Final answer:
The final velocity of the two football players after the collision is 0.738 m/s in the positive direction, calculated using the conservation of momentum.
Explanation:
The scenario described involves a conservation of momentum problem, where two football players collide and cling together. We can solve for the final velocity by using the principle of conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces are acting on it.
To find the final combined velocity of the players after the collision, we use the formula:
Momentum before collision = Momentum after collision
(m1 × v1) + (m2 × v2) = (m1 + m2) × v_final
Plugging in the given values:
(89.5 kg × 6.05 m/s) + (111 kg × (-3.55 m/s)) = (89.5 kg + 111 kg) × v_final
After calculating both sides,
541.475 kg·m/s - 393.55 kg·m/s = 200.5 kg · v_final
We get the final combined velocity,
v_final = 147.925 kg·m/s / 200.5 kg
v_final = 0.7377 m/s (to four significant figures)
The players will be moving with a velocity of 0.738 m/s in the positive direction.
Write a function to model the volume of a rectangular prism if the length is 24cm and the sum of the width and height is 28cm
[tex]w + h = 28[/tex]
[tex]l \times w \times h = v[/tex]
[tex]v = w \times h \times 24[/tex]
[tex]w = 28 - h[/tex]
[tex]h(28 - h) \times 24 = v[/tex]
[tex]24( - {h}^{2} + 28h) = v[/tex]
Describe the process of nuclear fission. How does it work and when is it used? What is meant by a chain reaction, and how do you control it?
A diffraction grating is illuminated with yellow light. The diffraction pattern seen on a viewing screen consists of three yellow bright spots, one at the central maximum (θ = 0°) and one on either side of it at θ = ±50°. Then the grating is simultaneously illuminated with red light. Where a red and a yellow spot overlap, an orange spot is produced. The new pattern consists of __________
Answer:
an orange fringe at 0°, yellow fringes at ±50° and red fringes farther out.
Explanation:
In the visible spectrum- red to violet, red has the highest wavelength.
The maximum internsity for diffraction grating is given by,
Sinθ[tex](_{max} )[/tex] = mλ/d
It is concluded that the angle of diffraction increases with increase in wavelength'λ' . So, red fringe will be farthest from the center, orange light will be at the center and yellow fringe will be at 50°.
Therefore, The new pattern consists of : an orange fringe at 0°, yellow fringes at ±50° and red fringes farther out.
When a rocket is traveling toward a mountain at 100 m/s, the sound waves from this rocket's engine approach the mountain at speed V.
If the rocket doubles its speed to 200 m/s, the sound waves from the engine will now approach the mountain at speed
A. 2–√V
B. 2V
C. V
D. 4V
Answer:
The correct option is C
Explanation:
From the question we are told that
The initial speed of the rocket is [tex]v_i = 100 m/s[/tex]
The speed of the rocket engine sound is [tex]V[/tex]
The final speed of the rocket is [tex]v_f = 200 \ m/s[/tex]
The speed of the sound at [tex]v_f[/tex] would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).
A clear example when lightning strikes you will first see (that is because it travels at the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the lightning
Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates
When you run around a track at 5 km/h, your velocity is constant. Please select the best answer from the choices provided T F
Answer:
T
Explanation:
Answer:
true
Explanation:
A 0.5 Kg pinball is initially at rest against a 120 N/m spring. The shooter is pulled back and has the spring compressed a distance of 0.2 m. The spring is released and the ball is shot up the ramp. It hits nothing and eventually comes to rest before it begins to roll down. We can ignore friction. The game board ramp is at an angle of 30o . How far did the ball travel on the board from the place of maximal compression to the first stop
Answer:
[tex]\Delta s = 0.978\,m[/tex]
Explanation:
The pinball-spring system is modelled after the Principle of Energy Conservation:
[tex]U_{g,1} + U_{k,1} + K_{1} = U_{g,2} + U_{k,2} + K_{2}[/tex]
[tex]-(0.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right) \cdot \Delta h + \frac{1}{2}\cdot \left(120\,\frac{N}{m}\right)\cdot (-0.2\,m)^{2} = 0[/tex]
The height reached by the pinball is:
[tex]\Delta h = 0.489\,m[/tex]
The distance travelled by the pinball is:
[tex]\Delta s =\frac{0.489\,m}{\sin 30^{\circ}}[/tex]
[tex]\Delta s = 0.978\,m[/tex]
For an experiment, a student wants to maximize the power output for a circuit. The student has three resistors of resistances 512, 1012, and 20 S, and a 5 V battery. Which of the following best describes how the students should arrange the resistors so that when they are connected to the battery it will maximize the power output for the circuit and explains why? a) The student should place all resistors in parallel, because the battery supplies more current to a parallel combination than to a series combination b) The student should place all resistors in parallel, because the potential difference across all the resistors is less for a parallel combination than for a series combination c) The student should place all resistors in series, because this creates the largest equivalent resistance d) The student should place all resistors in series, because the battery supplies more current to a series combination than to a parallel combination e) The student should place all resistors in series, because the potential difference across all the resistors is greater in series than in parallel
Final answer:
To maximize the power output of a circuit with a 5V battery and three resistors, the resistors should be placed in parallel because this arrangement supplies more current than a series combination, thereby maximizing power.
Explanation:
To maximize the power output for a circuit with a given voltage source, the resistors should be arranged to draw the highest current possible. For resistors connected in parallel, the total resistance of the circuit decreases, as opposed to when they are in series, where it increases. Since power (P) is calculated using the formula P = V^2 / R, where V is the voltage and R is the resistance, minimizing R will maximize P.
Because a parallel configuration guarantees that each resistor gets the full voltage of the battery, the correct answer is: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination. This is because in a parallel circuit, the voltage across each resistor is equal to the voltage of the source, which maximizes the current through each resistor due to Ohm's law (I = V/R), and therefore maximizes the total current in the circuit.
Answer option (a) is thus correct: The student should place all resistors in parallel because the battery supplies more current to a parallel combination than to a series combination.
A cart of mass 300 g is placed on a frictionless horizontal air track. A spring having a spring constant of 9.0 N/m is attached between the cart and the left end of the track. The cart is displaced 3.8 cm from its equilibrium position. (a) Find the period at which it oscillates. Correct: Your answer is correct. s (b) Find its maximum speed. Incorrect: Your answer is incorrect. Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m/s (c) Find its speed when it is located 2.0 cm from its equilibrium position. m/s
Answer:
Explanation:
the angular frequency ω of the pendulum is given by the formula
ω = [tex]\sqrt{\frac{k}{m} }[/tex] , k is spring constant , m is mass attached .
= [tex]\sqrt{\frac{9}{.3} }[/tex]
= 5.48 rad /s
time period = 2π / ω
= 2 x 3.14 / 5.48
= 1.146 s
b ) formula for speed
v = ω[tex]\sqrt{(a^2-\ x^2)}[/tex] , a is amplitude , x is displacement from equilibrium point.
for maximum speed x = 0
max speed = ωa
= 5.48 x 3.8 x 10⁻² ( initial displacement becomes amplitude that is 3.8 cm )
= .208 m /s
20.8 cm / s
c )
when x = .02 m , velocity = ?
v = ω[tex]\sqrt{(a^2-\ x^2)}[/tex]
= 5.48 [tex]\sqrt{(.038^2-\ .02^2)}[/tex]
= 5.48 x .0323109
= .177 m /s
17.7 cm /s .
Two identical disks, with rotational inertia I (= 1/2 MR2), roll without slipping across a horizontal floor and then up inclines. Disk A rolls up its incline without sliding. On the other hand, disk B rolls up a frictionless incline. Otherwise the inclines are identical. Disk A reaches a height 12 cm above the floor before rolling down again. Disk B reaches a height above the floor of:
Answer:
8cm
Explanation:
Here, two disc are identical and rolling on the horizontal surface
Also,while disc is in rolling motion its kinetic energy is sum of rotational kinetic energy and transnational kinetic energy.
Therefore,
KE = [tex]\frac{1}{2}[/tex]mv²+ [tex]\frac{1}{2}[/tex][tex]Iw[/tex]²
For pure rolling of disc we have: [tex]v=Rw[/tex]
[tex]I=\frac{1}{2} mR[/tex]²
By substituting in KE eq, we get
KE = [tex]\frac{1}{2}[/tex]mv²+ [tex]\frac{1}{2}[/tex]([tex]\frac{1}{2} mR[/tex]²)([tex]\frac{v^{2} }{R^{2} }[/tex])
KE= [tex]\frac{1}{2} mv^{2} + \frac{1}{4} mv^{2}[/tex]
The total kinetic energy will convert into gravitational potential energy when disc roll over the inclined surface.
mgH=[tex]\frac{1}{2} mv^{2} + \frac{1}{4} mv^{2}[/tex] =>[tex]\frac{3}{4} mv^{2}[/tex]
[tex]mv^{2}[/tex]= 4/3mgH
If another disc rolls up on frictionless inclined plane then it will lose all its translational kinetic energy but rotational kinetic energy will remain as it is as there is no torque on the disc
Therefore, mgh= 1/2mv²
1/2mv²= 4/3mgH
mgh=2/3mgH
h=2/3H
Height = 12cm is given
h= 8cm
Thus, Disk B reaches a height of 8cm above the floor.
An electron moves through a uniform electric field E = (2.60î + 5.90ĵ) V/m and a uniform magnetic field B = 0.400k T. Determine the acceleration of the electron when it has a velocity v = 6.0î m/s. (Give each component in m/s2.)
Answer:
The answer is "[tex]1.75 \cdot (2.60 \hat{i} + 3.5 \hat{j}) \times 10^{11} \ \ \frac{m}{s^2} \\[/tex] ".
Explanation:
Formula of acceleration =
[tex]\frac{F_e}{m_e} =\frac{-e(\underset{E}{\rightarrow} + \underset{V}{\rightarrow} \times \underset{B}{\rightarrow})}{m_e}[/tex]
values:
[tex]\underset{E}{\rightarrow} = (2.60 \hat{i} + 5.90 \hat{j}) \frac{V}{m} \\\\\underset{B}{\rightarrow} = 0.400 k \ T \\\\\underset{V}{\rightarrow} = 6.0 \hat {i} \ \ \frac{m}{s} \\\\\ apply \ value \ in \ above \ formula: \\\\ \frac{F_e}{m_e} =\frac{e}{m_e} \cdot (2.60 \hat{i} + 5.90 \hat{j}+6.0 \hat {i} \times 0.4\hat {k} ) \\\\ \frac{F_e}{m_e} =\frac{e}{m_e} \cdot (2.60 \hat{i} + 5.90 \hat{j}- 2.4 \hat {j}) \\\\\therefore \frac{e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}} \\\\[/tex]
[tex]\frac{F_e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}} \cdot (2.60 \hat{i} + 5.90 \hat{j}- 2.4 \hat {j}) \\\\\frac{F_e}{m_e} = \frac{1.6 \times 10^{-19}}{9.1 \times 10^{-21}} \cdot (2.60 \hat{i} + 3.5 \hat{j}) \\\\\frac{F_e}{m_e} = 1.75 \cdot (2.60 \hat{i} + 3.5 \hat{j}) \times 10^{11} \ \ \frac{m}{s^2} \\\\[/tex]
The acceleration of the electron can be found by using the net force formula and Newton's second law. The given values of velocity and magnetic field can be used to calculate the force experienced by the electron due to the magnetic field. The force can then be divided by the mass of the electron to find the acceleration.
Explanation:The acceleration of the electron can be found using the formula for the net force on a charged particle moving in a magnetic field.
The force experienced by the electron due to the magnetic field is given by the equation F = qvB, where F is the force, q is the charge of the particle, v is the velocity, and B is the magnetic field. Plugging in the given values, we have F = (1.60 x 10-19 C)(6.00 x 107 m/s)(0.500 T).
To determine the acceleration, we can use Newton's second law, F = ma. Rearranging the equation, we have a = F/m, where a is the acceleration and m is the mass of the electron. Plugging in the values, we get a = [(1.60 x 10-19 C)(6.00 x 107 m/s)(0.500 T)] / (9.11 x 10-31 kg).
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Your high-fidelity amplifier has one output for a speaker of resistance 8 Ω. How can you arrange two 8-Ω speakers, one 4-Ω speaker, and one 12-Ω speaker so that the amplifier powers all speakers and their equivalent resistance when connected together in this way is 8 Ω? Compare the power output of your arrangement with the power output of a single 8-Ω speaker.
Answer:
(a) 8Ω (b) Ratio = Parra/P8 ohm = 1
Explanation:
Solution
Recall that,
An high-fidelity amplifier has one output for a speaker of resistance of = 8 Ω
Now,
(a) How can two 8-Ω speakers be arranged, when one = 4-Ω speaker, and one =12-Ω speaker
The Upper arm is : 8 ohm, 8 ohm
The Lower arm is : 12 ohm, 4 ohm
The Requirement is = (16 x 16)/(16 + 16) = 8 ohm
(b) compare your arrangement power output of with the power output of a single 8-Ω speaker
The Ratio = Parra/P8 ohm = 1
Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjacent wall. The plane of the incident and reflected beams is perpendicular to the wall. The beam from the laser strikes the mirror at a distance a=33.7 cm from the wall. The reflected beam strikes the wall at a height b=36.7 cm above the surface of the mirror. Find the angle of incidence θi at which the laser beam strikes the mirror.
Answer:
The angle of incidence is [tex]\theta_i =42.6^o[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The distance between the mirror and the wall is [tex]a = 33.7 cm[/tex]
The height of the above the mirror is [tex]b = 36.7 cm[/tex]
Generally the angle which the reflected ray make with the mirror is mathematically evaluated as
[tex]\alpha =tan ^{-1} (\frac{b}{a})[/tex]
substituting values
[tex]\alpha = tan ^{-1}( \frac{36.7}{33.7})[/tex]
[tex]\alpha =47.4^o[/tex]
From the diagram we can deduce that the angle of incidence is
[tex]\theta_i = 90 - \alpha[/tex]
So [tex]\theta_i = 90 - 47.4[/tex]
[tex]\theta_i =42.6^o[/tex]
A 150 g pinball rolls towards a springloaded launching rod with a velocity of 2.0 m/s
to the west. The launching rod strikes the pinball and causes it to move in the
opposite direction with a velocity of 10.0 m/s. What impulse was delivered to the
pinball by the launcher?
A 0.75 kg•m/s to the east
B 1.2 kg•m/s to the east
C 1.8 kg•m/s to the east
D 3.0 kg•m/s to the east
The impulse delivered to the pinball by the launcher is 1.8 kg•m/s to the east
Impulse:Let us consider east as positive direction and west as negative direction.
Then, from the question, we get that the initial velocity of the pinball is
2m/s towards the west
or u = - 2 m/s
and the mass of the pinball is m = 150g = 0.15 kg
So, the initial momentum of the pinball is:
[tex]P_i=mu\\\\P_i=0.15\times(-2)\;kgm/s\\\\P_i=-0.3\;kgm/s[/tex]
Now, the final velocity of the pinball after being struck by the rod is 10 m/s towards the east,
or v = 10 m/s
So, the final momentum of the pinball is:
[tex]P_f=mu\\\\P_f=0.15\times(10)\;kgm/s\\\\P_f=1.5\;kgm/s[/tex]
Impulse is defined as the change in momentum, that is,
[tex]I=\Delta P\\\\I=P_f-P_i\\\\I=1.5-(-0.3)\\\\I=1.8\;kgm/s[/tex]
The impulse is 1.8 kgm/s towards the east.
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A stone is moving on a rough level surface. It has 24 J of kinetic energy, and the friction force on it is a constant 0.50 N. What is the maximum distance it can slide?
The energy an object has as a result of motion is known as kinetic energy.A force must be applied to an object in order to accelerate it.We must put in effort in order to apply a force.After the work is finished, energy is transferred to the item, which then moves at a new, constant speed.
Solve the problem ?
We have provided This is supposed to be a stone, and it is travelling in this way, on a rough, level surface.It has 24 Joel of initial kinetic energy.And 0.50 Newton is the major force of friction on it.The friction force is this.This equates to 0.50fresh banknotes.What is the longest it can slide before coming to a stop?Assume that it has a sliding distance of D before halting.Therefore, the final candidate energy will be 0.We will now use the idea of energy heroes, of which we are aware. This friction force must be equivalent to the change in candidate energy.Finally, without K. Initiative.As a result, friction force will multiply the work done by frictionAs a result, friction force will multiply the work done by friction.The displacement and force are at an angle of 1 80 degrees in this position.As a result, this has lost him 1 80 since his negative 24.Energy nil because of final counting.Therefore, the cost of this distance D. will be 24 divided by.0.50.Consequently, this is a 48-meter cost alternative.To learn more about kinetic energy refer
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Final answer:
To determine the maximum distance the stone can slide on a rough surface, divide the kinetic energy by the friction force. The maximum distance is 48 meters.
Explanation:
To determine the maximum distance the stone can slide on a rough surface, we need to calculate the work done by the friction force.
Work done = Force x Distance
Given that the friction force is 0.50 N and the stone has 24 J of kinetic energy, we can rearrange the equation to solve for distance:
Distance = Kinetic energy / Friction force
Plugging in the values, we get:
Distance = 24 J / 0.50 N = 48 meters
Therefore, the maximum distance the stone can slide on the rough surface is 48 meters.
Atomic physicists usually ignore the effect of gravity within an atom. To see why, we may calculate and compare the magnitude of the ratio of the electrical force and gravitational force between an electron and a proton separated by a distance of 1 m.
1. What is the magnitude of the electrical force?
The Coulomb constant is 8.98755 x 10^9 N*m^2/C^2 , the gravitational constant is 6.67259 x 10^11 m^3 /kg*s^2 , the mass of a proton is 1.67262 x 10^-27 kg, the mass of an electron is 9.10939 x 10^−31 kg, and the elemental charge is 1.602 x 10^-19 C. Answer in units of N.
Explanation:
The electrical force between charges is given by :
[tex]F_e=\dfrac{kq_eq_p}{r^2}[/tex]
[tex]q_e\ and\ q_p[/tex] are charge on electron and proton respectively.
[tex]F_e=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1^2}\\\\F_e=2.3\times 10^{-28}\ N[/tex]
The Gravitational force between masses is given by :
[tex]F_G=\dfrac{Gm_em_p}{r^2}[/tex]
[tex]m_e\ and\ m_p[/tex] are masses of electron and proton respectively.
[tex]F_G=\dfrac{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}{1^2}\\\\F_G=1.01\times 10^{-67}[/tex]
Ratio of electrical to the gravitational force is :
[tex]\dfrac{F_e}{F_G}=\dfrac{2.3\times 10^{-28}\ N}{1.01\times 10^{-67}\ N}\\\\\dfrac{F_e}{F_G}=2.27\times 10^{39}[/tex]
Hence, this is the required solution.
Final answer:
To find the magnitude of the electrical force between an electron and proton separated by 1 m, we use Coulomb's Law with given constants. The electrical force is calculated to be approximately 2.30 x 10⁻¹⁰ N.
Explanation:
To calculate the magnitude of the electrical force between an electron and proton separated by 1 meter, we use Coulomb's Law which is given by the formula:
F = k * |q¹ * q²| / r²
Where F is the force in Newtons (N), k is the Coulomb constant (8.98755 x 10⁹ N*m²/C²), q1 and q2 are the charges of the proton and electron respectively, and r is the separation distance in meters. Since both the proton and the electron have an elemental charge of 1.60² x 10⁻¹⁹ C, albeit with opposite signs, their charges can be multiplied to give their product in Coulomb's equation. The separation r is 1 meter.
So the magnitude of the electrical force F is calculated as:
F = (8.98755 x 109 N*m²/C²) * (1.60² x 10⁻¹⁹ C)² / 12m²
F ≈ 2.30 x 10¹⁰ N
Block B, with mass 5.00 kg, rests on block A, with mass 8.00 kg, which in turn is on a horizontal tabletop. There is no friction between block A and tabletop, but the coefficient of static friction between block A and block B is 0.750. A light string attached to block A passes over a frictionless, massless pulley, and block C is suspended from the other end of the string. What is the largest mass that block C can have so that blocks A and B still slide together when the system is released from rest
The largest mass that block C can have so that blocks A and B still slide together, given the values, is approximately 10.2 kg.
Explanation:The subject of this question is a physics problem involving static friction, mass, and gravity. The largest mass that block C can have so that blocks A and B still slide together can be calculated using the principle of static friction and the relevant equation:
fs_max = µs (mA + mB)g, with fs_max representing the maximum force of static friction, µs=0.75 being the coefficient of static friction, and g=9.8 m/s² being the acceleration due to gravity. Here mA=8 kg and mB=5 kg are the masses of block A and B respectively.
To keep A and B together, the tension (T) in the string must be less than or equal to fs_max. As T is also equal to the weight (mCg) of block C, from where we can find mC ≤ fs_max / g. Substituting the given values and calculations, we find that the maximum mass of block C should be around 10.2 kg for blocks A and B to slide together.
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Final answer:
The maximum mass that block C can have to prevent sliding between block A and B is calculated using the static friction coefficient [tex](\mu_s)[/tex] and the weight of block B [tex](m_B)[/tex], which results in a mass of 3.75 kg for block C.
Explanation:
To figure out the largest mass that block C can have without block A and B sliding relative to each other, we must use the static friction force and Newton's second law. The static friction force [tex](f_s)[/tex] is what keeps block B from sliding on block A.
This force is given by [tex]f_s = \mu_s \times N[/tex], where μ_s is the coefficient of static friction and N is the normal force. Since block B is at rest on block A, N is equal to the weight of block B, [tex]N = m_B \times g[/tex].
For block A and block B to accelerate together, the tension in the string (T) caused by the weight of block C must not exceed the static friction force. Therefore, the maximum force that can be applied by block C before block B starts sliding is the static friction force: T ≤ f_s.
So, the maximum weight (and hence mass) that block C can have is when [tex]T = f_s[/tex]. Since [tex]T = m_C \times g[/tex] for block C, we have m_C × g = μ_s × m_B × g. Canceling out g and solving for m_C gives us the formula [tex]m_C = \mu_s \times m_B[/tex].
Substituting the given values, we get [tex]m_C = 0.750 \times 5.00 kg = 3.75 kg[/tex]. Hence, the largest mass that block C can have without causing sliding between block A and B is 3.75 kg.
A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4 A is in the coil. The coil is set in a magnetic field of magnitude B = 1.1 T. The magnetic field points to right
The complete question is;
A circular coil consists of N = 410 closely winded turns of wire and has a radius R = 0.75 m. A counterclockwise current I = 2.4 A is in the coil. The coil is set in a magnetic field of magnitude B = 1.1 T.
a. Express the magnetic dipole moment μ in terms of the number of the turns N, the current I, and radius
R.
b. Which direction does μ go?
Answer:
A) μ = 1738.87 A.m²
B) The direction of the magnetic moment will be in upward direction.
Explanation:
We are given;
The number of circular coils;
N = 410
The radius of the coil;R = 0.75m
The current in the coils; I = 2.4 A
The strength of magnetic field;
B =1.1T
The formula for magnetic dipole moment is given as;
μ = NIA
Where;
N is number of turns
I is current
A is area
Now, area; A = πr²
So, A = π(0.75)²
Thus,plugging in relevant values, the magnetic dipole moment is;
μ = 410 * 2.4 * π(0.75)²
μ = 1738.87 A.m²
B) According to Fleming's right hand rule, the direction of the magnetic moment comes out to be in upward direction.
The circular coil with specific parameters immersed in the magnetic field is analyzed according to the principles of electromagnetism. The right-hand rule is used to determine the direction of the magnetic field, and formulas are used to calculate the magnetic force and torque on the coil given the counterclockwise current and magnetic field strength.
Explanation:In the described case, we are dealing with a circular coil of N=410 turns, with a radius R = 0.75m, carrying a counterclockwise current I = 2.4A. This coil is set in a magnetic field B = 1.1T, that points to the right. When a current flows in a wire, it creates a magnetic field around it. The direction of magnetic field can be determined by the right-hand rule, where your thumb points in the direction of the current and your fingers curl in the direction of the magnetic field.
In a circular loop, there is a simple formula for calculating the magnetic field strength at the center of the loop. If we consider the magnetic field created in this circular loop wire, its strength and directionality would vary. The magnetic force on this current-carrying conductors is given by F = I x B, where I is the current and B is the magnetic field.
The net torque on a current-carrying loop of any shape in a magnetic field is given by t = μ × B, where μ is the magnetic dipole moment and B is the magnetic field strength. The orientation and magnitude of magnetic field would cause varying effects in the coil.
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The potential energy of a 1.7 x 10-3 kg particle is described by U (x )space equals space minus (17 space J )space cos open square brackets fraction numerator x over denominator 0.35 space straight m end fraction close square brackets. What is the angular frequency of small oscillations around the point x = 0?
Answer:
Explanation:
Given a particle of mass
M = 1.7 × 10^-3 kg
Given a potential as a function of x
U(x) = -17 J Cos[x/0.35 m]
U(x) = -17 Cos(x/0.35)
Angular frequency at x = 0
Let find the force at x = 0
F = dU/dx
F = -17 × -Sin(x/0.35) / 0.35
F = 48.57 Sin(x/0.35)
At x = 0
Sin(0) =0
Then,
F = 0 N
So, from hooke's law
F = -kx
Then,
0 = -kx
This shows that k = 0
Then, angular frequency can be calculated using
ω = √(k/m)
So, since k = 0 at x = 0
Then,
ω = √0/m
ω = √0
ω = 0 rad/s
So, the angular frequency is 0 rad/s
Characteristics help scientists ________ objects.
Answer:Identify
Explanation:
Answer:
b
Explanation:
n optician is performing Young's double-slit experiment for her clients. She directs a beam of monochromatic light to a pair of parallel slits, which are separated by 0.134 mm from each other. The portion of this light that passes through the slits goes on to form an interference pattern upon a screen, which is 4.50 meters distant. The light is characterized by a wavelength of 553 nm. (a) What is the optical path-length difference (in µm) that corresponds to the fourth-order bright fringe on the screen? (This is the fourth fringe, not counting the central bright band, that one encounters moving from the center out to one side.)
Answer:
Explanation:
This is a problem based on interference pf light waves.
wavelength of light λ = 553 nm
slit separation d = .134 x 10⁻³ m
screen distance D = 4.5 m
for fourth order bright fringe, path- length difference = 4 x λ
= 4 x 553
= 2212 nm .
= 2.212 μm
To measure specific heat, the student flows air with a velocity of 20 m/s and a temperature of 25C perpendicular to the length of the tube, as illustrated below. In this time she heats the tube electrically, passing a current of 50 mA through the tube wall. The initial temperature of the tube is 25 C. After 10 min of heating, a thermocouple buried inside the tube wall measures a temperature of 31.2 C. The tube core is sealed and the heat lost from the internal surface of the tube is negligible during measurements. Furthermore, the tube is suspended on two thermally insulating supports. What is the specific heat of the tube?
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2. How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver
Determining the radius of the satellite TV dish to achieve a specific electric field requires knowing the intensity and power of the broadcast signal as well as the dish's area. The radius can be calculated from the formula for the area encompassed by the dish, but without additional information about the broadcast power or spread area, a specific radius cannot be provided.
Explanation:To determine how large the radius R of the satellite TV receiver dish must be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver, we need to relate the intensity of the electromagnetic wave to the electric field amplitude and the area of the dish. The intensity (I) is related to the electric field strength (E) by the relationship I = rac{1}{2} extZ_0[tex]E^2[/tex], where Z0 is the impedance of free space (approximately 377 ohms).
The power received by the dish (Pr) is the product of the intensity and the area of the dish, Ad: Pr = I imes Ad. Given that the receiver has an area of 5 cm2 and the required electric field amplitude is 0.1 mV/m, we would solve for the radius R of the dish using the formula for area of a circle, A = \\(pi)[tex]R^2[/tex].
However, to solve this problem, we would need additional information such as the power broadcast by the satellite and over what area this power is spread. With our current information, we cannot provide an exact answer, but typically residential satellite dishes have diameters a little less than half a meter to effectively receive TV signals.
Use the drop-down menus to complete the scenarios.
A patient has an ongoing history of cancer. She has a tumor in the abdominal region, and has been undergoing
treatment for it. There may be other tumors and a potential blockage in the surrounding area that need to be
investigated. The imaging technique that might provide the most information in this case is
Joe has ongoing issues with his throat and feels some sort of blockage or abnormality as he swallows. The doctor
decides to use X-ray imaging to visualize Joe's internal anatomy as he swallows to help determine the nature of the
problem.
will be used for this procedure.
First: CT Scan
Second: Fluoroscopy
Explanation:
Both correct
A CT scan can provide detailed images of a tumor and its surrounding area, which can be beneficial for a cancer patient. For ongoing throat issues, a Barium swallow study can be used, where the patient swallows a barium solution that is then visualized with an X-ray to identify abnormalities.
Explanation:In the first scenario, the patient with an ongoing history of cancer might benefit most from a Computed Tomography (CT) scan. CT scans are capable of creating detailed pictures of organs, bones, and other tissues, making it an excellent tool for capturing the size and position of a tumor and surrounding blockages in the abdominal area. Furthermore, it can reveal whether the cancer has spread to other parts of the body.
In the second scenario, Joe's doctor chooses to use X-ray imaging to detect any abnormalities in his throat. The specific procedure used for this is known as a Barium swallow study. This involves swallowing a barium solution that coats the esophagus, enabling the X-ray to capture clear images of the region as the patient swallows.
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Balloon like cell organelle in the cytoplasm that can store food, water, and wastes, also provides pressure against the cell wall for cellular support in plant cells
A
Vacuoles
B
Mitochondrion
C
Ribosomes
D
Nucleus
PLSSSSSS HELP MEEEEE!! :C
Answer:
Vacuoles......final answer
wo plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.70 nC . Ball 1 is negatively charged, and ball 2 is positively charged. The balls are held apart by a 500-mm stick stuck through the holes so that it runs from the center of one ball to the center of the other.
What is the magnitude of the dipole moment of the arrangement?
Answer:
the magnitude of the dipole moment is 3.5*10^-11Cm
Explanation:
The dipole moment is given by the following formula:
[tex]\mu=qr[/tex]
r: distance between the centers of the charges = 500mm
q: charges of the bowling balls = 0.70nC
By replacing you obtain:
[tex]\mu=(0.70*10^{-9}C)(500*10^{-4}m)=3.5*10^{-11}Cm[/tex]
hence, the magnitude of the dipole moment is 3.5*10^-11Cm
Learning Goal: To understand how to find the wavelength and diffraction patterns of electrons. An electron beam is incident on a single slit of width aaa. The electron beam was generated using a potential difference of magnitude VVV. After passing through the slit, the diffracted electrons are collected on a screen that is a distance LLL away from the slit. Assume that VVV is small enough so that the electrons are nonrelativistic. Ultimately, you will find the width of the central maximum for the diffraction pattern.
Final answer:
The question involves calculating the wavelength and diffraction patterns of electrons in a single slit experiment. The student would use the de Broglie wavelength formula and the principles of single slit diffraction to find the dimensions of the central maximum. It demonstrates the wave-particle duality exhibited by electrons.
Explanation:
The student is seeking to understand how to find the wavelength and diffraction patterns of electrons. The context involves an electron beam passing through a single slit, creating a diffraction pattern on a distant screen. From the given potential difference, one could calculate the de Broglie wavelength of the electrons, as their velocity can be determined under the assumption that they are nonrelativistic. The central maximum's width on the diffraction pattern can be deduced using the slit width and the wavelength.
The phenomenon of diffraction and interference highlighted in the question is a demonstration of the wave-like properties of electrons, referred to as wave-particle duality. The experimental setup often includes narrow slits whose sizes are comparable to the wavelength of electrons, resulting in observable wave effects such as constructive and destructive interference.
The angular positions of the minima and maxima in the diffraction pattern are crucial for determining the dimensions of the pattern. The de Broglie wavelength plays a significant role in these calculations, linking the microscopic quantum world to observable macroscopic patterns.
A sinusoidal electromagnetic wave is propagating in a vacuum in the z-direction. If at a particular instant and at a certain point in space the electric field is in the x-direction and has a magnitude of 4.50 V/m, what is the magnitude of the magnetic field of the wave at this same point in space and instant in time
Answer:
Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction
Explanation:
From maxwell's equations;
B = E/v
Where;
B is maximum magnitude of magnetic field
E is maximum electric field
v is speed of light which has a constant value of 3 x 10^(8) m/s
We are given, E = 4.5 V/m
Thus; B = 4.5/(3 x 10^(8))
B = 1.5 x 10^(-8) T
Now, for Electric field, vector E to be in the positive x-direction, the product of vector E and vector B will have to be in the positive z-direction when vector B is in the positive y-direction
Thus,
Magnitude of magnetic field is 1.5 x 10^(-8) T in the positive y-direction
Magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex] in the positive y-direction.
From Maxwell's equations,
[tex]\bold {B = \dfrac Ev}[/tex]
Where;
B - maximum magnitude of magnetic field = ?
E- maximum electric field = 4.5 V/m
v- speed of light = 3 x 10^(8) m/s
Put the values in the formula,
[tex]\bold {B = \dfrac {4.5}{3 x 10^8}}\\\\\bold {B = 1.5 x 10^{-8}\ T}[/tex]
When Electric field, is in the positive x-direction, vector B is in the positive y-direction and the product of vector E and vector B will have to be in the positive z-direction.
Therefore, magnitude of magnetic field in the space at given instant in time is [tex]\bold { 1.5 x 10^{-8}\ T}[/tex] in the positive y-direction.
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Follow these steps to solve this problem: Two identical loudspeakers, speaker 1 and speaker 2, are 2.0 m apart and are emitting 1700-Hz sound waves into a room where the speed of sound is 340 m/s. Consider a point 4.0 m in front of speaker 1, which lies along a line from speaker 1, that is perpendicular to a line between the two speakers. Is this a point of maximum constructive interference, a point of perfect destructive interference, or something in between?
Answer:
It is somewhere in between
Explanation:
Wave length of sound from each of the speakers = 340 / 1700 = 0.2 m = 20 cm
Distance between first speaker and the given point = 4 m.
Distance between second speaker and the given sound
D = √(4² + 2²)
D = √(16 + 4)
D = √20
D = 4.472 m
Path difference = 4.472 - 4 =
0.4722 m.
Path difference / wave length = 0.4772 / 0.2 = 2.386
This is a fractional integer which is neither an odd nor an even multiple of half wavelength. Hence this point is of neither a perfect constructive nor a perfect destructive interference.
To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(x,t)=Acos(kx−ωt). A transverse wave on a string is traveling in the +x direction with a wave speed of 8.50 m/s , an amplitude of 5.50×10−2 m , and a wavelength of 0.500 m . At time t=0, the x=0 end of the string has its maximum upward displacement. Find the transverse displacement y of a particle at x = 1.52 m and t = 0.150 s .
Answer:
0.0549 m
Explanation:
Given that
equation y(x,t)=Acos(kx−ωt)
speed v = 8.5 m/s
amplitude A = 5.5*10^−2 m
wavelength λ = 0.5 m
transverse displacement = ?
v = angular frequency / wave number
and
wave number = 2π/ λ
wave number = 2 * 3.142 / 0.5
wave number = 12.568
angular frequency = v k
angular frequency = 8.5 * 12.568
angular frequency = 106.828 rad/sec ~= 107 rad/sec
so
equation y(x,t)=Acos(kx−ωt)
y(x,t)= 5.5*10^−2 cos(12.568 x−107t)
when x =0 and and t = 0
maximum y(x,t)= 5.5*10^−2 cos(12.568 (0) − 107 (0))
maximum y(x,t)= 5.5*10^−2 m
and when x = x = 1.52 m and t = 0.150 s
y(x,t)= 5.5*10^−2 cos(12.568 (1.52) −107(0.150) )
y(x,t)= 5.5*10^−2 × (0.9986)
y(x,t) = 0.0549 m
so the transverse displacement is 0.0549 m
The air in a car tire la compressed when the car rolls over a rock. If the air
outside the tire becomes slightly warmer, what are the correct signs of heat
and work for this change?
Answer:
the signs of heat and work are; -Q and -W
Explanation:
The first law of thermodynamics is given by; ΔU = Q − W
where;
ΔU is the change in internal energy of a system,
Q is the net heat transfer (the sum of all heat transfer into and out of the system)
W is the net work done (the sum of all work done on or by the system).
Now, The system in this case is the tire and since the air gets warmer, heat must have left the system. Therefore Q is negative (-Q).
Since work is done by the system, W remains negative.
Thus, the signs of heat and work are; -Q and - W