is 2.044 meters.
What is equation of motion?
The equation of motion is the relation between the distance, velocity, acceleration and time of a moving body.
The second equation of the motion for distance can be given as,
[tex]y=ut+\dfrac{2}{2}gt^2[/tex]
Here, [tex]u[/tex] is the initial body, [tex]g[/tex] is the acceleration of the body due to gravity and [tex]t[/tex] is the time taken by it.
Given information-
The conveyor is tilted at an angle of 18° above the horizontal.
The Sand is moved without slipping at the rate of 2 m/s.
The sand is collected in a big drum 5 m below the end of the conveyor belt.
The horizontal component of the velocity is given as,
[tex]v_y=2\cos 18[/tex]
The vertical component of the velocity is given as,
[tex]v_y=2\sin18[/tex]
Put the value in the above equation as,
[tex]y-y_0=v_yt+\dfrac{1}{2}gt^2[/tex]
[tex]0-5=2\sin18 (t)+\dfrac{1}{2}\times9.8\tiems t^2\\t=1.075\rm sec[/tex]
The horizontal distance between the end of the conveyor belt and the middle of the collecting drum is,
[tex]d=v_xt\\d=2\cos18\times1.075\\d=2.044\rm m[/tex]
Thus, the horizontal distance between the end of the conveyor belt and the middle of the collecting drum is 2.044 meters.
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Two sinusoidal waves traveling in opposite directions interfere to produce a standing wave with the wave function y = (2.00) sin(0.500x) cos(300t) where x and y are in meters and t is in seconds. (a) Determine the wavelength of the interfering waves. m (b) What is the frequency of the interfering wave? Hz (c) Find the speed of the interfering waves.
Answer:
Part a)
[tex]\lambda = 4\pi[/tex]
Part b)
[tex]f = 47.7 Hz[/tex]
Part c)
[tex]v = 600 m/s[/tex]
Explanation:
Part a)
As we know that angular wave number is given as
[tex]k = \frac{2\pi}{\lambda}[/tex]
[tex]k = 0.500[/tex]
[tex]0.500 = \frac{2\pi}{\lambda}[/tex]
[tex]\lambda = \frac{2\pi}{0.500}[/tex]
[tex]\lambda = 4\pi[/tex]
Part b)
As we know that angular frequency is given as
[tex]\omega = 300 rad/s[/tex]
[tex]\omega = 2\pi f[/tex]
[tex]300 = 2\pi f[/tex]
[tex]f = \frac{300}{2\pi}[/tex]
[tex]f = 47.7 Hz[/tex]
Part c)
Speed of the wave is given as
[tex]v = \lambda \times frequency[/tex]
so we have
[tex]v = 4\pi \times 47.7[/tex]
[tex]v = 600 m/s[/tex]
A 7.0-kg bowling ball experiences a net force of 5.0 N.
Whatwill be its acceleration?
Answer:
It will have an acceleration of [tex]0.71 m/s^{2}[/tex].
Explanation:
Newton’s second law is defined as:
[tex]F = ma[/tex] (1)
Equation 1 can be rewritten for the case of the acceleration:
[tex]a = \frac{F}{m}[/tex]
The force and the acceleration are directly proportional (if one increase the other will increase too).
[tex]a = \frac{5.0 N}{7.0 Kg}[/tex]
But 1 N is equivalent to [tex]1 Kg.m/s^{2}[/tex]
[tex]a = \frac{5.0 Kg.m/s^{2}}{7.0 Kg}[/tex]
[tex]a = 0.71 m/s^{2}[/tex]
So a body will have an acceleration as a consequence of the action of a force.
The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of 539-nm photons. What is the energy gap between the ground state and excited state in the laser material?
Answer:
The answer is 2.3 eV
Explanation:
As energy should be conserved, we have to understand that the energy gap between the ground state and excited state is equal to the energy of the emitted photons.
The wavelength of the emitter photons is
[tex]\lambda = 539\ nm = 539 \times 10^{-9}\ m[/tex],
then their energy is
[tex]E = \frac{h c}{\lambda} = 36.855 \times 10^{-20}\ J[/tex]
where h is the Planck constant,
[tex]h = 6.626 \times 10^{-34}\ Js[/tex]
and c is the speed of light,
[tex]c = 2.998 \times 10^{8}\ m/s[/tex].
If we want our answer in electron-volts instead of joules...
[tex]1 J = 6.242 \times 10^{18}\ eV[/tex]
then
[tex]E = 36.855 \times 10^{-20} J = 2.3\ eV[/tex].
The energy gap between the ground state and excited state in the laser material is approximately 3.69 x 10^-19 Joules. This was calculated using the energy equation of photons, considering the given wavelength of 539 nm.
Explanation:The energy gap between the ground state and the excited state in a laser material can be calculated using the energy equation of photons, E=hc/λ, where h is Planck's constant (6.626 x 10^-34 Js), c is the speed of light (3 x 10^8 m/s), and λ is the wavelength (539 x 10^-9 m in this case). Upon calculation, we find the energy to be about 3.69 x 10^-19 Joules. Hence, the energy gap between the ground state and the excited state is 3.69 x 10^-19 Joules.
This calculation entails quantum physics because it is based on the principle that when an electron in an atom moves from an excited state to a ground state, it emits energy in the form of a photon. This emitted photon has a certain wavelength and frequency, in this case, 539 nm.
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A positron has a mass of 9.11 x 10^-31 kg, and charge qp = +e = +1.60 x 10^-19 C. It is moving towards an α particle (qα = +2e, mα = 6.66 x 10^-27 kg) with a speed of 3.00 x 10^6 m/s. At this instant the separation between the two is 2.00 x 10^-10 m. Assume α particle stays at rest. (a) Calculate the speed of positron at 1.00 x 10^-10 m from α particle. (b) What is the separation between the two when positron comes to rest?
Answer:
Part a)
[tex]v = 2.25 \times 10^6 m/s[/tex]
Part b)
[tex]r = 0.72 \times 10^{-10} m[/tex]
Explanation:
As we know that alpha particle remain at rest always so the energy of system of positron and alpha particle will remain constant
So we will have
[tex]\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2[/tex]
here we know that
[tex]q_1 = 1.6 \times 10^{-19} C[/tex]
[tex]q_2 = 2(1.6 \times 10^{-19}) C[/tex]
also we have
[tex]r_1 = 2.00 \times 10^{-10} m[/tex]
[tex]r_2 = 1.00 \times 10^{-10} m[/tex]
[tex]v_1 = 3.00 \times 10^6 m/s[/tex]
[tex]m = 9.11 \times 10^{-31} kg[/tex]
now from above equation we have
[tex]2.304 \times 10^{-18} + 4.0995\times 10^{-18} = 4.608 \times 10^{-18} + \frac{1}{2}(9.11 \times 10^{-31})v^2[/tex]
[tex]v = 2.25 \times 10^6 m/s[/tex]
Part b)
Now when it come to rest then again by energy conservation we can say
[tex]\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2[/tex]
now here final speed will be zero
[tex]2.304 \times 10^{-18} + 4.0995\times 10^{-18} = \frac{(9/times 10^9)(1.6\times 10^{-19})(2\times 1.6 \times 10^{-19})}{r_2}[/tex]
[tex]r = 0.72 \times 10^{-10} m[/tex]
Final answer:
To find the speed of the positron at a certain distance from the alpha particle and the separation distance when the positron comes to rest.
Explanation:
The speed of a positron at a certain distance from an α particle can be found using the conservation of energy. At 1.00 x 10^-10 m from the α particle, the positron's kinetic energy will be equal to the potential energy at that distance. Using this information, we can calculate the speed of the positron as it approaches the α particle.
To find the separation between the two when the positron comes to rest, we can set the final kinetic energy of the positron equal to zero and solve for the separation distance. This will give us the point at which the positron stops moving towards the α particle.
A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and the number of revolutions it makes before stopping. An athlete is holding a 2.5 meter pole by one end. The pole makes an angle of 60 with the horizontal. The mass of the pole is 4 kg. Determine the torque exerted by the pole on the athlete's hand. (The mass of the pole can be assumed to be concentrated at the center of mass.)
Answer:
(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
Explanation:
Given that,
Initial spinning = 50.0 rad/s
Time = 20.0
Distance = 2.5 m
Mass of pole = 4 kg
Angle = 60°
We need to calculate the angular acceleration
Using formula of angular velocity
[tex]\omega=-\alpha t[/tex]
[tex]\alpha=-\dfrac{\omega}{t}[/tex]
[tex]\alpha=-\dfrac{50.0}{20.0}[/tex]
[tex]\alpha=-2.5\ rad/s^2[/tex]
The angular acceleration is -2.5 rad/s²
We need to calculate the number of revolution
Using angular equation of motion
[tex]\theta=\omega_{0}t+\dfrac{1}{2}\alpha t[/tex]
Put the value into the formula
[tex]\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2[/tex]
[tex]\theta=500\ rad[/tex]
The number of revolution is 500 rad.
(II). We need to calculate the torque
Using formula of torque
[tex]\vec{\tau}=\vec{r}\times\vec{f}[/tex]
[tex]\tau=r\times f\sin\theta[/tex]
Put the value into the formula
[tex]\tau=2.5\times4\times 9.8\sin60[/tex]
[tex]\tau=84.87\ N-m[/tex]
Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.
(II). The torque is 84.87 N-m.
What is the average speed in km/h for a car that travels 50.0 km in 40.0 min?
Answer: Speed = 74.63 km/h
Explanation:
We are provided with the following information:
Distance traveled = 50 km
Time taken = 40 minutes = [tex]\frac{40}{60} \ hours[/tex]
Time taken = 0.67 hours
Since, we know that;
[tex]Speed = \frac{Distance}{Time}[/tex]
Speed = [tex]\frac{50}{0.67}[/tex]
Speed = 74.63 km/h
The average speed of the car is 74.63 km/h.
Explanation:The average speed in km/h for a car that travels 50.0 km in 40.0 min can be calculated by converting the time from minutes to hours and then dividing the distance by the time.
Conversion: 40.0 min × (1 hr/60 min) = 0.67 hr
Average speed = distance/time = 50.0 km/0.67 hr = 74.63 km/h
Therefore, the average speed of the car is 74.63 km/h.
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A plane flies 464 km east from city A to city B in 50.0 min and then 831 km south from city B to city C in 1.20 h. For the total trip, what are the (a) magnitude and (b) direction of the plane's displacement, the (c) magnitude and (d) direction of its average velocity, and (e) its average speed? Give your angles as positive or negative values of magnitude less than 180 degrees, measured from the +x direction (east).
Answer:
Explanation:
Displacement = minimum distance between initial and final point
Here displacement = Hypotenuse AC of triangle ABC
= [tex]\sqrt{464^2+831^2}[/tex]
= 951.76 km
Direction of AC
If θ be the angle AC makes with east
Tan θ = 831 / 464 = 1.79
θ = 60.8°
South of east.
c ) average velocity= Displacement / time
Total time =( 50/60 + 1.2 ) hour
2.0333 hour
average velocity = 951.76 km / 2.0333 hour
= 468 km / h
Direction of velocity will be same as the direction of displacement.
e ) speed = distance / time
distance travelled
= 464 + 831 = 1295 km
speed = 1295 / 2.0333
= 636.9 km/ h .
If an object falls twice as far, how do you think the energy acquired changes? halves
doesn't change
doubles
quadruples
Answer:
Doubles
Explanation:
Before starting to fall, the object will have energy in the form of potential energy. This will be the energy that will be transformed into kinetic energy, once the object starts its fall. And the potential energy that the object had before starting to fall, will be equal to the kinetic energy acquired during the fall. The potential energy is given by:
[tex]E_p = m*g*h[/tex]
Where m is the mass of the object, g is the gravity, and h is the distance to the floor. As you can see, The potential energy is proportional to the height of the fall, so if this height doubles, the potential energy and therefore, the kinetic energy acquired will doubles.
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:1.44 s,10.17 m
Explanation:
Given
two balls are separated by a distance of 41.1 m
One person drops the ball from a height of 41.1 and another launches a ball with velocity of 41.1 m exactly at the same time.
Let the ball launches from ground travels a distance of x m in t sec
For Person on window
[tex]41.1-x=ut+\frac{1}{2}gt^2[/tex]
[tex]41.1-x=0+\frac{1}{2}\times 9.81\times t^2--------1[/tex]
For person at ground
[tex]x=v_ft-\frac{1}{2}gt^2---------2[/tex]
add (1) & (2)
[tex]41.1=v_ft-\frac{1}{2}gt^2+\frac{1}{2}gt^2[/tex]
[tex]41.1=v_ft[/tex]
and [tex]v_f[/tex] is given by
[tex]v_f=\sqrt{2\times 9.81\times 41.1}=28.39 m/s[/tex]
[tex]t=\frac{41.1}{28.39}=1.44 s[/tex]
Substitute value of t in equation 1
[tex]41.1-x=0+\frac{1}{2}\times 9.81\times 1.44^2[/tex]
41.1-x=10.171 m
Thus the two ball meet at distance of 10.17 m below the window.
Two rockets are flying in the same direction and are side by side at the instant their retrorockets fire. Rocket A has an initial velocity of +5300 m/s, while rocket B has an initial velocity of +8700 m/s. After a time t both rockets are again side by side, the displacement of each being zero. The acceleration of rocket A is -11 m/s^2. What is the acceleration of rocket B?
Answer:[tex]-18.05 m/s^2[/tex]
Explanation:
Given
Initial velocity of rocket A ([tex]v_A[/tex]) 5300m/s
Initial velocity of rocket B([tex]v_B[/tex]) 8700 m/s
after time t they have displacement=0
[tex]s_A=5300\times t+\frac{1}{2}\times \left ( -11\right )t^2[/tex]
[tex]0=5300\times t+\frac{1}{2}\times \left ( -11\right )t^2[/tex]
[tex]t=\frac{5300\times 2}{11}=963.63 s[/tex]
for Rocket B
[tex]s_B=8700\times t+\frac{1}{2}\times \left ( a_B\right )t^2[/tex]
time is same for A & B
[tex]0=8700\times 963.63+\frac{1}{2}\times \left ( a_B\right )963.63^2[/tex]
[tex]a_B=\frac{2\times 8700}{963.63}=-18.05 m/s^2[/tex]
The mass of a muon is 207 times the electron mass; the average lifetime of muons at rest is 2.21 μs. In a certain experiment, muons moving through a laboratory are measured to have an average lifetime of 7.04 μs. For the moving muons, what are (a) β (b) K, and (c) p? The rest energy of the electron is 0.511 MeV.
Answer:
Given, average lifetime of muons at rest is To = 2.21μs
Average lifetime of moving muons, T = 7.04μs
mass of muon = 207 mass of electron
a)
rest mass energy of an electron is 0.511 MeV.
[tex] T =\frac{T_o}{\sqrt{1-\beta^2}}[/tex]
Substitute the values:
[tex]7.04 =\frac{2.21}{\sqrt{1-\beta^2}}\\ \beta^2 = 0.90146\\ \beta = 0.95[/tex]
b) K = (γ-1) m₀c²
[tex]\gamma = \frac{1}{\sqrt{1-\beta^2}} = 3.185[/tex]
K = (3.185-1) (207 ×9.1×10⁻³¹) (3×10⁸)² = 3.7×10⁻¹¹J
c) p = γmv = (3.185)((207 ×9.1×10⁻³¹)(0.95×3×10⁸)= 1.71×10⁻¹⁹ N/s
A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2 = 80 m at t2 = 5.0 s . Answer the following by considering ratios, without computing the car's velocity. Part A What is the car's position at
t = 2.5 s ?
Part B What will be its position at
t = 15 s ?
Answer:
Part A) x=40m at t= 2.5s
Part B) x=240m at t=15 s
Explanation:
Data:
x₁=0 , t₁=0
x₂=80m, t₂=5 s
t₃= 2.5 s
t₄= 15 s
problem development
We set the ratios x / t:
Part A)
[tex]\frac{x_{3} }{t_{3} } =\frac{x_{2} }{t_{2} }[/tex]
[tex]\frac{x_{3} }{2.5} =\frac{80}{5 }[/tex]
x₃=16*2.5
x₃=40m
Part B)
[tex]\frac{x_{4} }{t_{4} } =\frac{80 }{5 }[/tex]
[tex]\frac{x_{4} }{15} =\frac{80}{5 }[/tex]
x₄=16*15
x₄=240m
Final answer:
The car's position at t = 2.5 s will be 40 m as it is half of the interval to reach 80 m. At t = 15 s, three times the interval, the car will be at 240 m assuming constant velocity.
Explanation:
Given that the car moves from x1 = 0 m at t1 = 0 s to x2 = 80 m at t2 = 5.0 s, we can determine its position at other times under the assumption of constant velocity. Because the velocity is constant, the ratios of times to positions are constant as well.
Part A: Car's Position at t = 2.5 s
The time t = 2.5 s is exactly half of the time interval given (5.0 s), so the position should be half of 80 m, which is 40 m.
Part B: Car's Position at t = 15 s
To find the position at t = 15 s, let's first determine the time ratio. The ratio of 15 s to 5.0 s is 3:1. Since the velocity is constant, the positions scale with time. So, the position x at t = 15 s is three times 80 m, giving us 240 m.
A protein molecule in an electrophoresis gel has a negative charge. The exact charge depends on the pH of the solution, but 30 excess electrons is typical. What is the magnitude of the electric force on a protein with this charge in a 1700 N/C electric field?
Express your answer in newtons.
Answer:
Electric force, [tex]F=8.16\times 10^{-15}\ N[/tex]
Explanation:
A protein molecule in an electrophoresis gel has a negative charge. Electric field, E = 1700 N/C
There are 30 excess electrons. The charge on 30 electrons is, q = 30e
[tex]q=30\times 1.6\times 10^{-19}\ C=4.8\times 10^{-18}\ C[/tex]
The electric force in terms of electric field is given by :
[tex]F=q\times E[/tex]
[tex]F=4.8\times 10^{-18}\ C\times 1700\ N/C[/tex]
[tex]F=8.16\times 10^{-15}\ N[/tex]
So, the magnitude of the electric force on a protein is [tex]8.16\times 10^{-15}\ N[/tex]. Hence, this is the required solution.
Answer:
8.16 x 10^-15 N
Explanation:
Number of excess electrons = 30
Electric field, E = 1700 N/C
Charge of one electron, e = 1.6 x 10^-19 C
Total charge, q = charge of one electron x number of electrons
q = 30 x 1.6 x 10^-19 = 48 x 10^-19 C
The relation between the electric field and the force is given by
F = q E
F = 48 x 10^-19 x 1700
F = 8.16 x 10^-15 N
Over a time interval of 1.71 years, the velocity of a planet orbiting a distant star reverses direction, changing from +17.3 km/s to -22.8 km/s. Find (a) the total change in the planet's velocity (in m/s) and (b) its average acceleration (in m/s2) during this interval. Include the correct algebraic sign with your answers to convey the directions of the velocity and the acceleration.
Answer:
a)40100m/s
b)-4.348x10^- m/s^2
Explanation:
to calculate the change in the planet's velocity we have to rest the speeds
ΔV=-22.8-17.3=-40.1km/s=40100m/s
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)\\\\
{Vf^{2}-Vo^2}/{2.a} =X(2)\\\\
X=Xo+ VoT+0.5at^{2} (3)\\
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem we have to convert the time interval ins seconds, we know that a year has 53926560s
t=1.71years=53926560*1.71=92214417.6
then we can use the ecuation number 1 to calculate the aceleration
Vf=-22.8km/s
Vo=17.3km/s
Vf=Vo+at
a=(vf-vo)/t
a=(-22.8-17.3)/92214417.6
a=-4.348x10^-7 km/s^2=-4.348x10^- m/s^2
Final answer:
The total change in velocity is -40,100 m/s, and the average acceleration is calculated using the formula a = (final velocity - initial velocity) / time.
Explanation:
The total change in the planet's velocity:
To find the total change in velocity, we subtract the initial velocity from the final velocity: -22.8 km/s - 17.3 km/s = -40.1 km/s. Converting this to m/s, we have -40,100 m/s.
The average acceleration of the planet:
Acceleration is the rate of change of velocity over time. Using the given time interval, we can calculate the average acceleration: a = (final velocity - initial velocity) / time. Plugging in the values, we get a = (-40,100 m/s - 17,300 m/s) / 1.71 yrs.
A football quarterback throws a football. After it leaves his hand, what forces are acting on the football? Choose all that apply: The drag force from the air The normal force from the ground A thrust force The weight force from the Earth The force from the quarterback's hand
Answer:
The force of weight and drag force of the air
Explanation:
This is because when the ball leaves the hand the only force acting on it is the gravity of the earth, therefore its weight, since the force applied by the quarterback is no longer being applied to the ball. You can also consider the drag force of the air in the ball but in some cases it is minimal and can be neglected.
The forces acting on the ball as it leaves the hand of the quarterback are;
the drag force from the air, and the weight force from the Earth.The upward motion of an object thrown upwards is reduced by the opposing force. These opposing forces act in opposite direction to the motion of the object.
When the football is thrown upward, the force from the quarterback provides the thrust force with the ball moves upwards, after the ball leaves the hand of the quarterback the upward motion is reduced by opposing forces.
These opposing forces include the following;
the drag force from the air the weight of the ball due to gravitational force of EarthThus, form the given options, the forces acting on the ball as it leaves the hand of the quarterback are the drag force from the air and the weight force from the Earth.
Learn more here: https://brainly.com/question/14698012
A block of 250-mm length and 48 × 40-mm cross section is to support a centric compressive load P. The material to be used is a bronze for which E = 95 GPa. Determine the largest load that can be applied, knowing that the normal stress must not exceed 80 MPa and that the decrease in length of the block should be at most 0.12% of its original length.
Answer:
153.6 kN
Explanation:
The elastic constant k of the block is
k = E * A/l
k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m
0.12% of the original length is:
0.0012 * 0.25 m = 0.0003 m
Hooke's law:
F = x * k
Where x is the change in length
F = 0.0003 * 729.6*10^6 = 218.88 kN (maximum force admissible by deformation)
The compressive load will generate a stress of
σ = F / A
F = σ * A
F = 80*10^6 * 0.048 * 0.04 = 153.6 kN
The smallest admisible load is 153.6 kN
In this exercise we have to use the knowledge of force to calculate that:
[tex]153.6 kN[/tex]
Then from the formula of the elastic constant we get that:
[tex]k = E * A/l\\k = 95*10^9 * 0.048*0.04/0.25 = 729.6 MN/m[/tex]
With that, now using Hooke's law we find that:
[tex]F = x * k\\F = 0.0003 * 729.6*10^6 = 218.88 kN[/tex]
The compressive load will generate a stress of
[tex]\sigma = F / A\\F = \sigma * A\\F = 80*10^6 * 0.048 * 0.04 = 153.6 kN[/tex]
See more about force at brainly.com/question/26115859
A coin is tossed vertically upward from the edge of the top of a tall building. The coin reaches its maximum height above the top of the building 1.29 s after being launched. After barely missing the edge of the building as it falls downward, it strikes the ground 6.30 s after it was launched. How tall is the building? We assume an answer in meters
Answer:
115 m
Explanation:
The coin is in free fall, subject only to the acceleration of gravity. We set up a frame of reference with the 0 at ground and the X axis pointing upwards. We use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
In this case X0 will be the height of the building as it is the starting point of the coin.
We have one equation with 2 unknowns (V0 and X0), we need another equation.
The equation for speed under constant acceleration is:
V(t) = V0 + a * t
We know that at t = 1.29 s it will reach the highest point, at this point speed is zero.
V0 = V(t) - a * t
V0 = 0 - (-9.81) * 1.29 = 12.65 m/s.
Now we can go back to the other equation:
X(t) = X0 + V0 * t + 1/2 * a * t^2
X0 = X(t) - V0 * t - 1/2 * a * t^2
X0 = 0 - 12.65 * 6.3 - 1/2 * (-9.81) * 6.3^2 = 115 m
Calculate the magnitude of the electric field at one corner of a square 2.42 m on a side if the other three corners are occupied by 4.25×10^−6 C charges. Express your answer to three significant figures and include the appropriate units.
Ans:
12500 N/C
Explanation:
Side of square, a = 2.42 m
q = 4.25 x 10^-6 C
The formula for the electric field is given by
[tex]E = \frac{Kq}{r^2}[/tex]
where, K be the constant = 9 x 10^9 Nm^2/c^2 and r be the distance between the two charges
According to the diagram
BD = [tex]\sqrt{2}\times a[/tex]
where, a be the side of the square
So, Electric field at B due to charge at A
[tex]E_{A}=\frac{Kq}{a^2}[/tex]
[tex]E_{A}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}[/tex]
EA = 6531.32 N/C
Electric field at B due to charge at C
[tex]E_{C}=\frac{Kq}{a^2}[/tex]
[tex]E_{C}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2.42^2}[/tex]
Ec = 6531.32 N/C
Electric field at B due to charge at D
[tex]E_{D}=\frac{Kq}{2a^2}[/tex]
[tex]E_{D}=\frac{9\times10^{9}\times 4.25 \times 10^{-6}}{2\times 2.42^2}[/tex]
ED = 3265.66 N/C
Now resolve the components along X axis and Y axis
Ex = EA + ED Cos 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C
Ey = Ec + ED Sin 45 = 6531.32 + 3265.66 x 0.707 = 8840.5 N/C
The resultant electric field at B is given by
[tex]E=\sqrt{E_{x}^{2}+E_{y}^{2}}[/tex]
[tex]E=\sqrt{8840.5^{2}+8840.5^{2}}[/tex]
E = 12500 N/C
Explanation:
Answer:
[tex]1.250\times 10^4\ N/C.[/tex]
Explanation:
Given:
Charge on each corner of the square, [tex]q=4.25\times 10^{-6}\ C.[/tex] Length of the side of the square, [tex]a = 2.42\ m.[/tex]According to the Coulomb's law, the strength of the electric field at a point due to a charge [tex]q[/tex] at a point [tex]r[/tex] distance away is given by
[tex] E = \dfrac{kq}{r^2}[/tex]
where,
[tex]k[/tex] = Coulomb's constant = [tex]9\times 10^9\ Nm^2/C^2[/tex].
The direction of the electric field is along the line joining the point an d the charge.
The electric field at the point P due to charge at A is given by
[tex]E_A = \dfrac{kq}{a^2}[/tex]
Since, this electric field is along positive x axis direction, therefore,
[tex]\vec E_A = \dfrac{kq}{a^2}\ \hat i.[/tex]
[tex]\hat i[/tex] is the unit vector along the positive x-axis direction.
The electric field at the point P due to charge at B is given by
[tex]E_B = \dfrac{kq}{a^2}[/tex]
Since, this electric field is along negative y axis direction, therefore,
[tex]\vec E_B = \dfrac{kq}{a^2}\ (-\hat j).[/tex]
[tex]\hat j[/tex] is the unit vector along the positive y-axis direction.
The electric field at the point P due to charge at C is given by
[tex]E_C = \dfrac{kq}{r^2}[/tex]
where, [tex]r=\sqrt{a^2+a^2}=a\sqrt 2[/tex].
Since, this electric field is along the direction, which is making an angle of [tex]45^\circ[/tex] below the positive x-axis direction, therefore, the direction of this electric field is given by [tex]\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j)[/tex].
[tex]\vec E_C = \dfrac{kq}{2a^2}\ (\cos(45^\circ)\hat i+\sin(45^\circ)(-\hat j))\\=\dfrac{kq}{2a^2}\ (\dfrac{1}{\sqrt 2}\hat i-\dfrac{1}{\sqrt 2}\hat j)\\=\dfrac{kq}{2a^2}\ \dfrac{1}{\sqrt 2}(\hat i-\hat j).\\[/tex]
Thus, the total electric field at the point P is given by
[tex]\vec E = \vec E_A+\vec E_B +\vec E_C\\=\dfrac{kq}{a^2}\hat i+\dfrac{kq}{a^2}(-\hat j)+\dfrac{kq}{2a^2}\ \dfrac{1}{\sqrt 2}(\hat i-\hat j).\\=\left ( \dfrac{kq}{a^2}+\dfrac{kq}{2\sqrt 2\ a^2} \right )\hat i+\left (\dfrac{kq}{a^2}+\dfrac{kq}{2\sqrt 2\ a^2} \right )(-\hat j)\\=\dfrac{kq}{a^2}\left [\left ( 1+\dfrac{1}{2\sqrt 2} \right )\hat i+\left (1+\dfrac{1}{2\sqrt 2} \right )(-\hat j)\right ]\\=\dfrac{kq}{a^2}(1.353\hat i-1.353\hat j)[/tex]
[tex]=\dfrac{(9\times 10^9)\times (4.25\times 10^{-6})}{2.42^2}\times (1.353\hat i-1.353\hat j)\\=(8.837\times 10^3\hat i-8.837\times 10^3\hat j)\ N/C.[/tex]
The magnitude of the electric field at the given point due to all the three charges is given by
[tex]E=\sqrt{(8.837\times 10^3)^2+(-8.837\times 10^3)^2}=1.250\times 10^4\ N/C.[/tex]
You are taking a picture of a giraffe that is standing far away from you. The image is just too small, so you swap the 60-mm-focal-length lens in your camera for a 960 mm telephoto lens. By what factor does this increase the size of the image?
Explanation:
Given that, you are taking a picture of a giraffe that is standing far away from you. The image is just too small, so you swap the 60-mm-focal-length lens in your camera for a 960 mm telephoto lens such that,
[tex]m_1=960\ mm[/tex]
[tex]m_2=60\ mm[/tex]
We need to find the the factor with which the size of the image increases. It can be calculated by taking ratios of both [tex]m_1\ and\ m_2[/tex]
So, [tex]\dfrac{m_1}{m_2}=\dfrac{960}{60}[/tex]
[tex]\dfrac{m_1}{m_2}=16[/tex]
So, the size of image increases by a factor of 16. Hence, this is the required solution.
An alert physics student stands beside the tracks as a train rolls slowly past. He notes that the frequency of the train whistle is 496 Hz when the train is approaching him and 478 Hz when the train is receding from him. Using these frequencies, he calculates the speed of the train. What value does he find? (Assume the speed of sound in air is 343 m/s.)
Answer:
6.455 m/s
Explanation:
Given;
Frequency of train whistle, f = 496 Hz
Frequency observed, f' = 478 Hz
Speed of sound in air = 343 m/s
Now,
From the instant Doppler's effect, we have the relation
[tex]f = \frac{v+v_s}{v-v_s}\times f'[/tex]
here,
v is the speed of the sound
[tex]v_s[/tex] is the speed of the train
[tex]496 = \frac{343+v_s}{343-v_s}\times478[/tex]
or
1.037 × ( 343 - [tex]v_s[/tex] ) = 343 + [tex]v_s[/tex]
or
355.91 - [tex]v_s[/tex] = 343 + [tex]v_s[/tex]
or
12.91 = [tex]2v_s[/tex]
or
[tex]v_s[/tex] = 6.455 m/s
Abus moves to the cast with a speed 10 ms, and a student in the bus moves with a speed 9 m/s to the cust. What is the velocity of the . relative to the bus A. 19 m/s to the cast B. 19 ms to the west c. 1 m/s to the east D. L us to the west.
Answer : The velocity of the student relative to the bus is 1 m/s to the east.
Explanation :
Relative speed : It is defined as the speed of a moving object when compared to the another moving object.
There are two condition of relative speed.
(1) It two bodies are moving in same direction then the relative speed will be the difference of two bodies.
[tex]\text{Relative speed}=|V_2-V_1|[/tex]
(2) It two bodies are moving in opposite direction then the relative speed will be the sum of two bodies.
[tex]\text{Relative speed}=V_1+V_2|[/tex]
As per question,
The speed of bus, [tex]V_1[/tex] = 10 m/s to east
The speed of student, [tex]V_2[/tex] = 9 m/s to east
In this problem, the two bodies are moving in same direction then the relative speed will be the difference of two bodies.
[tex]\text{Relative speed}=|V_2-V_1|[/tex]
[tex]\text{Relative speed}=|9-10|m/s[/tex]
[tex]\text{Relative speed}=1m/s[/tex]
Therefore, the velocity of the student relative to the bus is 1 m/s to the east.
A car traveling at 45 m/s can brake to a stop within 10m. Assume that the braking force is effective for stop and constant with speed. If the car speeds up to 90 m/s, what is its stopping distance? Provide your explanation using principle of work and energy.
Answer:
40 m
Explanation:
A car travelling at 45 ms has a kinetic energy of
Ec = 1/2 * m * v^2
We do not know the mass, so we can use specific kinetic energy:
ec = 1/2 * v^2
In this case
ec = 1/2 * 45^2 = 1012 J/kg
If it stops in 10 m, the braking force performed a sppecific work of 1012 J/kg in 10 m
L = F * d
F = L / d
F = 1012 / 10 = 101.2 N/kg
If the car is running at 90 m/s, the specific kinetic energy of:
ec2 = 1/2 * 90^2 = 4050 J/kg
With the same braking force the braking distance is:
d2 = L2 / F
d2 = 4050 / 101.2 = 40 m
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) while returning to base camp. He was supposed to travel due north for 5.3 km, but when the snow clears, he discovers that he actually traveled 8.3 km at 50° north of due east. (a) How far and (b) in what direction (south of due west) must he now travel to reach base camp?
Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.