Answer:
The force exerted by three charges on the fourth is [tex]F_{resultant}=2.74\times10^{-5}\ \rm N[/tex]
Explanation:
Given:
The magnitude of three identical charges, [tex]q=10\ \rm nC[/tex]Length of the edge of the square a=3 cmMagnitude of fourth charge ,Q=3 nCAccording to coulombs Law the force F between any two charge particles is given by
[tex]F=\dfrac{kQq}{r^2}[/tex]
where r is the radial distance between them.
Since the force acting on the charge particle will be in different directions so according to triangle law of vector addition
[tex]F_{resultant}=\sqrt ((\dfrac{kQq}{L^2})^2+(\dfrac{kQq}{L^2} })^2)+\dfrac{kQq}{(\sqrt{2}L)^2}\\F_{resultant}=\dfrac{kQq}{L^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=\dfrac{9\times10^9\times10\times10^{-10}\times3\times10^{-9}}{0.03^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=2.74\times 10^{-5}\ \rm N[/tex]
The force on the fourth charge is calculated by first determining the individual forces exerted by each of the three other charges separately using Coulomb's Law and then adding these forces as vectors. This involves resolving each force into its x and y components, combining them separately, and then determining the resultant force's magnitude and direction.
Explanation:The problem here involves Coulomb's Law and the superposition principle in physics. Coulomb's Law defines the force between two point charges as directly proportional to the product of their charges, and inversely proportional to the square of the distance between them.
First, you need to calculate the forces exerted on the fourth charge by each of the three other charges separately. This involves calculating the distance from each existing charge to the fourth charge, then subbing these distances, along with the relevant charge values, into the Coulomb's Law formula. Remember that if the charge is positive (like in the case of charge +q), the force vector points directly from the charge, while if the charge is negative, the force vector points towards the charge.
After calculating the force vectors resulting from each charge, you add these vectors together to get the resultant force vector which is the force exerted on the fourth charge. This problem also involves trigonometry as when you add the force vectors, you have to take into account the direction which each force vector is pointing.
Force due to the positive charge at the lower left: F1 is in the first quadrant
Force due to the positive charge at the lower right: F2 is in the fourth quadrant
Force due to the negative charge at the upper left: F3 is in the third quadrant
In each case, you'll need to resolve each force into its x and y components, and then add up all the x and y components separately to get the x and y components of the total force. Finally, calculate the magnitude of the total force using the Pythagorean theorem.
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student drove to the university from her home and noted that the odometer on her car increased by 14.0 km. The trip took 18.0 min. (a) What was her average speed? (b) If the straight-line distance from her home to the university is 10.3 km in a direction 25.0° south of east, what was magnitude of her average velocity?
The student's average speed on her trip to university was 0.78 km/min or 46.67 km/h, while the magnitude of her average velocity was 0.57 km/min or 34.3 km/h.
Explanation:In this scenario, we have to find the average speed and the magnitude of her average velocity. Average speed is calculated by dividing the total distance travelled by the total time taken. So, her average speed is 14.0 km / 18.0 min = 0.78 km/min or 46.67 km/h (note: convert minutes to hours for speed in km/h).
The next part of the question involves average velocity, which is the displacement (straight-line distance) divided by time. It can be different from the average speed when the path travelled is not a straight line. In this case, the magnitude of her average velocity is 10.3 km / 18.0 min = 0.57 km/min or 34.3 km/h.
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A rock is rolled in the sand. It starts at 5.0 m/s, moves in
astraight line for a distance of 3.0 m, and then stops. What is
themagnitude of the average acfeleration?
Answer:
magnitude of the average acceleration is 4.16 m/s²
Explanation:
given data
initial velocity u = 5 m/s
distance s = 3 m
final velocity v = 0
to find out
magnitude of the average acceleration
solution
we will apply here linear motion equation that is
v²-u² = 2×a×s ..............1
put here all these value
v is final velocity and u is initial velocity and s is distance
0²-5² = 2×a×3
a = -4.16
so magnitude of the average acceleration is 4.16 m/s²
Answer: The magnitude of the average acceleration is 4.16m/s^2
A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 27 m/s^2. If he reaches the ground with a speed of 14 m/s, how long was he in the air (in seconds) ?
Answer:
3.83 s
Explanation:
Initially for 50 m A falls with acceleration equal to g ie 9.8 m /s².In this journey his initial velocity u = 0 , a = 9.8 , v = ? , t = ?
h = ut + 1/ 2 g t²
50 = .5 x 9.8x t²
t = 3.19 s
v = u +gt
= o + 9.8 x 3.19
= 31.26 m /s
For motion under deceleration
initial speed u = 31.26 m/s
Final speed v = 14 m/s
deceleration a = - 27 m/s²
v = u - at
14 = 31.26 - 27 t
t = 0.64 s
So total time in the air
= 3.19 + .64 = 3.83 s
You have put a 1,000 μF capacitor and a 2,000 ohm resistor in series with an AC voltage source with amplitude of 45 V and frequency of 4,000 Hz. What is Vout across the capacitor? What is Vout across the resistor?
Answer:
voltage across capacitor [tex]V_C=i\times X_C=0.022\times 0.04=9\times 10^{-4}V[/tex]
Voltage across resistor [tex]V_R=i\times R=0.022\times 2000=80V[/tex]
Explanation:
We have given resistance R = 2000 OHM
Capacitance [tex]C=1000\mu F=0.001F[/tex]
Voltage V = 45 volt
Frequency = 4000 Hz
Capacitive reactance [tex]X_C=\frac{1}{\omega C}=\frac{1}{2\times \pi \times 4000\times 0.001}=0.04ohm[/tex]
Impedance [tex]Z=\sqrt{R^2+X_C^2}=\sqrt{2000^2+0.04^2}=2000ohm[/tex]
Current [tex]i=\frac{V}{Z}=\frac{45}{2000}=0.022A[/tex]
Now voltage across capacitor [tex]V_C=i\times X_C=0.022\times 0.04=9\times 10^{-4}V[/tex]
Voltage across resistor [tex]V_R=i\times R=0.022\times 2000=80V[/tex]
Two small particles of mass m1 and mass m2 attract each other with a force that varies inversely with the cube of their separation. At time t0,m1 has a velocity of magnitude v0, directed towards m2, which is at rest a distance d away. At time t1, the particles collide. Calculate L, the distance traveled by particle 1 during the time interval t1 − t0. Express your answer using some or all of the following variables: m1, m2, t0, t1, v0, and d.
Answer:
[tex]r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}[/tex]
Explanation:
Given,
mass of the first particle = [tex]m_1[/tex]velocity of the first particle = [tex]v_o[/tex]mass of the second particle = [tex]m_2[/tex]velocity of the second particle = [tex]v_2 = 0[/tex]Time interval = [tex](t_1\ -\ t_o)[/tex]Let [tex]v_{cm}[/tex] be the initial velocity of the center of mass of the system of particle at time [tex]t_o[/tex]
[tex]\therefore v_{cm}\ =\ \dfrac{m_1v_1\ +\ m_2v_2}{m_1\ +\ m_2}\\\Rightarrow v_{cm}\ =\ \dfrac{m_1v_0}{m_1\ +\ m_2}[/tex]
Assuming that the first particle is at origin, distance of the second particle from the origin is 'd'
[tex]x_1\ =\ 0[/tex][tex]x_2\ =\ d[/tex]Center of mass of the system of particles
[tex]x_{cm}\ =\ \dfrac{m_1x_1\ +\ m_2x_2}{m_1\ +\ m_2}\\\Rightarrow x_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\\[/tex]
Hence, at time [tex]t_0[/tex], the center of mass of the system is at [tex]x_0\ =\ \dfrac{m_2d}{m_1\ +\ m_2}[/tex] at an initial speed of [tex]v_{cm}[/tex]
Both the particles are assumed to be the point masses, therefore at the time [tex]t_1[/tex] the center of mass is at the position of the second particle which should be equal to the total distance traveled by the first particle because the second particle is at rest.
Let [tex]r_{cm}[/tex] be the distance traveled by the center of mass of the system of particles in the time interval [tex](t_1\ +\ t_0)[/tex]
From the kinematics,
[tex]s\ =\ x_0\ +\ vt\\\Rightarrow r_{cm}\ =\ x_{cm}\ +\ v_{cm}{t_1\ -\ t_0}\\\Rightarrow r_{cm}\ =\ \dfrac{m_2d}{m_1\ +\ m_2}\ +\ \left ( \dfrac{m_1v_0}{m_1\ +\ m_2}\ \right )\times (t_1\ -\ t_0)\\\Rightarow r_{cm}\ =\ \dfrac{m_2d\ +\ m_1v_0 (t_1\ -\ t_0)}{m_1\ +\ m_2}[/tex]
Hence, this is the required distance traveled by the first mass to collide with the second mass which is at rest.
In a 770 kW hydroelectric plant, 300 m^3 of water passes through the turbine each minute. Assuming complete conversion of the water's initial gravitational potential energy to electrical energy, what distance does the water fall? Assume two significant figures.
Answer:
16 m
Explanation:
given,
power of hydraulic plant = 770 kW
volume of water pass through the turbine = 300 m³
density of water = 1000 kg/m³
m =ρ × V
mass of water pass each minute = 300 × 1000 = 3 × 10⁵
assume height of the fall be h
potential head of the water = mgh
[tex]\dfrac{mgh}{60}= 770 \times 10^3[/tex]
[tex]3\times 10^5 \times 9.81\times h= 770 \times 10^3\times 60[/tex]
[tex]h = \dfrac{770 \times 10^3\times 60}{3\times 10^5 \times 9.81}[/tex]
h = 15.69 m ≈ 16 m
the distance of the water fall is equal to 16 m.
A brick is thrown upward from the top of a building at an angle of 25.7° above the horizontal and with an initial speed of 15.1 m/s. The acceleration of gravity is 9.8 m/s^2 If the brick is in flight for 3.4 s, how tall is the building? Answer in units of m
Answer:
the building is 34.408 m tall
Explanation:
given,
initial velocity of brick = 15.1 m/s
at an angle of = 25.7°
vertical component of the velocity
Vy = 15.1 sin 25.7°
= 6.54 m/s
we know
[tex]s = u t + \dfrac{1}{2} a t^2[/tex]
[tex]s = 6.54\times 3.4 + \dfrac{1}{2}\times -9.8 \times 3.4^2[/tex]
s = -34.408 m
hence, the building is 34.408 m tall .
Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. That is, a = F / mWhat are the dimensions of force?
Answer:
[tex][F]=[MLT^{-2}][/tex]
Explanation:
Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :
F = m × a
F is the applied force
m is the mass of the object
a is the acceleration due to gravity
We need to find the dimensions of force. The dimension of force m and a are as follows :
[tex][m]=[M][/tex]
[tex][a]=[LT^{-2}][/tex]
So, the dimension of force F is, [tex][F]=[MLT^{-2}][/tex]. Hence, this is the required solution.
Final answer:
In physics, the dimensions of force are derived from Newton's second law of motion, which can be represented as F=ma. Force has dimensions of mass (M), length (L), and time (T), leading to the dimensional formula [F] = [M][L][T]^-2. The SI unit for force is the Newton (N).
Explanation:
According to Newton's second law of motion, the acceleration a of an object is directly proportional to the net external force Fnet acting upon it and inversely proportional to its mass m. This law is mathematically represented by the equation Fnet = m × a, where Fnet is the net force, m is the mass, and a is the acceleration of an object.
The weight w of an object is another type of force, which is defined as the gravitational force acting on an object with mass m. The object experiences an acceleration due to gravity g, and this is represented by the equation w = m × g. In the International System of Units (SI), force is measured in Newtons (N), and one Newton is defined as the force required to accelerate a one-kilogram mass at a rate of one meter per second squared (1 N = 1 kg × m/s2).
Therefore, the dimensions of force in terms of the base physical quantities are mass (M), length (L), and time (T), and the dimensional formula for force is [F] = [M][L][T]-2.
A railroad car with a mass of 11 000 kg collides and couples with a second car of mass 19 000 kg that is initially at rest. The first car is moving with a speed of 4.5 m/s prior to the collision. a. What is the initial momentum of the first car? b. If external forces can be ignored, what is the final velocity of the two railroad cars after they couple?
Answer:
a) 49500 Kg.m/s
b) 1.65 m/s
Explanation:
Given:
Mass of the car, m₁ = 11000 kg
Mass of the second car, m₂ = 19000 kg
Initial Speed of the first car, u₁ = 4.5 m/s
Initial velocity of the second car , u₂ = 0
Now,
Momentum = Mass × Velocity
Initial momentum = m₁u₁
Thus
Initial momentum P₁ = 11000 × 4.5 = 49500 kg-m/sec
b)By using the concept of momentum conservation
Initial momentum = Final momentum
m₁u₁ + m₂u₂ = ( m₁ + m₂ )v
Where, v is the velocity after collision
thus,
49500 + 19000 × 0 = ( 11000 + 19000 ) × v
or
V = [tex]\frac{\textup{49500}}{\textup{30000}}[/tex]
or
v = 1.65 m/s
a) The initial momentum of the first car is 49,500 kg·m/s. b) The final velocity of the coupled cars is 1.65 m/s.
The question involves a physics concept called the conservation of momentum, which is particularly applicable to collisions, such as the one described between two railroad cars.
a) The initial momentum (p) of the first car can be calculated using the formula p = mv, where m is the mass and v is the velocity of the car. For the first car with a mass m of 11,000 kg moving with a velocity v of 4.5 m/s, the initial momentum is: p = 11,000 kg × 4.5 m/s = 49,500 kg·m/s.
b) For the final velocity, we apply the principle of conservation of linear momentum. As external forces are ignored, the total momentum before the collision is equal to the total momentum after the collision. The combined mass of the coupled cars is 11,000 kg + 19,000 kg = 30,000 kg. The final velocity (v_f) is calculated by setting the initial total momentum equal to the final total momentum: (11,000 kg × 4.5 m/s) = 30,000 kg × v_f, hence v_f = 49,500 kg·m/s / 30,000 kg = 1.65 m/s.
A surveillance satellite has a camera which detects 550 nm light and is equipped with a 35 cm diameter lens. If the satellite is at an altitude of 160 km, what is the minimum spacing between two objects on the ground that the camera can barely resolve (in m)?
Answer:
0.3067 m
Explanation:
Resolving power of a lens is given by the expression
R.P = 1.22λ / D
where λ is wavelength of light used, D is diameter of the lens
Substituting the data given
R.P in radian =
[tex]\frac{1.22\times550\times10^{-9}}{35\times10^{-2}}[/tex]
R P = 19.17 X 10⁻⁷ radian
If d be the minimum spacing between two objects on the ground that is resolvable by lens
[tex]\frac{d}{160\times10^3} =19.17\times10^{-7}[/tex]
d = 0.3067 m
Interstellar space (far from any stars) contains atomic hydrogen (H) with a density of 1 atom/cm3 and at a temperature of about 2.7 K. Determine (a) the pressure in interstellar space, (b) root-mean square speed of the atoms and (c) The kinetic energy stored in 1 km3 of space.
Explanation:
Given that,
Number density [tex]n= 1\ atom/cm^{3} =10^{6}\ atom/m^3[/tex]
Temperature = 2.7 K
(a). We need to calculate the pressure in interstellar space
Using ideal gas equation
[tex]PV=nRT[/tex]
[tex]P=\dfrac{nRT}{V}[/tex]
[tex]P=\dfrac{10^{6}\times8.314\times2.7}{6.023\times10^{23}}[/tex]
[tex]P=3.727\times10^{-17}\ Pa[/tex]
[tex]P=36.78\times10^{-23}\ atm[/tex]
The pressure in interstellar space is [tex]36.78\times10^{-23}\ atm[/tex]
(b). We need to calculate the root-mean square speed of the atom
Using formula of rms
[tex]v_{rms}=\sqrt{\dfrac{3RT}{Nm}}[/tex]
Put the value into the formula
[tex]v_{rms}=\sqrt{\dfrac{3\times8.314\times2.7}{1.007\times10^{-3}}}[/tex]
[tex]v_{rms}=258.6\ m/s[/tex]
The root-mean square speed of the atom is 258.6 m/s.
(c). We need to calculate the kinetic energy
Average kinetic energy of atom
[tex]E=\dfrac{3}{2}kT[/tex]
Where, k = Boltzmann constant
Put the value into the formula
[tex]E=\dfrac{3}{2}\times1.38\times10^{-23}\times2.7[/tex]
[tex]E=5.58\times10^{-23}\ J[/tex]
The kinetic energy stored in 1 km³ of space is [tex]5.58\times10^{-23}\ J[/tex].
Hence, This is the required solution.
A person on a trampoline can go higher with each bounce. how is this possible? is there a maximum height to which this person can go?
Explanation:
when a person jumps on the trampoline he stores his potential energy in the form of elastic energy of trampoline. Potential Energy converts into Elastic energy when a person is at the bottom point during stretching of the trampoline. When he again going upwards, Elastic energy is converted to the potential energy of the person. This is the reason why a person goes higher each time. This process goes on until trampoline reaches its elastic limit and finally breaks or get permanently deformed.
So there is a limit up to which a person can be reached.
Final answer:
A person goes higher on a trampoline by landing and pushing off with their feet, making better use of leg strength to transform kinetic to potential energy. There is a maximum height due to energy losses like air resistance and imperfect energy transfers. The sweet spot in tennis rackets illustrates efficient energy transfer with reduced arm jarring.
Explanation:
A person on a trampoline goes higher with each bounce because they can harness the elastic potential energy stored in the trampoline material. When landing on the back or feet, the height reached can differ.
Typically, a person may reach greater heights when landing and launching off their feet because they can make use of the stronger leg muscles to add upward force, thus converting more kinetic energy to gravitational potential energy upon ascent. There is, indeed, a maximum height obtainable, as energy losses due to factors like air resistance and less-than-perfect energy transfer prevent infinite increases in height.
Addressing the professional application, the 'sweet spot' on a tennis racket is an example of an optimal point where energy transfer from the racket to the ball is the most efficient, and vibrating force transmitted to the player's arm is minimal, hence no jarring of the arm.
The physics of motion and energy conservation provides us the information that increasing the initial speed (kinetic energy) of the object would result in a higher ascent, yet this is bounded by practical limitations such as the strength of the jumper and the efficiency of the trampoline material.
For example, to double the impact speed of a falling object, one would have to quadruple the height from which it falls, due to the relationship between gravitational potential energy and kinetic energy.
Starting from one oasis, a camel walks 25 km in a direction 30° south of west and then walks 30 km toward the north to a second oasis. What distance separates the two oases?
Answer:
distance between both oasis ( 1 and 2) is 27.83 Km
Explanation:
let d is the distance between oasis1 and oasis 2
from figure
OC = 25cos 30
OE = 25sin30
OE = CD
Therefore BC = 30-25sin30
distance between both oasis ( 1 and 2) is calculated by using phytogoras theorem
in[tex]\Delta BCO[/tex]
[tex]OB^2 = BC^2 + OC^2[/tex]
PUTTING ALL VALUE IN ABOVE EQUATION
[tex]d^2 = 930-25sin30)^2 + (25cos30)^2[/tex]
[tex]d^2 = 775[/tex]
d = 27.83 Km
distance between both oasis ( 1 and 2) is 27.83 Km
Using vector addition and the Pythagorean theorem, the distance separating the two oases is found to be approximately 27.95 km after the camel walks a two-leg journey with specified directions and distances.
Explanation:To determine the distance that separates the two oases, we can use vector addition and Pythagorean theorem. The camel walks 25 km in a direction 30° south of west, which can be represented as a vector with components to the south and to the west.
Then, the camel walks 30 km towards the north. In terms of vectors, these two displacements will partly cancel each other out in the north-south direction.
First, let's resolve the initial 25 km walk into components. The southern component is 25 km * sin(30°) = 12.5 km, and the western component is 25 km * cos(30°) = 21.65 km. After the camel walks 30 km north, the remaining southward component is 30 km - 12.5 km = 17.5 km north. The westward component remains unchanged at 21.65km.
Now, we can use the Pythagorean theorem to find the resultant distance between the two oases, which is the hypotenuse of the right triangle formed by the northward and westward components.
The distance is √(17.5 km² + 21.65 km²), which is approximately 27.95 km. Therefore, the distance that separates the two oases is around 27.95 km.
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A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.959 g, q = 5.84 µC is located on the x axis at x = 20.7 cm, moving with a speed of 47.9 m/s in the positive ydirection. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)
Answer:
[tex]8.66\times 10^{-6}\ C[/tex] or [tex]8.66\ \mu C[/tex].
Explanation:
Given:
Charge on the particle at origin = Q.Mass of the moving charged particle, [tex]\rm m = 0.959\ g = 0.959\times 10^{-3}\ kg.[/tex]Charge on the moving charged particle, [tex]\rm q = 5.84\ \mu C = 5.84\times 10^{-6}\ C.[/tex]Distance of the moving charged particle from first at t = 0 time, [tex]\rm r=20.7\ cm = 0.207\ m.[/tex]Speed of the moving particle, [tex]\rm v = 47.9\ m/s.[/tex]For the moving particle to circular motion, the electrostatic force between the two must be balanced by the centripetal force on the moving particle.
The electrostatic force on the moving particle due to the charge Q at origin is given by Coulomb's law as:
[tex]\rm F_e = \dfrac{kqQ}{r^2}.[/tex]
where, [tex]\rm k[/tex] is the Coulomb's constant having value [tex]\rm 9\times 10^9\ Nm^2/C^2.[/tex]
The centripetal force on the moving particle due to particle at origin is given as:
[tex]\rm F_c = \dfrac{mv^2}{r}.[/tex]
For the two forces to be balanced,
[tex]\rm F_e = F_c\\\dfrac{kqQ}{r^2}=\dfrac{mv^2}{r}\\\Rightarrow Q = \dfrac{mv^2}{r}\times \dfrac{r^2}{kq}\\=\dfrac{mv^2r}{kq}\\=\dfrac{(0.959\times 10^{-3})\times (47.9)^2\times (0.207)}{(9\times 10^9)\times (5.84\times 10^{-6})}\\=8.66\times 10^{-6}\ C\\=8.66\ \mu C.[/tex]
For the moving particle to execute circular motion, the charge Q must be approximately 6.75 µC and have the same sign as the particle's charge (5.84 µC) to ensure attraction and centripetal motion.
To determine the value of charge Q that will allow the moving particle to execute circular motion, we must consider the forces acting on the charged particle due to the electric field created by Q.
Mass of the moving particle, m = 0.959 g = 0.000959 kg Charge of the moving particle, q = 5.84 µC = 5.84 x 10^-6 C Initial position on the x-axis, x = 20.7 cm = 0.207 m Speed in the positive y-direction, v = 47.9 m/sFor the charged particle to execute circular motion, the centripetal force required to keep it in circular motion must equal the electric force acting on it due to charge Q.
Centripetal Force:
The centripetal force (
F_c) is given by:
[tex]F_c = \frac{mv^2}{r}[/tex]
where r is the radius of the circular motion. Here, r = 0.207 m (distance from the origin).
Substituting the given values, we get:
[tex]F_c = \frac{0.000959 \text{ kg} \times (47.9 ext{ m/s})^2}{0.207 ext{ m}}[/tex]
[tex]F_c = \frac{0.000959 \times 2299.61}{0.207} \approx 0.008291 \text{ N}[/tex]
Electric Force:
The electric force (F_e) on the particle due to the charge Q at the origin is given by Coulomb's Law:
[tex]F_e = \frac{k |Q| |q|}{r^2}[/tex]
where k is Coulomb's constant [tex]k \approx 8.99 \times 10^9 \text{ N m}^2/ ext{C}^2[/tex].
Setting the electric force equal to the centripetal force for circular motion:
[tex]\frac{k |Q| |q|}{r^2} = F_c[/tex]
Plugging in the numbers:
[tex]\frac{(8.99 \times 10^9) |Q| (5.84 \times 10^{-6})}{(0.207)^2} = 0.008291[/tex]
Solving for Q:
Rearranging for Q:
[tex]|Q| = \frac{0.008291 \cdot (0.207)^2}{8.99 \times 10^9 imes (5.84 \times 10^{-6})}[/tex]
Calculate the right side:
[tex]|Q| = \frac{0.008291 \cdot 0.042849}{8.99 \times 10^9 imes 5.84 \times 10^{-6}}[/tex]
[tex]|Q| \approx \frac{0.000355}{0.05256} \approx 6.75 \times 10^{-6} \text{ C}[/tex] or 6.75 µC.
Suppose you walk 11 m in a direction exactly 24° south of west then you walk 21 m in a direction exactly 39° west of north. 1) How far are you from your starting point, in meters?
2) What is the angle of the compass direction of a line connecting your starting point to your final position measured North of West in degrees?
Answer:
(1) 42.94 m
(2) [tex]16.02^\circ[/tex]
Explanation:
Let us first draw a figure, for the given question as below:
In the figure, we assume that the person starts walking from point A to travel 11 m exactly [tex]24^\circ[/tex] south of west to point B and from there, it walks 21 m exactly [tex]39^\circ[/tex] west of north to reach point C.
Let us first write the two displacements in the vector form:
[tex]\vec{AB} = (-11\cos 24^\circ\ \hat{i}-11\sin 24^\circ\ \hat{j})\ m =(-10.05\ \hat{i}-4.47\ \hat{j})\ m\\\vec{BC} = (-21\sin 39^\circ\ \hat{i}+21\cos 39^\circ\ \hat{j})\ m =(-31.22\ \hat{i}+16.32\ \hat{j})\ m[/tex]
Now, the vector sum of both these vectors will give us displacement vector from point A to point C.
[tex]\vec{AC}=\vec{AB}+\vec{BC}\\\Rightarrow \vec{AC}=(-10.05\ \hat{i}-4.47\ \hat{j})\ m+(-31.22\ \hat{i}+16.32\ \hat{j})\ m\\\Rightarrow \vec{AC}=(-41.25\ \hat{i}+11.85\ \hat{j})\ m[/tex]
Part (1):
the magnitude of the shortest displacement from the starting point A to point the final position C is given by:
[tex]AC=\sqrt{(-41.25)^2+(11.85)^2}\ m= 42.94\ m[/tex]
Part (2):
As the vector AC is coordinates lie in the third quadrant of the cartesian vector plane whose angle with the west will be positive in the north direction.
The angle of the shortest line connecting the starting point and the final position measured north of west is given by:
[tex]\theta = \tan^{-1}(\dfrac{11.85}{41.27})\\\Rightarrow \theta = 16.02^\circ[/tex]
Automobile A starts from O and accelerates at the constant rate of 0.75 m/s2. A short time later it is passed by bus B which is traveling in the opposite direction at a constant speed of 6 m/s. Knowing that bus B passes point O 20 s after automobile A started from there, determine when and where the vehicles passed each other.
Answer:
they meet from point o at distance 50.46 m and time taken is 11.6 seconds
Explanation:
given data
acceleration = 0.75 m/s²
speed B = 6 m/s
time B = 20 s
to find out
when and where the vehicles passed each other
solution
we consider here distance = x , when they meet after o point
and time = t for meet point z
we find first Bus B distance for 20 s ec
distance B = velocity × time
distance B = 6 × 20
distance B = 120 m
so
B take time to meet is calculate by distance formula
distance = velocity × time
120 - x = 6 × t
x = 120 - 6t .................1
and
distance of A when they meet by distance formula
distance = ut + 1/2 × at²
here u is initial speed = 0 and t is time
x = 0 + 1/2 × 0.75 × t²
x = 0.375 × t² .............2
so from equation 1 and 2
0.375 × t² = 120 - 6t
t = 11.6
so time is 11.6 second
and
distance from point o from equation 2
x = 0.375 (11.6)²
x = 50.46
so distance from point o is 50.46 m
Automobile A and bus B pass each other approximately 11.6 seconds after automobile A started, at a position 50.5 meters from point O. This is determined by setting their displacement equations equal and solving the quadratic equation. The final position of the meeting point is found to be around 50.5 meters from point O.
To solve this problem, we need to determine the position and time at which automobile A and bus B pass each other.
Determine the position of automobile A:
Automobile A starts from rest with an acceleration of 0.75 m/s². We use the formula for displacement under constant acceleration: displacement = (1/2) * acceleration * time². Let t be the time in seconds after automobile A starts.For automobile A:xA = (1/2) * 0.75 * t² = 0.375t²Determine the position of bus B:
Bus B passes point O at t = 20 seconds (since bus B passed O 20 seconds after automobile A started). Thus, the time since automobile A started is t - 20 seconds for bus B.Bus B is traveling in the opposite direction at a constant speed of 6 m/s. The displacement of bus B relative to point O can be given by:xB = -6 * (t - 20) = -6t + 120Set the positions equal to find when they pass each other:
At the point where they meet: xA = xB0.375t2 = -6t + 120Solve the quadratic equation:0.375t2 + 6t - 120 = 0Multiplying through by 8 to clear the decimal:
3t2 + 48t - 960 = 0Using the quadratic formula, t = (-b ± √(b² - 4ac)) / 2a with a = 3, b = 48, and c = -960:t = [-48 ± √(482 + 4 * 3 * 960)] / 6t = [-48 ± √(2304 + 11520)] / 6t = [-48 ± √13824] / 6t = (-48 ± 117.62) / 6We have two potential solutions: t = (69.62 / 6) ≈ 11.6 seconds (since a negative time is not meaningful in this context)
Determine the meeting point:
Now substitute t = 11.6 back to find xA:xA = 0.375 * 11.62 ≈ 50.5 metersTwo hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 26.0 m . Simultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 2.30 m/s , and puck B moves with a speed of 3.90 m/s .
Your answer should satisfy common sense. For instance, can you decide which of the following values for the distance covered by puck A would definitely be wrong, regardless of the speed of the two pucks and considering that the two pucks are sliding toward each other?
(A) 5 m
(B) 33 m
(C) 27 m
(D) 1 m
(E) 21 m
Answer:
B) 33 m C) 27 m
Explanation:
considering that the two pucks are sliding toward each other we can understand that they are on a collision course.
Since the total distance between them is 26 m, the common sense dictates that the distance traveled by each puck must be less than 26 m regardless of the speed of the two pucks.
so the options B) 33 m and C) 27 m are definitely wrong since they are greater than 26 m.
We can also easily find the distance traveled by each pucks also.
let [tex]v_{A}[/tex] and [tex]v_{B}[/tex] be the velocity of the pucks A and B respectively
[tex]v_{A}t = 26- v_{B}t\\\\2.30t = 26 - 3.90t\\\\6.2t = 26\\\\t = \frac{26}{6.2} \\\\x_{A} =v_{A}t\\\\x_{A} =2.3 \times \frac{26}{6.2}\\\\x_{A} =9.65\\\\x_{B} =v_{B}t\\\\x_{B} =3.9 \times \frac{26}{6.2}\\\\x_{B} =16.35\\[/tex]
A power electronics package has been designed to handle 400W of power dissipation. Find out the change in output power handling capability of the power electronics package when efficiency is increased from 89% to 94%. Discuss briefly the importance of efficiency base on output power capability calculated.
Final answer:
To calculate the change in output power handling capability of a power electronics package, you can use the formula Power = Output power / Efficiency. By comparing the original power dissipation with the new power dissipation, you can determine the change in output power. Efficiency is important because it affects the amount of input power required and the output power capability of a system.
Explanation:
To find out the change in output power handling capability, we need to calculate the new power dissipation and compare it with the original value. The formula for power is given as:
Power = Output power / Efficiency
Using the formula, we can calculate the original power dissipation:
Original Power = 400W / 0.89 = 449.43W
Next, we can calculate the new power dissipation:
New Power = 400W / 0.94 = 425.53W
The change in output power handling capability is the difference between the original power and the new power:
Change in Output Power = 449.43W - 425.53W = 23.9W
The increase in efficiency from 89% to 94% leads to a decrease in the power dissipation by 23.9W. Efficiency is important because it determines the amount of input power required to achieve a certain output power. Higher efficiency means less power is wasted as heat, resulting in a higher output power capability for a given input power.
A ball is kicked with an initial velocity of 18.1 m/s in the horizontal direction and 15.8 m/s in the vertical direction. For how long does the ball remain in the air?
Answer:
3.22 s
Explanation:
The ball would describe a projectile motion, where the horizontal velocity will remain constant, as there are no forces that act on the x-axis, and the vertical velocity will vary because of gravity in the following way:
1. First, the ball will go up, but the vertical velocity will decrease until it has a value of 0.
2. After the vertical velocity has reached the value of 0, the ball will start to fall, with the vertical velocity increasing because of gravity.
You need to know that the time that the ball's verical velocity takes to reach 0 is exactly the same that it takes to go from 0 to its original vertical velocity:
[tex]a = \frac{v_f - v_o}{t}\\t = \frac{v_f - v_o}{a}[/tex]
And not only the time will be the same, but also the distance traveled. Therefore, we can conclude that the time that the ball remain in the air is simply two times the time it takes for the ball to desacelerate:
[tex]t_{desaceleration} = \frac{v_f - v_o}{a} = \frac{0m/s - 15.8m/s}{-9.81 m/s^2} =1.61 s\\t_{air} = 2* t_{desaceleration} = 2*1.61s = 3.22 s[/tex]
At time t=0, rock A is dropped from rest from a height of 90 m. At the same instant, rock B is launched straight up from the ground level with an initial speed of 30 m/s. Write an equation of motion for rock A and B, giving its position at all time.
Answer:
rock A: [tex]y=90-1/2*g*t^2[/tex]
rock B: [tex]y=30*t-1/2*g*t^2[/tex]
g=9.81m/s^2
Explanation:
Kinematics equation:
[tex]y=y_{o}+v_{oy}*t+1/2*a*t^2[/tex]
in our case the acceleration is the gravity and it has a negative direction.
a=-g
rock A, yo=90m, Voy=0m/s:
[tex]y=90-1/2*g*t^2[/tex]
rock B, yo=0m, Voy=30m/s:
[tex]y=30*t-1/2*g*t^2[/tex]
A rigid, insulated tank that is initially evacuated is connected through a valve to a supply line that carries steam at 1 MPa and 300 ºC. Now the valve is opened and steam is allowed to flow slowly into the tank until the pressure reaches 1 MPa, at which point the valve is closed. Determine the final temperature of the steam in the tank.
Answer:
Explanation:
The steam was earlier at 300 ° C. and pressure of 1 MPa. When the gas is allowed to expand against vacuum , work done by the gas is nil because there is no external pressure against which it has to work . Therefore there will not be any change in its internal energy. Since the tank is insulated therefore there is no possibility of external heat to increase its internal energy.
Hence the temperature of gas will remain unaffected. It will remain stagnant at 300
Coco serves a tennis ball at vs = 50 m/s and charges the net at vc = 10 m/s. The opponent,x = 25 m away on the other side of the court, returns the ball with a speed half that of the serve. How close does Coco get to the net (x/2 away) before she meets the return?
Answer:
Distance from the net when she meets the return: 14,3 m
Explanation:
First we need to know how much time it takes the ball to reach the net, the kinematycs general equation for position:
(1) [tex]x = x_{0} + V_{s} t + \frac{1}{2} a t^{2}[/tex]
Taking the net as the origin (x = 0), [tex]x_{0} = 25m[/tex], velocity will be nagative [tex]V_{s} = - 50m/s[/tex] and assuming there is no friction wit air acceleration would be 0, so:
(2) [tex]x = x_{0} + V_{s} t [/tex]
we want to know the time when it reaches the net so when x=0, replacing de values:
(3) [tex]0 = 25m - 50 m/s t_{net} [/tex]
So: [tex]t_{net} = 0,5 s [/tex]
The opponent will return the ball at [tex]V_{ret} = 25m/s[/tex], the equation for the return of the ball will be:
(4) [tex]y = y_{0} + V_{ret} t + \frac{1}{2} a t^{2}[/tex]
Note that here it start from the origin, [tex]y_{0} = 0[/tex], as in the other case acceleration equals 0, and here we have to consider that the time starts when the ball reaches the net ([tex]t_{net} [/tex]) so the time for this equiation will be [tex]t - t_{net} [/tex], this is only valid for [tex]t >= t_{net} [/tex]:
(5) [tex]y = V_{ret} (t - t_{net}) [/tex]
Coco starts running as soon as he serves so his equiation for position will be:
(6) [tex]z = z_{0} + V_{c} t + \frac{1}{2} a t^{2}[/tex]
As in the first case it starts from 25m, [tex]z_{0} = 25m[/tex], acceleration equals 0 and velocity is negative [tex]V_{c} = - 10m/s[/tex]:
(7) [tex]z = z_{0} + V_{c} t[/tex]
To get the time when they meet we have that [tex]z = y[/tex], so from equiations (5) and (7):
[tex]V_{ret} (t - t_{net}) = z_{0} + V_{c} t[/tex]
[tex]t = \frac{z_{0} + V_{ret}* t_{net}}{V_{ret}-V_{c}}[/tex]
Replacing the values:
[tex]t = 1,071 s[/tex]
Replacing t in either (5) or (7):
[tex]z = y = 14,3 m[/tex]
This is the distance to the net when she meets the return
What power in Kw is required to transfer a charge of 9000 coulomb through a potential difference of 220 volts in 45 min?
0.9 kW
0.73 kW
0.5 kW
1.6 kW
Answer:
The power is 0.73 kW.
(B) option is correct.
Explanation:
Given that,
Charge = 9000 C
Potential difference = 220 V
Time = 45 min
We need to calculate the energy required to transfer charge Q through V
Using formula of energy
[tex]E =QV[/tex]
Put the value into the formula
[tex]E=9000\times220[/tex]
[tex]E=1980000\ J[/tex]
We need to calculate the power
Using formula of power
[tex]P=\dfrac{1980000}{45\times60}[/tex]
[tex]P=733.33\ W[/tex]
[tex]P=0.73\ kW[/tex]
Hence, The power is 0.73 kW.
A car is decelerating at the rate of 2 km/s square. If its initial speed is 66 km/s, how long will it take the car to come to a complete stop?
Answer:
It will take 33 seconds to stop the car.
Explanation:
Using the first equation of kinematics we have
[tex]v=u+at[/tex]
where
'v' is final speed of object
'u' is initial speed of object
'a' is acceleration of object
't' is time of acceleration of object
Now since it is given that [tex]a=-2km/s^{2}[/tex] since acceleration is negative and [tex]u=66km/s[/tex]
We know that the object will stop when it's velocity reduces to zero hence in the equation above setting v = 0 we get
[tex]0=66-2\times t\\\\\therefore t=\frac{66}{2}=33seconds[/tex]
Final answer:
Using the kinematic equation for velocity and acceleration, it is calculated that the car, decelerating at a rate of 2 km/s² from an initial speed of 66 km/s, will take 33 seconds to come to a complete stop.
Explanation:
To find out how long it will take the car to come to a complete stop, we need to use the kinematic equation that relates initial velocity, acceleration, and time without displacement:
Final velocity (v) = Initial velocity (u) + (Acceleration (a) × Time (t))
Here, the final velocity (v) is 0 km/s, since the car is coming to a stop, the initial velocity (u) is 66 km/s, and the acceleration (a) is -2 km/s² (negative because it is deceleration).
The equation becomes:
0 = 66 + (-2 × t)
We can solve for t as follows:
0 = 66 - 2t
2t = 66
t = 33 seconds
So it will take the car 33 seconds to come to a complete stop.
an automobile travels on a straight road for 42 km at 45 km/h. it then continues in the same direction for another 42 km at 90 km/h. (assume that it moves in the positive x direction.) a. what is the average velocity of the car during this 84 km trip?
b. what is it’s average speed?
C. Graph x versus t and indicate how the average velocity is found on the graph?
Answer:
a) The average velocity is v = (60 km/h ; 0)
b) The average speed is 60 km/h
Explanation:
The velocity is a vector that has a magnitude and direction. The average speed is the distance traveled over time without taking into account the direction of the motion.
a)The average velocity is calculated as the displacement over time:
v = Δx/Δt
where
v = velocity
Δx = final position - initial position = traveled distance relative to the center of the reference system.
Δt = final time - initial time (initial time is usually = 0)
We know that the displacement is 84 km but we do not know the time. It can be calculated from the two parts of the trip.
In part 1:
v = 45 km/h = 42 km / t
t = 0.93 h
In part 2:
v = 90 km/h = 42 km / t
t = 42 km / 90 km/h
t = 0.47 h
The time of travel is 0.47 h + 0.93 h = 1.4 h
The average velocity will be:
v = 84 km / 1.4 h = 60 km/h
Expressed as a vector in a 2-dimension plane:
v = (60 km/h; 0)
b) The average speed is calculated as the distance traveled over time. Note that in this case, the distance is equal to the displacement since the direction of the motion is always in one direction. But if the direction of the second part of the trip would have been the opposite to the direction of the first part, the displacement would have been 0 (final position - initial position = 0, because final position = initial position), then, the average velocity would have been 0. In change, the average speed would have been the distance traveled (84 km, 42 km in one direction and 42 km in the other) over time.
Then:
average speed = 84 km / 1.4 h = 60 km/h
c) see attached figure.
What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicle traveling 70 mi/hr?
Answer:
[tex]276.74\times 10^8Mg/m^3[/tex]
31.29 m/sec
Explanation:
We have given density of substance [tex]0.14lb/in^3[/tex]
We have convert this into [tex]Mg/m^3[/tex]
We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg
We know that 1 kg = 1000 g ( 1000 gram )
So 0.06349 kg = 63.49 gram
And we know that 1 gram = 1000 milligram
So 63.49 gram [tex]=63.49\times 10^3\ Mg[/tex]
We know that [tex]1 in^3=1.6387\times 10^{-5}m^3[/tex]
So [tex]0.14in^3=0.14\times 1.6387\times 10^{-5}=0.2294\times 10^{-5}m^3[/tex]
So [tex]0.14lb/in^3[/tex] =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]
In second part we have to convert 70 mi/hr to m/sec
We know that 1 mi = 1609.34 meter
So 70 mi = 70×1609.34 = 112653.8 meter
1 hour = 3600 sec
So 70 mi/hr [tex]=\frac{70\times 1609.34meter}{3600sec}=31.29m/sec[/tex]
Final answer:
The density of the substance is 3.87118626 Mg/m³. The velocity of the vehicle is 31.29222 m/s.
Explanation:
The density of a substance is the ratio of its mass to its volume. In order to convert the density from lb/in³ to Mg/m³, we need to use conversion factors. 1 lb/in³ is equal to 27679.9 Mg/m³. Therefore, the density of the substance is 0.14 lb/in³ * 27679.9 Mg/m³/lb/in³ = 3.87118626 Mg/m³.
For the second question, to convert mi/hr to m/s, we can use the conversion factor of 1 mi = 1609.34 m and 1 hr = 3600 s. Therefore, the velocity of the vehicle is 70 mi/hr * 1609.34 m/mi * 1 hr/3600 s = 31.29222 m/s.
A charge 4.96 nC is placed at the origin of an xy-coordinatesystem, and a charge -1.99 nC is placed on the positive x-axis at x = 4.01 cm . A third particle, of charge 5.99 nC is now placed at the point x = 4.01 cm , y = 2.98 cm . Find the x-component of the total force exerted on the third charge by the other two. Find the y-component of the total force exerted on the third charge by the other two. Find the magnitude of the total force acting on the third charge. Find the direction of the total force acting on the third charge.
Answer:
B) x-component of the total force exerted on the third charge by the other two (Fn₃x)
Fn₃x =8,56*10⁻⁵ N (+x)
B) y-component of the total force exerted on the third charge by the other two (Fn₃y)
Fn₃y = 5.7*10⁻⁵ N (-y)
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Equivalences:
1nC= 10⁻⁹C
1cm = 10⁻²m
Data
q₁=4.96 nC = +4.96*10⁻⁹C
q₂=-1.99 nC = -1.99*10⁻⁹C
[tex]d_{1} =\sqrt{ 4.01^{2}+2.98^{2} } = 4.996 cm = 4.996*10^{-2} m=49.96*10^{-3} m[/tex]
d₂= 2.98 cm
Graphic attached
The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs.
the force F₂₃ of q₂ and q₃ is attractive because the charges have opposite signs.
Calculation of the forces exerted on the charge q₁ and q₂ on q₃
To calculate the magnitudes of the forces exerted by the charges q₁, q₂, on q₃ we apply Coulomb's law:
F₁₃=k*q₁*q₃/d₁² =9*10⁹*4.96*10⁻⁹*5.99*10⁻⁹/(49.96*10⁻³)²=10.7**10⁻⁵ N
F₂₃=k*q₂*q₃/d₂² =9*10⁹*1.99*10⁻⁹*5.99*10⁻⁹/(2.98*10⁻²)²=12.08*10⁻⁵ N
F₁₃x=F₁₃cosβ=10.7*10⁻⁵* (4.01/4.996)=8,56*10⁻⁵ N (+x)
F₁₃y=F₁₃sinβ= 10.7*10⁻⁵* (2.98/4.996)=6,38*10⁻⁵ N (+x)
F₂₃x= 0
F₂₃y= F₂₃=12.08*10⁻⁵ N (-y)
A) x-component of the total force exerted on the third charge by the other two (Fn₃x)
Fn₃x= F₁₃x + F₂₃x= 8,56*10⁻⁵ N + 0
Fn₃x =8,56*10⁻⁵ N (+x)
B) y-component of the total force exerted on the third charge by the other two (Fn₃y)
Fn₃y = F₁₃y + F₂₃y= 6,38*10⁻⁵ N - 12.08*10⁻⁵ N
Fn₃y = 5.7*10⁻⁵ N (-y)
The charges, 4.96 nC, -1.99 nC, and 5.99 nC, forming a triangle gives
the following approximate values of the force at the 5.99 nC charge.
x-component is [tex]\underline{8.6 \times 10^{-5} \, \mathbf{\hat i} \ N}[/tex]y-component is [tex]\underline{-5.69 \times 10^{-5} \, \mathbf{\hat j} \ N}[/tex]The magnitude of the total force is 1.03 × 10⁻⁴ NThe direction of the force is 33.5° clockwise from the horizontal x-axisHow can the force acting at the 5.99 nC charge be resolved?The given charges are;
Q₁ = 4.96 nC, at point (0, 0)
Q₂ = -1.99 nC, at point (4.01, 0)
Q₃ = 5.99 nC at point (4.01, 2.98)
According to Coulomb's Law, we have;
[tex]F_{13} = \mathbf{\dfrac{9 \times 10^9 \times \left(4.96 \times 10^{-9} \right)\times \left(5.99\times 10^{-9} \right)}{4.01^2 + 2.98^2} } }\approx 1.07 \times 10^{-4}[/tex]
F₁₃ ≈ 1.07 × 10⁻⁴ N
The components of the force are;
[tex]cos\left(arctan \left(\dfrac{2.96}{4.01} \right) \right) \times 1.07 \times 10^{-4} \, \mathbf{\hat i} + sin\left(arctan \left(\dfrac{2.96}{4.01} \right) \right) \times 1.07 \times 10^{-4} \, \mathbf{\hat j}[/tex]
Which gives;
[tex]\vec{F_{13}} \approx \mathbf{8.6 \times 10^{-5} \, \mathbf{\hat i} + 6.39 \times 10^{-5} \, \mathbf{\hat j}}[/tex]
Therefore;
[tex]F_{23}= \dfrac{9 \times 10^9 \times \left((-1.99)\times 10^{-9} \right)\times \left(5.99\times 10^{-9} \right)}{2.98^2} } \approx\mathbf{ 1.208 \times 10^{-4}}[/tex]
Which gives;
[tex]\vec{F_{23}} \approx \mathbf{-1.208 \times 10^{-4} \, \mathbf{\hat j}}[/tex]
The components of the force at Q₃ is therefore;
x-component = [tex]8.6 \times 10^{-5} \, \mathbf{\hat i} + 0 = \underline{8.6 \times 10^{-5} \, \mathbf{\hat i}}[/tex]y-component = [tex]6.39 \times 10^{-5} \, \mathbf{\hat j} + -1.208 \times 10^{-4} \, \mathbf{\hat j} \approx \underline{-5.69 \times 10^{-5} \, \mathbf{\hat j}}[/tex]The magnitude of the total force is therefore;
|F| ≈ √((8.6×10⁻⁵)² + (-5.69 × 10⁻⁵)²) ≈ 1.03 × 10⁻⁴
The magnitude of the total force, |F| ≈ 1.03 × 10⁻⁴ NThe direction of the total force is found as follows;
[tex]The \ direction, \ \theta \approx \mathbf{arctan \left(\dfrac{-5.69}{8.6} \right)} \approx -33.5 ^{\circ}\alpha[/tex]
The force acta in a direction of approximately 33.5° clockwise from the horizontal x-axisLearn more about Coulomb's Law on electric force here
https://brainly.com/question/12472287
What is the energy of the photon that, when absorbed by a hydrogen atom, could cause the following? (a) an electronic transition from the n = 3 state to the n = 6 state
(b) an electronic transition from the n = 3 state to the n = 8 state
Answer:
(a): [tex]\rm 1.133\ eV\ \ \ or\ \ \ 1.8128\times 10^{-19}\ J.[/tex]
(b): [tex]\rm 1.298\ eV \ \ \ or \ \ \ 2.077\times 10^{-19}\ J.[/tex]
Explanation:
The energy of the photon that absorbed by a hydrogen atom causes a transition is equal to the difference in energy levels of the hydrogen atom corresponding to that transition.
According to Rydberg's formula, the energy corresponding to [tex]\rm n^{th}[/tex] level in hydrogen atom is given by
[tex]\rm E_n = -\dfrac{E_o}{n^2}.[/tex]
where,
[tex]\rm E_o=13.6\ eV.[/tex]
Part (a): For the electronic transition from the n = 3 state to the n = 6 state.
The energy of the photon which cause this transition is given by
[tex]\rm \Delta E=E_6-E_3.\\\\where,\\E_6=-\dfrac{E_o}{6^2}=-\dfrac{13.6}{36}=-0.378\ eV.\\E_3=-\dfrac{E_o}{3^2}=-\dfrac{13.6}{9}=-1.511\ eV.\\\\\therefore \Delta E = (-0.378)-(-1.511)=1.133\ eV\\or\ \ \ \Delta E = 1.133\times 1.6\times 10^{-19}\ J=1.8128\times 10^{-19}\ J.[/tex]
Part (b): For the electronic transition from the n = 3 state to the n = 8 state.
The energy of the photon which cause this transition is given by
[tex]\rm \Delta E=E_8-E_3.\\\\where,\\E_8=-\dfrac{E_o}{8^2}=-\dfrac{13.6}{64}=-0.2125\ eV.\\E_3=-\dfrac{E_o}{3^2}=-\dfrac{13.6}{9}=-1.511\ eV.\\\\\therefore \Delta E = (-02125)-(-1.511)=1.298\ eV\\or\ \ \ \Delta E = 1.298\times 1.6\times 10^{-19}\ J=2.077\times 10^{-19}\ J.[/tex]
Final answer:
The energy of the photon that can cause an electronic transition in a hydrogen atom from n = 3 to n = 6 or n = 8 is -3.04 x 10^-19 J. The frequency of the photon is -4.60 x 10^14 Hz.
Explanation:
To calculate the energy of a photon that can cause an electronic transition in a hydrogen atom, we can use the equation:
E = Ef - Ei
where Ef is the energy of the final state and Ei is the energy of the initial state. The energy of a photon is given by:
E = hf
where h is Planck's constant (6.626 x 10-34 J·s) and f is the frequency of the photon. By substituting these equations, we can determine the frequency and energy of the photon.
(a) For an electronic transition from the n = 3 state to the n = 6 state, we have:
E = E6 - E3
E = (-3.4 eV) - (-1.5 eV)
E = -1.9 eV
Using the conversion factor 1 eV = 1.6 x 10-19 J, we can convert the energy to joules:
E = -1.9 eV x (1.6 x 10-19 J/eV)
E = -3.04 x 10-19 J
Now, we can find the frequency of the photon by rearranging the equation:
f = E/h
f = (-3.04 x 10-19 J) / (6.626 x 10-34 J·s)
f = -4.60 x 1014 Hz
(b) For an electronic transition from the n = 3 state to the n = 8 state, we follow the same steps:
E = E8 - E3
E = (-3.4 eV) - (-1.5 eV)
E = -1.9 eV
E = -3.04 x 10-19 J
f = E/h
f = (-3.04 x 10-19 J) / (6.626 x 10-34 J·s)
f = -4.60 x 1014 Hz
Therefore, the energy of the photon that can cause the electronic transition from the n = 3 state to the n = 6 state or the n = 8 state is -3.04 x 10-19 J and the frequency is -4.60 x 1014 Hz.
Why must we be careful when measuring current with a DMM?
Answer:
DMM should be placed in the series combination with the circuit.
Explanation:
DMM is the digital multi meter. It can measure the voltage, current and resistance at a time.
While measuring the current with the DMM you must be ensure that the DMM should be connected with the circuit in series combination. So that it will give the resultant current accurately.While measuring the voltage the observer should check the open probes.A team is building a ballistic ball launcher. The target is 18 ft above the ground, and it needs to catch the ball at the top of its trajectory. Your launcher throws balls from 0.50 m above the ground and must be located 6 m from the target. At what speed must the launcher toss the food in m/s? At what angle above the horizontal must the launcher toss the food? Explain.
Answer:
[tex]\theta = 58.98[/tex]°
Explanation:
given data:
h = 18 ft = 5.48 m
from figure
[tex]h_{max} = 5.48 - 0.50 = 4.98 m[/tex]
[tex]h_{max} = \frac{ v_y^2}{2g}[/tex]
[tex]v_y =\sqrt{2gh_{max}[/tex]
[tex]v_y = \sqrt{2*9.8* 4.98} m/s[/tex]
[tex]v_y = 9.88 m/s[/tex]
[tex]t = \frac{v_y}{g} =\frac{9.88}{9.8} = 1.01 s[/tex]
[tex]v_x = \frac{d}{t} = \frac{6}{1.01} = 5.49 m/s
[tex]v = \sqrt{v_x^2+v_y^2} = \sqrt{5.95^2+9.88^2}[/tex]
v = 11.53 m/s
[tex]tan\theta = \frac[v_x}{v_y}[/tex]
[tex]\theta = tan^{-1} \frac{9.88}{5.94}[/tex]
[tex]\theta = 58.98[/tex]°