That TIME is called the "period" of the wave.
(It's not one of the choices.)
Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to the magnetic field. The electric field strength is 4.0 mV/m at a point 1.5 m away from the center of the circle. At what rate is the magnetic field changing?
Answer:
The rate at which the magnetic field changes is [tex]\frac{\Delta B }{\Delta t } = - 5.33*10^{-3} \ T/ s[/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 4.0 mV/m = 4.0 *10^{-3} V/m[/tex]
The radius of the circular region where the electric field is induced is
[tex]d = 1.5 \ m[/tex]
Generally the induced electric field is mathematically represented as
[tex]E = - \frac{r}{2} * \frac{\Delta B }{\Delta t }[/tex]
The negative sign show that the induced electric field is acting in opposite direction to the change in magnetic field
Where [tex]\frac{\Delta B }{\Delta t }[/tex] is the change in magnetic field
So
[tex]\frac{\Delta B }{\Delta t } = - \frac{2 * E }{r}[/tex]
substituting values
[tex]\frac{\Delta B }{\Delta t } = - \frac{2 * 4.0 *10^{-3}}{ 1.5 }[/tex]
[tex]\frac{\Delta B }{\Delta t } = - 5.33*10^{-3} \ T/ s[/tex]
Rank the six combinations of electric charges on the basis of the electric force acting on q1. Define forces pointing to the right as positive and forces pointing to the left as negative. Rank positive forces as larger than negative forces.
1. q1=+1nC
q2=-1nC
q3 =-1nC
2. q1= -1nC
+ q2 = + 1nC
q3= +1nC
3. q1= +1nC
q2= +1nC
q3= +1nC
4. q1= +1nC
q2= + 1nC
q3= -1nC
5. q1= -1nC
q2= - 1nC
q3= -1nC
6. q1=+1nC
q2=-1nC
q3 =+1nC
Answer:
Plss see attached file
Explanation:
Consider the following spectrum where two colorful lines (A and B) are positioned on a dark background. The violet end of the spectrum is on the left and the red end of the spectrum is on the right. A B 5. (1 point) What is the name for this type of spectrum? 6. (1 point) Transition A is associated with an electron moving between the n= 1 and n= 3 levels. Transition B is associated with an electron moving between the n= 2 and n= 5 levels. Which transition is associated with a photon of longer wavelength?
Answer:
Explanation:
a )
This type of spectrum is called line emission spectrum . Because it consists of lines . It is emission spectrum because it is due to emission of radiation from a source .
b ) The wavelength of a photon is inversely proportional to its energy . Photon due to transition between n = 1 and n = 3 will have higher energy than
that due to transition between n = 2 and n = 5 . So the later photon ( B) will have greater wavelength or photon due to transition between n = 2 and n = 5 will have greater wavelength .
Two factors that regulate (control) glandular secretion.
Answer:
The factors include age and puberty
Explanation:
Glandular secretion release chemicals such as hormones in response to the body’s metabolic needs.
As an individual ages , the metabolic rate of the body also reduces . This is due to the stress and ageing of the cells of the body. This explains why glandular secretion is optimal with young people and Lower in older people. It also explains why the immune system of a young person is mostly stronger than older people.
Puberty is another factor which affects glandular secretion as during puberty there is usually a high amount of hormonal changes due to high levels of secretions of some hormones. These hormones could however inhibit the other glandular secretions.
A system is a group of objects that’s analyzed as one unit. Consider a car moving along a road that has a flat section and a hill. The energy of the car at any given time is equal to the energy that its engine provides minus the energy that the car. When the car moves along the flat section, all of its energy is , which is calculated from its velocity and . When the car moves uphill, some of its energy is transformed to , which is calculated from its gravity, height, and .
Answer:
a) Em= K +U, b) Em= K
Explanation:
The system in this case is formed by the mobilizes and the hill.
Let's write the expressions correctly and completely.
a) When the car moves in the path, the mechanical energy is the siua of the kinetic energy of the car and the potential energy of the car when going up the hill.
Em = K + U
be) when the car moves in the flat part all the mechanical energy is formed by its kinetic energy that is calculated with the mass and speed of the car
Em = K
c) When the car goes up the hill the energy the mechanical energy is conserved, but part of the kinetic energy is transformed into potential energy.
Answer:
leaves
kinetic energy
mass
potential energy
mass
Explanation:
Two long parallel wires separated by 4.0 mm each carry a current of 24 A. These two currents are in the same direction. What is the magnitude of the magnetic field at a point that is between the two wires and 1.0 mm from one of the two wires
Answer:
Explanation:
Magnetic field at a a point R distance away
B = μ₀ / 4π X 2I / R where I is current
Magnetic field due to one current
= 10⁻⁷ x 2 x 24 / 1 x 10⁻³
48 x 10⁻⁴ T
Magnetic field due to other current
= 10⁻⁷ x 2 x 24 / 3x 10⁻³
16 x 10⁻⁴ T
Total magnetic field , as they act in opposite direction, is
= (48 - 16 ) x 10⁻⁴
32 x 10⁻⁴ T .
A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid cons
Complete question:
A solenoid of length 2.40 m and radius 1.70 cm carries a current of 0.190 A. Determine the magnitude of the magnetic field inside if the solenoid consists of 2100 turns of wire.
Answer:
The magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.
Explanation:
Given;
length of solenoid, L = 2.4 m
radius of solenoid, R = 1.7 cm = 0.017 m
current in the solenoid, I = 0.19 A
number of turns of the solenoid, N = 2100 turns
The magnitude of the magnetic field inside the solenoid is given by;
B = μnI
Where;
μ is permeability of free space = 4π x 10⁻⁷ m/A
n is number of turns per length = N/L
I is current in the solenoid
B = μnI = μ(N/L)I
B = 4π x 10⁻⁷(2100 / 2.4)0.19
B = 4π x 10⁻⁷ (875) 0.19
B = 2.089 x 10⁻⁴ T
Therefore, the magnitude of the magnetic field inside the solenoid is 2.089 x 10⁻⁴ T.
A person standing 180m from the foot of a high building claps hi
hand and hears the echo 0.03minutes later. What is the speed
sound in air at that temperature?
A) 331m/s
B) 240m/s C) 200m/s D) 300m/s
Answer:
C) 200 m/s
Explanation:
The sound travels a total distance of 360 m in 0.03 minutes.
v = (360 m) / (0.03 min × 60 s/min)
v = 200 m/s
An electromagnetic standing wave in air of frequency 750 MHz is set up between two conducting planes 80.0 cm apart. At how many positions between the planes could a point charge be placed at rest so that it would remain at rest?
Answer:
The positions the point charge will be placed at rest and still remain at rest are 20 cm, 40 cm, and 60 cm between the ends.
Explanation:
Given;
distance between the conducting planes, d = 80 cm
frequency of the electromagnetic wave, f = 750 MHz
speed of light, c = 3 x 10⁸ m/s²
Determine the wavelength
λ = C/f
where;
λ is the wavelength
C is the speed of light
f is the frequency
λ = C/f
λ = (3 x 10⁸) / (750 x 10⁶)
λ = 0.4 m = 40 cm
One complete cycle = one wavelength = 40 cm
half of the wavelength ( λ / 2) = 20 cm
one wavelength + half wavelength (3λ / 2) = 60 cm
The positions of the wave at zero amplitude (between 0 and 80 cm) = 20 cm, 40 cm, 60 cm
Thus, the positions the point charge will be placed at rest and still remain at rest are 20 cm, 40 cm, and 60 cm between the ends.
A capacitor is charged to a potential of 12.0 V and is then connected to a voltmeter having an internal resistance of 3.40 MΩ. After a time of 4.00 s, the voltmeter reads 3.0 V.
A) What are the capacitance?
B) What is the time constant of the circuit?
Answer:
a. 0.849 micro farad
b. 2.89 s
Explanation:
a) V=V0 e^-t/RC
3=12*e^-4/3.4*10^6*C
3/12=e^-4/3.4*10^6*C
-1.3863 =-4/3.4*10^6*C
C=8.49*10^-7 F
=0.849 micro farad
B) time constant= R*C
=3.4*10^6*8.49*10^-7
=2.89 S
a. The capacitance is 0.849 micro farad
b. The time constant of the circuit is 2.89 s
Calculation of capacitance & time constant:a)
We know that
V=V0 e^-t/RC
3=12*e^-4/3.4*10^6*C
3/12=e^-4/3.4*10^6*C
-1.3863 =-4/3.4*10^6*C
C=8.49*10^-7 F
=0.849 micro farad
B)
Now
time constant= R*C
=3.4*10^6*8.49*10^-7
=2.89 S
Learn more about capacitor here: https://brainly.com/question/24708010
Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true about them as they move apart? (There could be more than one correct choice.)a. Their electrical potential energy keeps decreasing.b. Their acceleration keeps decreasing.c. Their kinetic energy keeps increasing.d. Their kinetic energy keeps decreasing.e. Their electric potential energy keeps increasing.
Answer:
Explanation:
correct options
a ) Their electrical potential energy keeps decreasing
Actually as they move apart , their electrical potential energy decreases due to increase of distance between them and kinetic energy increases
so a ) option is correct
b ) Their acceleration keeps decreasing
As they move apart , their mutual force of repulsion decreases due to increase of distance between them so the acceleration decreases .
c ) c. Their kinetic energy keeps increasing
Their kinetic energy increases because their electrical potential energy decreases . Conservation of energy law will apply .
The moving apart should be true statements:
a. The electrical potential energy should be reduced.
b. The acceleration should be reduced.
c. The kinetic energy should be increased.
True statements related to moving apart:At the time when the moving part, there is the reduction of the electric potential energy because there is a rise in the distance due to which the increment of the kinetic energy. The reduction of the mutual force of repulsion because of increment in the distance due to this the acceleration should be reduced. There is the increase in the kinetic energy due to the reduction of the electrical potential energy. here the law of conversation of energy should be applied.
Learn more about energy here: https://brainly.com/question/10658188
Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheel's energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg. Its maximum angular velocity is 1200 rpm.
How long does it take the flywheel to reach top angular speed of 1200 rpm?
Answer:
t = 2.95 min
Explanation:
Given that,
The diameter of flywheeel, d = 1.5 m
Mass of flywheel, m = 250 kg
Initial angular velocity is 0
Final angular velocity, [tex]\omega_f=1200\ rpm = 126\ rad/s[/tex]
We need to find the time taken by the flywheel to each a speed of 1200 rpm if it starts from rest.
Firstly, we will find the angular acceleration of the flywheel.
The moment of inertia of the flywheel,
[tex]I=\dfrac{1}{2}mr^2\\\\I=\dfrac{1}{2}\times 250\times (0.75)^2\\\\I=70.31\ kg-m^2[/tex]
Now,
Let the torque is 50 N-m. So,
[tex]\alpha =\dfrac{\tau}{I}\\\\\alpha =\dfrac{50}{70.31}\\\\\alpha =0.711\ rad/s^2[/tex]
So,
[tex]t=\dfrac{\omega_f-\omega_i}{\alpha }\\\\t=\dfrac{126-0}{0.711}\\\\t=177.21\ s[/tex]
or
t = 2.95 min
A stunt driver rounds a banked, circular curve. The driver rounds the curve at a high, constant speed, such that the car is just on the verge of skidding to the outside of the curve. A front view of a car driving on a banked curve. The cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car. Which forces are directly responsible for producing the car’s centripetal acceleration? Coriolis force centripetal force frictional force normal force gravitational force
Answer: C
Frictional force
Explanation:
The description of the question above is an example of a circular motion.
For a car travelling in a curved path, the frictional force between the tyres and the road surface will provide the centripetal force.
Since the road is banked, and the cross section of the banked road is constructed like a ramp. The car drives transversely to the slope of the ramp, so that the wheels of one side of the car are lower than the wheels on the other side of the car, for cornering the banked road, the car will not rely only on the frictional force.
Therefore, the correct answer is option C - the frictional force.
If the magnetic field steadily decreases from BBB to zero during a time interval ttt, what is the magnitude III of the induced current
Answer:
Using ohms law
The current is found from Ohm's Law.
I = V /R = E /R = Bxy /Rt.
A circular loop of wire has radius of 9.50 cm. A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is 0.0295 W/m^2, and the wavelength of the wave is 6.40 m.
Required:
What is the maximum emf induced in the loop?
Answer:
The maximum emf induced in the loop is 0.132 Volts
Explanation:
Given;
radius of the circular loop, r = 9.5 cm
intensity of the wave, I = 0.0295 W/m²
wavelength, λ = 6.40 m
The intensity of the wave is given as;
[tex]I = \frac{B_o^2*c }{2\mu_o}[/tex]
where;
B₀ is the amplitude of the field
c is the speed of light = 3 x 10⁸ m/s
μ₀ is permeability of free space = 4π x 10⁻⁵ m/A
[tex]I = \frac{B_o^2*c }{2\mu_o}\\\\B_o^2 = \frac{I*2\mu_o}{c} \\\\B_o^2 = \frac{0.0295*2*4\pi*10^{-7}}{3*10^8} \\\\B_o^2 = 2.472 *10^{-16}\\\\B_o = \sqrt{2.472 *10^{-16}}\\\\ B_o = 1.572*10^{-8} \ T[/tex]
Area of the circular loop;
A = πr²
A = π(0.095)²
A = 0.0284 m²
Frequency of the wave;
f = c / λ
f = (3 x 10⁸) / (6.4)
f = 46875000 Hz
Angular velocity of the wave;
ω = 2πf
ω = 2π(46875000)
ω = 294562500 rad/s
The maximum induced emf is calculated as;
emf = B₀Aω
= (1.572 x 10⁻⁸)(0.0284)(294562500)
= 0.132 Volts
Therefore, the maximum emf induced in the loop is 0.132 Volts
The horizontal surface on which the objects slide is frictionless. If F = 6.0 N and M = 1.0 kg, what is the magnitude of the force exerted on the large block by the small block?
The image is missing, so i have attached it.
Answer:
The force exerted on the large block by the small block = 8.4 N
Explanation:
From the image attached, the mass of the small block = 2M while the mass of the large block = 3M
Also,Force on small block = F and force on large block = 2F
Equilibrium of forces on the left gives;
2F - N = 3Ma
Thus,
Ma = (2F - N)/3 - - - - eq1
Also, on right hand side, Equilibrium of forces gives;
N - F = 2Ma
Ma = (N - F)/2 - - - - eq2
Equating eq(1) and eq(2) gives us;
(2F - N)/3 = (N - F)/2
Where N is the force exerted on the large block by the small block.
Making N the subject gives;
4F - 2N = 3N - 3F
5N = 7F
N = 7F/5
We are given F = 6N
Thus;
N = 7(6)/5
N = 8.4 N
We repeat the experiment from the video, but this time we connect the wires in parallel rather than in series. Which wire will now dissipate the most heat?
Both wires will dissipate the same amount of heat.
A. The Nichrome wire (resistance 2.7)
B. The copper wire (resistance 0.1)
Answer: B. The copper wire (resistance 0.1)
Explanation: When resistance is in parallel, voltage (V) is the same but current is different for every resistance. Current (i) is related to voltage and resistance (R) by Ohm's Law
i = [tex]\frac{V}{R}[/tex]
So, since both wires are in parallel, they have the same voltage but because the copper wire resistance is smaller than Nichrome wire, the first's current will be bigger.
Every resistor in a circuit dissipates electrical power (P) that is converted into heat energy. The dissipation can be found by:
P = [tex]i^{2}*R[/tex]
As current for copper wire is bigger than nichrome, power will be bigger and it will dissipate more heat.
In conclusion, the copper wire will dissipate more heat when connected in parallel.
A metal ring 4.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.200 T/s.
(a) What is the magnitude of the electric field induced in the ring?
(b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?
Explanation:
a) d[phi]/dt = (dB/dt)*Acos(0) = (-0.20)*(pi(2.25*10^-2)^2) = -3.98*10^-4 Wb
E = (1/2r*pi)*(d[phi]/dt) = -2.8*10^-3 N/C
b) Clockwise because The induced magnetic field will be in the direction to oppose the change. Since the magnetic flux from the magnets is decreasing, the induced magnetic field will be in the same direction as the magnet's field.
0.25-kg block oscillates on the end of a spring with a spring constant of 200 N/m. If the oscillations is started by elongating the spring 0.15 m and giving the block a speed of 3.0 m/s, then the maximum speed of the block is A :
Answer:
5.2m/s
Explanation:
Plss see attached file
A flat loop of wire consisting of a single turn of cross-sectional area 8.60 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.40 T in 1.02 s. What is the resulting induced current if the loop has a resistance of 2.80
Answer:
The induced current is [tex]I = 5.72*10^{-4 } \ A[/tex]
Explanation:
From the question we are told that
The cross-sectional area is [tex]A = 8.60 \ cm^2 = \frac{8.60 }{10000} = 8.60 *10^{-4} \ m[/tex]
The initial value of magnetic field is [tex]B_1 = 0.500 \ T[/tex]
The value of magnetic field at time t is [tex]B_f = 2.40 \ T[/tex]
The number of turns is N = 1
The time taken is [tex]dt[/tex]= 1.02 \ s
The resistance of the loop is [tex]R = 2.80\ \Omega[/tex]
Generally the induced emf is mathematically represented as
[tex]e = - \frac{d \phi}{dt }[/tex]
Where [tex]d \phi[/tex] is the change n the magnetic flux which is mathematically represented as
[tex]d \phi = N *A * d B[/tex]
Where [tex]dB[/tex] is the change in magnetic field which is mathematically represented as
[tex]d B = B_f - B_i[/tex]
substituting values
[tex]d B = 2.40 - 0.500[/tex]
[tex]d B = 1.9 \ T[/tex]
So
[tex]d \phi = 1 * 1.9 * 8.60 *10^{-4}[/tex]
[tex]d \phi = 1.63*10^{-3} \ T[/tex]
So
[tex]e = - \frac{1.63 *10^{-3}}{ 1.02 }[/tex]
[tex]e = - 1.60*10^{-3} \ V[/tex]
Here the negative only indicates that the emf is acting in opposite direction of the motion producing it so the magnitude of the emf is
[tex]e = 1.60*10^{-3} \ V[/tex]
Now the induced current is evaluated as follows
[tex]I = \frac{e}{R }[/tex]
substituting values
[tex]I = \frac{1.60 *10^{-3}}{2.80 }[/tex]
[tex]I = 5.72*10^{-4 } \ A[/tex]
greater than: The electric potential energy of a proton at point A is _____ the electric potential energy of an proton at point B.
Answer:
[similar to]
Explanation:
it is the missing word
In each of the following situations, a wave passes through an opening in an absorbing wall. Rank the situations in order from the one in which the wave is best described by the ray approximation to the one in which the wave coming through the opening spreads out most nearly equally in all directions in the hemisphere beyond the wall.
(a) The sound of a low whistle at 1 kHz passes through a doorway 1 m wide.
(b) Red light passes through the pupil of your eye.
(c) Blue light passes through the pupil of your eye.
(d) The wave broadcast by an AM radio station passes through a doorway 1 m wide. (e) An x-ray passes through the space between bones in your elbow joint.
Answer:
A) geometric optics, B) geometric optics , c) geometric optics ,
e) geometric optics, f) geometric optics
Explanation:
For this exercise we must use the condition for interference and diffraction so that these phenomena are relevant the wavelength must be comparable to the gap spacing
λ> = a
Lam when the spacing is much greater than the wavelength, the description of geometric optics is more and more exact
let's analyze each situation
a) let's find the wavelength
v = λ f
λ= v / f
λ= 343/1000
λ = 0.343 m
0.343 << 1m
therefore the description of the geometric optics of
b) red light passes through the pupil of the eye
red light has a wavelength of 700 num or more, the lojo pupil has a maximum of 8 me
λ = 700 10⁻⁹ m = 7 10⁻⁷ m
a = 8 mm 10⁻³
longitudinal is much less therefore the geometric optics is correct
c) luz azul lam = 450 nm = 450 10⁻⁹ m = 4.5 10⁻⁷ m
again the wavelength is much less than the diameter of the pupil, for which the description with the optics is generally sufficient
d) a radio A transmits up to a maximum of f = 1400 Khz = 1,400 10⁶ Hz
let's find the wavelength
c = λf
λ = c / f
λ= 3 108 / 1,400 106
λ= 2.14 102 m
in this case the wavelength is greater than the width of the gate, so the description of diffraction should be used to explain the phenomenon
e) X-rays have wavelength lam = 10-10 m
the separation of the elbow bones is of the order of a few millimeters, for local the wavelength is much less than the separation, therefore with the relations of geometric optics it is sufficient
An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.
What is the emf of the coil? (in V)
Answer:
The induced emf is [tex]\epsilon =7.68 \ V[/tex]
Explanation:
From the question we are told that
The number of turns is [tex]N = 1300 \ turns[/tex]
The diameter is [tex]d = 2.2 \ cm = 2.2*10^{-2}[/tex]
The initial magnetic field is [tex]B_i = 0.14 \ T[/tex]
The final magnetic field is [tex]B_f = 0 \ T[/tex]
The time taken is [tex]dt = 9.0ms = 9.0*10^{-3} \ s[/tex]
The radius is mathematically evaluated as
[tex]r = \frac{d}{2 }[/tex]
substituting values
[tex]r = \frac{2.2 *10^{-2}}{2 }[/tex]
[tex]r = 1.1*10^{-2} \ m[/tex]
The induced emf is mathematically represented as
[tex]\epsilon =- N * \frac{d\phi }{dt }[/tex]
Where [tex]d\phi[/tex] is the change in magnetic field which is mathematically represented as
[tex]d\phi = dB * A * cos\theta[/tex]
=> [tex]d\phi = [B_f - B_i ] * A * cos\theta[/tex]
Here [tex]\theta = 0[/tex] given that the axis of the coil is parallel to the field
Also A is the cross-sectional area which is mathematically represented as
[tex]A = \pi r^2[/tex]
substituting values
[tex]A = 3.142 * [1.1*10^{-2}]^2[/tex]
[tex]A = 3.8 *10^{-4] \ m^2[/tex]
So
[tex]d\phi = [0 - 0.14 ] * 3.8*10^{-4}[/tex]
[tex]d\phi = -5.32*10^{-5} \ weber[/tex]
So
[tex]\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }[/tex]
[tex]\epsilon =7.68 \ V[/tex]
Determine the position in the oscillation where an object in simple harmonic motion: (Be very specific, and give some reasoning to your answer.) has the greatest speed has the greatest acceleration experiences the greatest restoring force experiences zero restoring force g
Answer:
Explanation:
The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .
When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .
The greatest acceleration is attained at maximum displacement or at one of the two extreme end .
Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .
Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .
Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .
A proton moving at 4.80 106 m/s through a magnetic field of magnitude 1.74 T experiences a magnetic force of magnitude 7.00 10-13 N. What is the angle between the proton's velocity and the field? (Enter both possible answers from smallest to largest. Enter only positive values between 0 and 360.)
Answer:
31.55° and 148.45°
Explanation:
Formula for calculating the force experiences by the proton placed in a magnetic field is as expressed below;
F = qvBsinθ where;
F is the magnetic force experienced by the proton
q is the charge on the proton
v is the velocity of the proton
B is the magnetic field
θ is the angle between the proton's velocity and the field (Required)
Given parameters
F = 7.00 * 10⁻¹³N
q = 1.602*10⁻¹⁹C
v = 4.80 * 10⁶ m/s
B = 1.74 T
θ =?
From the formula F = qvBsinθ;
sinθ = F/qvB
sinθ = 7.00 * 10⁻¹³/1.602*10⁻¹⁹* 4.80 * 10⁶*1.74
sinθ = 7.00 * 10⁻¹³/13.38*10⁻¹³
sinθ = 0.5231689 * 10⁰
sinθ = 0.5231689
θ = sin⁻¹0.5231689
θ = 31.55°
The following are the positive values of the angle between 0° and 360°
Sin is positive in the first and second quadrant. In the second quadrant the angle is equal to 180°-31.55° = 148.45°.
Hence the possible values of the angle from smallest to largest are 31.55° and 148.45°
a 5 charge is locataed 1.25 m to the left of a -3 charge. What is the magnitude and direction of the electrostatic force on the postive charge
Answer:
The force is 86.5×10^9 N towards the negative charge (to the right)
Explanation:
The electrostatic force on the charges is given by Coulomb's law;
F= Kq1q2/r^2
This an inverse square law.
F= electrostatic force on the charges
K= constant of Coulomb's law
q1 and q2= magnitude of the charges
Since K= 9.0×10^9Nm^2C^2
F= 9.0×10^9 × 5 × 3/(1.25)^2 = 135×10^9/1.56
F= 86.5×10^9 N
The force is 86.5×10^9 N towards the negative charge.
A skater spins at 3rev/s when she stretches her arms outward. If she keeps her fists on her chest she can spin at 4.5rev/s and her body inertia is 3kg.m2. What is her body inertia when she stretches her arms outward?
Answer:
Body inertia I = 4.5 kg/m^2
Explanation:
Here, we want to calculate the body inertia when the arms are stretched outwards.
We know from the question that angular momentum is conserved
Thus;
I * 3 = 4.5 * 3
I = 4.5 kg/m^2
The voltage across the terminals of an ac power supply varies with time according to V=V0cos(t). The voltage amplitude is V0 = 41.0V .
A. What is the root-mean-square potential difference Vrms?
B. What is the average potential difference Vav between the two terminals of the power supply?
Answer:
A) V_rms = 29 V
B) Vav = 0 V
Explanation:
A) We are told that;
V = V_o cos ωt
voltage amplitude; V = V_o = 41.0V
Now, the formula for the root-mean-square potential difference Vrms is given as;
V_rms = V/√2
Thus plugging in relevant values, we have;
V_rms = 41/√2
V_rms = 29 V
B) Due to the fact that the voltage is sinusoidal from the given V = V_o cos ωt, we can say that the average potential difference Vav between the two terminals of the power supply would be zero.
Thus; Vav = 0 V
A. The root-mean-square potential difference ([tex]V_{rms}[/tex]) is equal to 28.99 Volts.
B. For this voltage with a sinusoidal waveform (sine wave), the average potential difference ([tex]V_{ave}[/tex]) between the two terminals of the power supply is equal to zero (0).
Given the following data:
Voltage amplitude = 41.0 Volts.The voltage across the terminals of an alternating current (AC) power supply varies directly with time according to the equation:
[tex]V_0 = V_0cos(t)[/tex]
A. To find the root-mean-square potential difference ([tex]V_{rms}[/tex]):
Mathematically, root-mean-square for voltage in an alternating current (AC) power supply (circuit) is given by the formula:
[tex]V_{rms} = \frac{V}{\sqrt{2} }[/tex]
Substituting the given parameter into the formula, we have;
[tex]V_{rms} = \frac{41}{\sqrt{2} }\\\\V_{rms} = \frac{41}{1.4142 }\\\\V_{rms} = 28.99\; Volts[/tex]
B. To find the average potential difference ([tex]V_{ave}[/tex]) between the two terminals of the power supply:
For this voltage with a sinusoidal waveform (sine wave), the average potential difference ([tex]V_{ave}[/tex]) between the two terminals of the power supply is equal to zero (0).
Read more: https://brainly.com/question/15121836
A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g and a volume of 4,064 cm3. The density of seawater is 1.03 g/mL.
a. What is the weight of the statuette?
b. What is the mass of displaced water?
c. What is the weight of displaced water?
d. What is the buoyant force on the statuette?
e. What is the net force on the statuette?
f. How much force would be required to lift the statuette?
Answer:
A) W = 103.55 N
B) mass of displaced water = 4186 g
C) W_displaced water = 41.06 N
D) Buoyant force = 41.06 N.
E) ZERO
F) 62.54 N
Explanation:
We are given;
mass of statuette;m = 10,566 g = 10.566 kg
volume = 4,064 cm³
Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³
A) The dry weight of the statuette can be calculated as;
W = mg
So;
W = 10.556 × 9.81
W = 103.55 N
B) Mass of displaced water is calculated from;
Density = mass/volume
So, mass = Density × Volume
m = 1.03 × 4,064 = 4186 g
C) Weight of displaced water is given by;
W_displaced water = (m_displaced water) × g
W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N
D) The buoyant force is the same as the weight of the displaced water.
Thus, Buoyant force = 41.06 N.
E) The apparent weight of the statuette is calculated from;
Apparent weight = Dry weight - Weight of displaced water
Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.
F. From E above, The Force required to lift the statuette = 62.54 N
A 50 g ball of clay traveling at speed v0 hits and sticks to a 1.0 kg brick sitting at rest on a frictionless surface.
Required:
a What is the speed of the block after the collision?
b. What percentage of the ball's initial energy is "lost"?
Answer:
(a) The speed of the block after the collision is 0.0476v0.
(b) The percentage of the ball's initial energy lost, is 0 % (energy is conserved)
Explanation:
Given;
mass of ball of clay, m₁ = 50 g = 0.05 kg
mass of brick, m₂ = 1 kg
initial velocity of the ball of clay, u₁ = v0
initial velocity of the brick, u₂ = 0
Since the clay ball sticks with the brick after collision, it is inelastic collision.
Therefore, let their final velocity = v
(a) What is the speed of the block after the collision?
Apply the principle of conservation linear momentum
m₁u₁ + m₂u₂ = v (m₁ + m₂)
0.05v₀ + 1(0) = v( 0.05 + 1)
0.05v₀ = 1.05v
v = 0.05v₀ / 1.05
v = 0.0476v₀
Thus, the speed of the block after the collision is 0.0476 of its initial velocity.
(b). What percentage of the ball's initial energy is "lost"?
Initial kinetic energy of the ball = ¹/₂mv₀²
= ¹/₂ x 0.05 x v₀²
= 0.025v₀²
Final kinetic energy of the ball, = ¹/₂(m₁ + m₂)v²
= ¹/₂ x 1.05 x 0.0476v₀²
= 0.025v₀²
Change in kinetic energy = 0.025v₀² - 0.025v₀²
= 0
percentage change in the initial kinetic energy of the ball;
= (0 / 0.025v₀²) x 100%
= 0 x 100%
= 0 %
Therefore, the percentage of the ball's initial energy lost, is 0 % (energy is conserved)