Answer:
B.
Explanation:
I'm taking the test right now.
The final momentum of the system is 0.4kgm/s
According to the law conservation of momentum, the momentum of the system before the collision and after the collision remains consevred.
if [tex]m_{1}[/tex] is the mass of one ball and its initial and final velocities be [tex]v_{1}[/tex] and [tex]v_{1f}[/tex] respectively, and
if [tex]m_{2}[/tex] is the mass of one ball and its initial and final velocities be [tex]v_{2}[/tex] and [tex]v_{2f}[/tex] respectively
then, [tex]m_{1}v_{1} + m_{2}v_{2}=m_{1}v_{1f} + m_{2}v_{2f}[/tex]
Momentum after collision = [tex]m_{1}v_{1} + m_{2}v_{2}[/tex][tex]=0.6*0.5+0.5*0.2[/tex]
Momentum after collision = 0.4 kgm/s
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Suppose we take a 1 m long uniform bar and support it at the 21 cm mark. Hanging a 0.40 kg mass on the short end of the beam results in the system being in balance. Find the mass of the beam.
Answer:
The mass of the beam is = 29 kg.
Explanation:
A beam with mass 40 kg is shown in figure. Point S is the support point. Point B is the middle point on the beam where mass of the beam acts.
Taking moment about Point S
40 × 21 = [tex]M_{beam}[/tex] × 29
[tex]M_{beam}[/tex] = 29 kg
Therefore the mass of the beam is = 29 kg.
An open-delta three-phase transformer system has one transformer center-tapped to provide a neutral for single-phase voltages. If the voltage from line to center tap is 277 V, what is the high-leg voltage?
Answer:
The high-leg voltage is [tex]V_h= 497.76V[/tex]
Explanation:
A diagram showing the arrangement of this three phase transformer is shown on the first uploaded image
The high - leg terminal is in the diagram is from [tex]Z_1 - n[/tex]
Generally the high-leg terminal is 1.732 times of the the voltage from line to center tap which is given as 277 V
The high -voltage can be computed as follows
[tex]V_h = 1.732 *277[/tex]
[tex]V_h= 497.76V[/tex]
A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a speed equal to one third the original speed of the 7.7 kg sphere. What is the mass of the second sphere
Answer:
15.4 kg.
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').................... Equation 1
Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.
Given: m = 7.7 kg, u' = 0 m/s (at rest)
Let: u = x m/s, and V = 1/3x m/s
Substitute into equation 1
7.7(x)+m'(0) = 1/3x(7.7+m')
7.7x = 1/3x(7.7+m')
7.7 = 1/3(7.7+m')
23.1 = 7.7+m'
m' = 23.1-7.7
m' = 15.4 kg.
Hence the mass of the second sphere = 15.4 kg
Answer:
The mass of the second sphere is 15.4 kg
Explanation:
Given;
mass of the first sphere, m₁ = 7.7 kg
initial velocity of the second sphere, u₂ = 0
let mass of the second sphere = m₂
let the initial velocity of the first sphere = u₁
final velocity of the composite system, v = ¹/₃ x u₁ = [tex]\frac{u_1}{3}[/tex]
From the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ + m₂)
Substitute the given values;
[tex]7.7u_1 + 0=\frac{u_1}{3} (7.7+m_2)[/tex]
Divide through by u₁
7.7 = ¹/₃(7.7 + m₂)
multiply both sides by 3
23.1 = 7.7 + m₂
m₂ = 23.1 - 7.7
m₂ = 15.4 kg
Therefore, the mass of the second sphere is 15.4 kg
If given a device that has unknown circuitry, and you measure that the voltage across the device leads the voltage across a resistor at 500 Hz, can you know what type of elements are in the device?If so, how?
Final answer:
A leading voltage across a device at 500 Hz compared to a resistor suggests inductive elements within the device. Voltage leads current in inductors, which can be determined by comparing phase differences using a voltmeter parallel to the device.
Explanation:
When measuring the voltage across a device and comparing it to the voltage across a resistor at a specific frequency, such as 500 Hz, the phase difference provides information about the type of elements in the device. If the voltage across the device leads the voltage across the resistor, this indicates the presence of inductive reactance, suggesting that there are inductive elements within the unknown circuitry.
Reactance and impedance play key roles in AC circuits. The voltage drop across a resistor is in phase with the current, meaning it neither leads nor lags the current. However, in an inductor, the voltage leads the current, and in a capacitor, the voltage lags the current. Therefore, measuring a leading voltage implies inductive characteristics. To evaluate this, a voltmeter would be placed in parallel with the device, ensuring minimal disturbance to the circuit as high resistance in the voltmeter minimizes current flow through it.
For components like inductors and capacitors, the relationship between voltage and current is not linear; thus, they are not ohmic devices like resistors. This is evident in an RLC series circuit, where the impedance at a given frequency depends on the resistance (R), inductance (L), and capacitance (C) of the circuit.
The voltage leading the resistor's voltage at 500 Hz suggests the device contains capacitive elements. A phase difference measurement can confirm this. Capacitors cause the voltage to lead the current in AC circuits.
If you observe that the voltage across a device leads the voltage across a resistor at 500 Hz, it is likely that the device includes a capacitive element. In AC circuits, capacitors cause the voltage to lead the current, which means that the voltage across the capacitor will peak before the voltage across a purely resistive element.
To confirm the presence of a capacitive component, you can perform a more detailed analysis:
Measure the phase difference between the voltage across the device and the voltage across the resistor.If the phase difference is close to 90 degrees, it indicates the presence of a pure capacitor.If the phase difference is less than 90 degrees but still significant, it may indicate a combination of resistive and capacitive elements, known as an RC circuit.This phase difference can be analyzed using an oscilloscope to visualize the waveform and confirm the capacitive nature of the device.
Interactive LearningWare 22.2 reviews the fundamental approach in problems such as this. A constant magnetic field passes through a single rectangular loop whose dimensions are 0.46 m x 0.68 m. The magnetic field has a magnitude of 3.0 T and is inclined at an angle of 67o with respect to the normal to the plane of the loop. (a) If the magnetic field decreases to zero in a time of 0.49 s, what is the magnitude of the average emf induced in the loop
Explanation:
The dimension of a single rectangular loop is 0.46 m x 0.68 m.
Magnetic field, B = 3 T
The loop is inclined at an angle of 67 degrees with respect to the normal to the plane of the loop.
It is required to find the magnitude of the average emf induced in the loop if the magnetic field decreases to zero in a time of 0.49 s.
Te induced emf in the loop is given by :
[tex]\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA)}{dt}\\\\\epsilon=A\dfrac{dB}{dt}\cos\theta\\\\\epsilon=0.46\times 0.68\times \dfrac{3}{0.49}\times \cos(67)\\\\\epsilon=0.74\ V[/tex]
So, the magnitude of the average emf induced in the loop is 0.74 V.
Vector A has a magnitude of 6.0 m and points 30° north of east. Vector B has a magnitude of 4.0 m and points 30° west of north. The resultant vector A + B is given by:
(a) 9.8 m at an angle of 26° north of east,
(b) 3.3 m at an angle of 26° north of east,
(c) 7.2 m at an angle of 26° east of north,
(d) 3.3 m at an angle of 64° east of north or
(e) 9.8 m at an angle of 64° east of north
Answer:
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,
C
Explanation:
Resolving the vectors to vertical and horizontal component;
Vertical;
Vector A = 6sin30
Vector B = 4sin60
Resultant vertical = 6sin30 + 4sin60 = 6.464m
Horizontal;
Vector A = 6cos30
Vector B = -4cos60
Resultant horizontal = 6cos30 - 4cos60 = 3.196m
Resultant R = √(6.464^2 + 3.196^2) = 7.2m
Tanθ = 6.464/3.196
θ = taninverse (6.464/3.196) BN
θ = 64° north of East.
Or
26° east of north
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north,
The resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.
Given data:
The magnitude of vector A is, A = 6.0 m.
The direction of vector A is, 30° north of east.
The magnitude of vector B is, B = 4.0 m.
The direction of vector B is, 30° west of north.
The quantity having both the magnitude as well as the magnitude are known as vector quantities.
Resolving the vectors to vertical and horizontal component;
Along the Vertical direction;
Vector A = 6sin30
Vector B = 4sin60
Resultant vertical vector = 6sin30 + 4sin60 = 6.464 m
Along the Horizontal direction;
Vector A = 6cos30
Vector B = -4cos60
Resultant horizontal vector = 6cos30 - 4cos60 = 3.196 m
Now, the resultant vector is calculated as,
[tex]R=\sqrt{6.464^{2}+3.196^{2}}\\\\R = 7.2 \;\rm m[/tex]
And the resultant direction of vectors is,
[tex]tan \theta = 6.464/3.196\\\\\theta = tan^{-1} (6.464/3.196) \\\theta =64 ^{\circ}[/tex]( North of East)
θ = 64° north of East.
Or
26° east of north
Thus, we can conclude that the resultant vector A + B is given by 7.2 m at an angle of 26° east of north. Hence, option (c) is correct.
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g A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.462 of the escape speed from Earth and (b) its initial kinetic energy is 0.462 of the kinetic energy required to escape Earth
Answer:
a.)r=4RE/4
b.)r=2RE
c.)ZERO
Explanation:
a.) given v= 0.462 which is V= 0.466Ve
Since the projectile is shot directly away from earth surface the speed it escape with;
v=√2GM/RE......................eqn(1)
M is the Earth's Mass
RE is the Radius
From the Law of conservation of Energy
K+U=0...............................eqn(2)
K₁+U₁=K₂+U₂....................eqn(3)
where K₁ and U₁ is initial kinetic and potential energy
K₂ and U₂ are final kinetic and potential energy
Kinetic Energy (K.E) decrease with time as the projectile moves up and there is decrease in Potential Energy (P.E) , it will let to a point where K.E will turn to zero i.e K₂=0
U₂=K₁U₂ ..........................eqn(3)
From Gravitational Law
U₁= -GMm/RE ..................(5)
U₂= -GMm/r .....................(6)
Where "r" is the distance
v= 0.462√2GM/RE
v= √GM/2RE
GM/4RE - GM/RE = -GM/r
r= 4RE/3
b.) "r" is calculated by this equation;
K₁=0.466
K₁= 1/4MVe².............................eqn(9)
substitute eqn(1) into eqn (9) then
1/4m2GM/RE=0.466GM/2Re
GM/2RE - GM/RE =-GM/r
r=2RE
c.)The potential energy and kinetic energy is the same in terms of their size both in different directions, while the potential energy face outward, the kinetic energy face inward therefore the least initial mechanical energy
required at launch if the projectle is to escape is ZERO
A closed loop conductor that forms a circle with a radius of is located in a uniform but changing magnetic field. If the maximum emf induced in the loop is what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies
Final answer:
The maximum rate at which the magnetic field strength is changing in a closed loop conductor can be determined using Faraday's law. The rate is equal to the maximum induced emf divided by the area of the loop.
Explanation:
The maximum rate at which the magnetic field strength is changing can be determined using Faraday's law of electromagnetic induction. According to Faraday's law, the induced emf (voltage) in a closed loop conductor is equal to the negative rate of change of magnetic flux through the loop. In this case, the magnetic field is perpendicular to the plane of the loop, so the magnetic flux is given by the product of the magnetic field strength and the area of the loop. Therefore, the maximum rate of change of magnetic field strength is equal to the maximum induced emf divided by the area of the loop.
So, the maximum rate at which the magnetic field strength is changing is given by: (d(B)/dt) = (Emax / A)
where (d(B)/dt) is the rate of change of magnetic field strength, Emax is the maximum induced emf, and A is the area of the loop.
An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 1.80 T field with his fingers pointing in the direction of the field. His wedding ring has a diameter of 2.23 cm, and it takes 0.320 s to move it into the field.
1. What average current is induced in the ring in A if its resistance is 0.0100 Ω?
2. What average power is dissipated in mW?
3. What magnetic field is induced at the center of the ring in T?
Given that,
Magnetic field strength is
B = 1.8T
The wedding ring has a diameter of
d = 2.23 cm = 0.023m
Time take t = 0.32 secs
A. Current induced
From ohms law
V= iR, given that R = 0.01Ω
So, we need to get the induced emf
Using
ε = -NdΦ / dt
Where Φ = BA
ε = -A ∆B / ∆t
ε = -¼πd²(B2-B1) / (t2-t1)
ε = -¼ × π × 0.023² × -1.8 / 0.32
ε = 0.0023371 V
Then
I = ε / R
I = 0.002337 / 0.01
I =0.2337 A
B. Power discippated?
Power is given as
P = iV
P = 0.2337 × 0.002337
P = 0.0005462 W
P = 0.56 mW
C. The magnetic field at the centre of the ring.
The electric field at the centre of the ring is zero because each part of the ring will cause a symmetrical opposite magnitude at every point,
Then, B = 0 T
As a dilligent physics student, you carry out physics experiments at every opportunity. At this opportunity, you carry a 1.11-m-long rod as you jog at 3.07 m/s, holding the rod perpendicular to your direction of motion. What is the strength of the magnetic field that is perpendicular to both the rod and your direction of motion and that induces an EMF of 0.287 mV across the rod
Answer:
The strength of the magnetic field is 0.0842 mT
Explanation:
Given:
Velocity of rod [tex]v = 3.07[/tex] [tex]\frac{m}{s}[/tex]
Length of rod [tex]l = 1.11[/tex] m
Induced emf across the rod [tex]\epsilon = 0.287 \times 10^{-3}[/tex] V
According to the faraday's law
We have a special case for moving rod in magnetic field.
Induced emf in moving rod is given by,
[tex]\epsilon = Blv[/tex]
Where [tex]B =[/tex] strength of magnetic field
[tex]B = \frac{\epsilon}{lv}[/tex]
[tex]B = \frac{0.287 \times 10^{-3} }{3.07 \times 1.11}[/tex]
[tex]B = 0.0842 \times 10^{-3}[/tex] T
[tex]B = 0.0842[/tex] mT
Therefore, the strength of the magnetic field is 0.0842 mT
In astronomy, distances are often expressed in light-years. One light-year is the distance traveled by light in one year. If the distance to a star is 3.6 light-years, what is this distance in meters? (There are 365.25 days in one year.)
Answer:
The distance of the star is [tex]3.40x10^{16}[/tex] meters
Explanation:
It is known that the speed of light has a value of [tex]3x10^{8}m/s[/tex] in vacuum. That is, it travels [tex]3x10^{8]m[/tex] in one second, according with the following equation:
[tex]v = \frac{x}{t}[/tex]
Where v is the speed, x is the distance and t is the time.
[tex]x = v\cdot t[/tex] (1)
Equation 1 can be used to determine the distance that the light travels in 1 year:
It is necessary to find how many seconds are in 1 year (365.25 days).
[tex]365.25 days \cdot \frac{86400s}{1 day}[/tex] ⇒ [tex]31557600s[/tex]
[tex]x = (3x10^{8}m/s)(31557600s)[/tex]
[tex]x = 9.46x10^{15}m[/tex]
Therefore, in 1 year, light travels [tex]9.46x10^{15}[/tex] meters.
If the distance to a star is 3.6 light-years, what is this distance in meters?
A simple conversion between units can be used to get the distance in meters
[tex]x_{star} = 3.6ly \cdot \frac{9.46x10^{15}m}{1ly}[/tex] ⇒ [tex]3.40x10^{16}m[/tex]
Hence, the distance of the star is [tex]3.40x10^{16}[/tex] meters.
A light-year, a distance unit used in astronomy, is approximately 9.46 x [tex]10^{15[/tex] meters. Hence, a distance of 3.6 light years would convert to approximately 3.4 x [tex]10^{16[/tex] meters.
Explanation:A light-year is a unit of distance used in astronomy, which represents the distance light travels in one year. Light travels at a speed of 186,000 miles per second. Over the course of a year, this distance adds up significantly.
To calculate a light-year in meters, we would carry out the following calculation - light travels at 299,792 kilometers per second. This converts to exactly 299,792,000 meters per second. If we multiply this by the number of seconds in a year (60 seconds/minute, 60 minutes/hour, 24 hours/day, 365.25 days/year) we find that one light-year is approximately 9.46 x [tex]10^{15[/tex] meters.
Therefore, if the distance to a star is 3.6 light-years, then this would convert to about 3.4 x [tex]10^{16[/tex] meters.
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A man pushes a shopping cart at a constant speed. What happens if the man increases the force on the cart?
Answer:
The cart would speed up.
Explanation:
According to Newton's 1st law, object subjected to no force, or net force 0, would have a constant speed. In our case the cart is initially at constant speed, meaning the man exerts a force that is equal to friction force. If he increases the force on the cart, the net force would no longer be 0. The cart would gain an acceleration and increases its speed.
Answer:
A. The cart accelerates
Explanation:
A smooth circular hoop with a radius of 0.600 m is placed flat on the floor. A 0.350-kg particle slides around the inside edge of the hoop. The particle is given an initial speed of 10.00 m/s. After one revolution, its speed has dropped to 5.50 m/s because of friction with the floor.
(a) Find the energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution.
(b) What is the total number of revolutions the particle makes before stopping? Assume the friction force remains constant during the entire motion.
Answer:
a. The energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules
b. The total number of revolutions the particle makes before stopping is 1.43 revolutions
Explanation:
a.
Given
m = mass of particle = 0.350-kg
u = initial speed of 10.00 m/s
v = final speed = 5.50 m/s
r = radius = 0.600 m
We assume that the floor is horizontal;
This means that F = mg.
We also assume the rotational kinetic energy to be negligable.
Having listed the assumptions, we proceed as follows;
Let ∆E represent The energy transformed from mechanical to internal in the particle hoop.
This is given by
∆E = KE1 - KE2
Where KE1 = ½mu²
KE2 = ½mv²
So, ∆E = KE1 - KE2 becomes
∆E = ½mu² - ½mv²
∆E = ½m(u² - v²)
∆E = ½ * 0.350 * (10² - 5.5²)
∆E = 12.20625
∆E = 12.21J (Approximated)
Hence, the energy transformed from mechanical to internal in the particle—hoop is 12.21 Joules
b.
Let n = number of revolutions
The relationship between n and the energy is
1/n = (KE1 - KE2)/KE1
Make n the subject of formula
n = KE1 / (KE1 - KE2)
n = ½mu² / (½mu² - ½mv²) --- Simplify
n = ½mu² / (½m(u² - v²)) ---- Divide through by ½m
n = u² / (u² - v²)
n = 10² / (10² - 5.5²)
n = 1.433691756272401
n = 1.43 rev
Hence, the total number of revolutions the particle makes before stopping is 1.43 revolutions
Final answer:
The energy lost to friction in one revolution is 11.9875 J, and the particle makes a total of 2 revolutions before coming to a stop.
Explanation:
Energy Loss Due to Friction
The energy transformed from mechanical to internal in the particle—hoop—floor system as a result of friction in one revolution (a) is determined by the change in kinetic energy of the particle. The initial kinetic energy is given by ½ mv², where ‘m’ is the mass of the particle and ‘v’ is the initial velocity. The final kinetic energy is likewise given by ½ mv² using the final velocity. The difference in these energies gives the energy lost to friction.
Kinetic Energy Initial = ⅓ (0.350 kg)(10.00 m/s)² = 17.5 J
Kinetic Energy Final = ⅓ (0.350 kg)(5.50 m/s)² = 5.5125 J
Energy Transformed = Kinetic Energy Initial - Kinetic Energy Final = 17.5 J - 5.5125 J = 11.9875 J
Total Number of Revolutions
For part (b), to find the total number of revolutions the particle makes before stopping, assuming a constant frictional force, we can use a deceleration approach or an energy approach. The work done against friction for each subsequent revolution will be the same; thus each revolution results in the same amount of energy loss until the particle comes to a stop. By dividing the total initial kinetic energy by the energy lost per revolution, we find the number of revolutions.
Revolutions = Total Initial Kinetic Energy / Energy Transform per Revolution = 17.5 J / 11.9875 J ≈ 1.46 revolutions
Since it cannot complete a partial revolution after losing all kinetic energy, we round down to get 1 full revolution. Therefore, the particle makes a total of 1 + 1 = 2 revolutions before stopping.
A bothersome feature of many physical measurements is the presence of a background signal (commonly called "noise"). In Part 2.2.4 of the experiment, some light that reflects off the apparatus or from neighboring stations strikes the photometer even when the direct beam is blocked. In addition, due to electronic drifts, the photometer does not generally read 0.0 mV even in a dark room. It is necessary, therefore, to subtract off this background level from the data to obtain a valid measurement. Suppose the measured background level is 5.1 mV. A signal of 20.7 mV is measured at a distance of 29 mm and 15.8 mV is measured at 32.5 mm. Correct the data for background and normalize the data to the maximum value. What is the normalized corrected value at 32.5 mm?
Answer:
0.685
Explanation:
the background-corrected light measurement at 29mm: 20.7mv- 5.1mv⇒15.6mv
at 32.5mm: 15.8mv- 5.1mv⇒ 10.7mv
So, Normalize the data to the maximum value probably means to set the peak value to unity and scale the remaining data by the same factors.
Therefore, the normalized corrected value at 32.5mm is ⇒ 10.7mv/15.6mv ⇒0.685
If the vertical displacement from crest to trough is 0.50 cm, what is the amplitude?
cm
0.50
0.25
1.0
Answer:
0.25
Explanation:
Since there is a vertical displacement of 0.5cm from the crest to the trough, there is half of that displacement to the midline, which is also known as the amplitude. Therefore, the amplitude is 0.5/2=0.25cm. Hope this helps!
A sphere of radius 5.00 cm c m carries charge 3.00 nC n C . Calculate the electric-field magnitude at a distance 4.00 cm c m from the center of the sphere and at a distance 6.00 cm c m from the center of the sphere if the sphere is a solid insulator with the charge spread uniformly throughout its volume.
To calculate the electric-field magnitude at points inside and outside a uniformly charged insulating sphere, we apply Gauss's Law, considering the proportion of enclosed charge within a Gaussian surface for points inside, and treating the sphere as a point charge for points outside.
Explanation:We apply Gauss's Law to understand the electric-field magnitude around a uniformly charged insulating sphere, which states that the electric flux through a closed surface is proportional to the enclosed charge. We'll calculate the electric field (E) at two distances from the centre of the sphere: 4.00 cm, which is inside the sphere, and 6.00 cm, which is outside the sphere.
For the inside point (r = 4.00 cm), we only consider the charge enclosed by a Gaussian surface of radius 4.00 cm. Using the fact that charge is distributed uniformly throughout the volume of the sphere, we calculate the enclosed charge (q_enclosed) by using the ratio of volumes: q_enclosed = (Q * (4/3 * π * r^3)) / (4/3 * π * R^3), where Q is the total charge of the sphere. R is the radius of the sphere.
The electric field inside the sphere at 4.00 cm is then E = (1 / (4 * π * ε_0)) * (q_enclosed / r^2). We substitute the values using the given radius and charge and solve for E.
For the outside point (r = 6.00 cm), we use the formula for the electric field outside a spherically symmetric charge distribution, which is E = (1 / (4 * π * ε_0)) * (Q / r^2), and solve for E using the sphere's total charge and the distance of 6.00 cm.
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Why are the peaks opposite in direction?
a. The peaks are opposite in direction because the change in magnetic field at one end of the coil is opposite to the change in magnetic field at the other end. Faraday's law predicts that the direction of the induced voltage is dependent on the nature of the change in magnetic field.
b. The peaks are opposite in direction because the change in magnetic field at one end of the coil has the same direction to the change in magnetic field at the other end. Faraday's law predicts that the direction of the induced voltage is dependent on the nature of the change in magnetic field.
c. The peaks are in the same direction because the change in magnetic field at one end of the coil has the same direction to the change in magnetic field at the other end. Faraday's law predicts that the direction of the induced voltage is dependent on the nature of the change in magnetic field
Answer:
A
Explanation:
Peaks are in opposite direction because change in magnetic field at one end of the coil is opposite to the change in magnetic field at other end of the coil. Faraday's law predict that the direction of induced voltage is dependent on the nature of change in magnetic field
For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. The distance the worker’s eyes must travel is 20 in. The perpendicular distance from her eyes to the line of travel is 24 in. No refocus is required. What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element?
Answer:
The answer is 12.67 TMU
Explanation:
Recall that,
worker’s eyes travel distance must be = 20 in.
The perpendicular distance from her eyes to the line of travel is =24 in
What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?
Now,
We solve for the given problem.
Eye travel is = 15.2 * T/D
=15.2 * 20 in/24 in
so,
= 12.67 TMU
Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU
Newton's law of gravity and Coulomb's law are both inverse-square laws. Consequently, there should be a "Gauss's law for gravity." The electric field was defined as E⃗ =F⃗ onq/q , and we used this to find the electric field of a point charge. Using analogous reasoning, what is the gravitational field g⃗ of a point mass? Write your answer using the unit vector r^ , but be careful with signs; the gravitational force between two "like masses" is attractive, not repulsive. Express your answer in terms of the variables G , M , r , and r^ .
The correct answer for the gravitational field of a point mass is [tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
The gravitational force between two masses is given by Newton's law of gravitation:
[tex]F = \dfrac{GM_1M_2}{r^2}[/tex]
[tex]F[/tex] is the gravitational force between [tex]M_1[/tex] and [tex]M_2[/tex]
[tex]G[/tex] is the gravitational constant,
[tex]r^2[/tex] is the square of the distance between two masses.
The gravitational force is a vector quantity, and it is directed along the line connecting the two masses.
[tex]g= \dfrac{F}{m}[/tex]
[tex]g[/tex] is the gravitational field.
[tex]F[/tex] is the gravitational force.
Let [tex]m[/tex] be the mass test object.
Substitute the value of gravitational force from Newton's law of gravity:
[tex]g = \dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
[tex]r[/tex]is the unit vector pointing from the mass M to the test mass, which represents the direction of the gravitational field.
[tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
The negative sign indicates the direction of the gravitational force is attractive.
The gravitational field is [tex]g = -\dfrac{GM_1M_1}{r^2} * \dfrac{r}{m}[/tex]
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The gravitational field of a point mass is given by -GM/(r^2)*r^, where the negative sign indicates the attractive nature (opposite direction to r^) of gravity.
Explanation:The gravitational field g of a point mass using the same reasoning as for the electric field would be analogous but with some slight modifications to account for the attractive nature of gravity, unlike the repulsive nature of electric charges of the same sign. The formula is given by -GM/(r^2)*r^, where G is the gravitational constant, M is the mass of the object, and r is the distance to the object. The negative sign indicates that the force is attractive and acts in the direction opposite to r (the vector pointing directly away from the mass).
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In a Millikan oil-drop experiment, the condenser plates are spaced 2.00 cm apart, the potential across the plates is 4000 V, the rise or fall distance is 4.00 mm, the density of the oil droplets is 0.800 g/cm3 , and the viscosity of the air is 1.81 105 kg m1 s 1 . The average time of fall in the absence of an electric field is 15.9 s. The following different rise times in seconds are observed when the field is turned on: 36.0, 17.3, 24.0, 11.4, 7.54. (a) Find the radius and mass of the drop used in this experiment. (b) Calculate the charge on each drop, and show that charge is quantized by considering both the size of each charge and the amount of charge gained (lost) when the rise time changes. (c) Determine the electronic charge from these data. You may assume that e lies between 1.5 and 2.0 1019 C. 7.
Answer:
Answer is in the following attachment
Explanation:
A copper rod is sliding on two conducting rails that make an angle of 19o with respect to each other, as in the drawing. The rod is moving to the right with a constant speed of 0.60 m/s. A 0.63-T uniform magnetic field is perpendicular to the plane of the paper. Determine the magnitude of the average emf induced in the triangle ABC during the 7.5-s period after the rod has passed point A.
Answer:
0.2923 V
Explanation:
Given that
Angle between the rails, θ = 19°
Speed of the rod, v = 0.6 m/s
Magnetic field present, B = 0.63 T
Time used, t = 7.5 s
E = -ΔΦ/Δt
where, Φ = BA, so
E = -BΔA / Δt
To get the area, if we assume the rails are joined in a triangular fashion(see attachment)
E = -B(1/2 * AC * BC) / Δt
E = -B(vΔt * vΔt tanθ) / 2Δt
E = -(B * v² * Δt² * tanθ) / 2Δt
E = -Bv²Δt.tanθ/2
E = -(0.63 * 0.6² * 7.5 * tan 19) / 2
E = -0.5857 / 2
E = -0.2923
Thus, the magnitude of average emf induced if 0.2923 V
By considering the magnetic force in the second region, develop a mathematical expression that relates the mass of the particle to the other variables. Do not include the velocity in your expression. You can use the condition that the particle passed through the region of electric and magnetic fields undeflected to eliminate v from your expression. Your expression will also contain the radius of the circular path. i. Your expression for m should depend on B, E, r, and q
Answer:
M = [tex]\frac{qrB^{2} }{E}[/tex]
Explanation:
considering the magnetic force in the second region derive a mathematical expression that equates the mass of the particle to other variables
In a magnetic field
q = charge, M = mass of particle, E = electric field,B= magnetic field
qvb = [tex]\frac{mv^{2} }{r}[/tex] = therefore m = [tex]\frac{qrb}{v}[/tex] (equation 1)
note : the particle passes through the region undeflected
therefore : qvb = qE therefore (E = VB)
hence v = [tex]\frac{E}{B}[/tex] ( equation 2 )
insert equation 2 into equation 1
m = [tex]\frac{qrB^{2} }{E}[/tex]
What is equilibrium????
Answer:
Explanation:
An equilibrium is a state in which opposing forces or influences are banned.
An example of equilibrium is in economics when supply and demand are equal. An example of equilibrium is when you are calm and steady. An example of equilibrium is when hot air and cold air are entering the room at the same time so that the overall temperature of the room does not change at all.
Answer:
a state in which opposing forces or influences are balanced.
Explanation:
In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time, so that there is no observable change in the properties of the system.
A step-up transformer is connected to a generator that is delivering 141 V and 110 A. The ratio of the turns on the secondary to the turns on the primary is 1110 to 5. What voltage is across the secondary
Final answer:
To find the voltage across the secondary winding of a step-up transformer with a primary voltage of 141 V and a turns ratio of 1110 to 5, the secondary voltage is calculated to be 31302 V.
Explanation:
Calculating the Voltage Across the Secondary in a Step-Up Transformer
To calculate the voltage across the secondary winding of a step-up transformer, we use the transformer equation that relates the primary and secondary voltages with the number of turns on the primary (Np) and secondary (Ns) windings:
Vs / Vp = Ns / Np
Given that the primary voltage (Vp) is 141 V, and the ratio of the number of turns is 1110 to 5 (Ns / Np), we can rearrange the equation to solve for the secondary voltage (Vs) as follows:
Vs = Vp × (Ns / Np)
Vs = 141 V × (1110 / 5)
Vs = 141 V × 222
Vs = 31302 V
Thus, the voltage across the secondary winding is 31302 V.
The rotor of an electric motor has rotational inertia Im = 4.36 x 10-3 kg·m2 about its central axis. The motor is used to change the orientation of the space probe in which it is mounted. The motor axis is mounted along the central axis of the probe; the probe has rotational inertia Ip = 6.07 kg·m2 about this axis. Calculate the number of revolutions of the rotor required to turn the probe through 30.6° about its central axis.
Answer:
The number of revolutions of the rotor required to turn the probe is 118 revolutions
Explanation:
Given;
rotational inertia of the electric motor, Im = 4.36 x 10⁻³ kg·m²
rotational inertia of the probe, Ip = 6.07 kg·m²
the angular position of the probe, θ = 30.6°
From the principle of conservation of angular momentum;
[tex]I_m \omega _m = I_p \omega _p \\\\Also;\\\\I_m \theta _m = I_p \theta _p[/tex]
where;
[tex]\omega _m[/tex] is the angular velocity of the electric motor
[tex]\omega _p[/tex] is the angular velocity of the probe
[tex]\theta _m[/tex] is the angular position of the electric motor
[tex]\theta _p[/tex] is the angular position of the probe
[tex]\theta _m = \frac{I_p \theta_p}{I_m} \\\\\theta _m = \frac{6.07* 30.6^o}{4.36*10^{-3}} = 42601.4^o[/tex]
360° = One revolution
42601.4° = ?
Divide 42601.4° by 360°
= 118 revolutions
Therefore, the number of revolutions of the rotor required to turn the probe is 118 revolutions
One end of a steel rod of radius R = 10 mm and length L = 80cm is held in a vise. A force of magnitude F = 60kN is then applied perpendicularly to the end face (uniformly across the area) at the other end, pulling directly away from the vise. What is the stress on the rod?
Answer: 1.91*10^8 N/m²
Explanation:
Given
Radius of the steel, R = 10 mm = 0.01 m
Length of the steel, L = 80 cm = 0.8 m
Force applied on the steel, F = 60 kN
Stress on the rod, = ?
Area of the rod, A = πr²
A = 3.142 * 0.01²
A = 0.0003142
Stress = Force applied on the steel/Area of the steel
Stress = F/A
Stress = 60*10^3 / 0.0003142
Stress = 1.91*10^8 N/m²
From the calculations above, we can therefore say, the stress on the rod is 1.91*10^8 N/m²
In an interference pattern, the intensity is Group of answer choices smaller in regions of constructive interference than in regions of destructive interference. unchanged in regions of destructive interference but smaller in regions of constructive interference. the same in both the regions of constructive interference and the regions of destructive interference. unchanged in regions of destructive interference but greater in regions of constructive interference. greater in regions of constructive interference than in regions of destructive interference.
Answer:
The same in both the regions of constructive interference and the regions of destructive interference.
Explanation:
Interference is a phenomenon which occurs when two waves meet while moving along the same medium . The amplitude formed as a result of the interference could be greater, lower, or the same amplitude.
Constructive and destructive interference result from the interaction of waves that are correlated or coherent with each other. This is because arose from the same source or they have the same or nearly the same frequency.
The waves being coherent, arising from the same source and having the same frequency explains why it’s the same in both the regions of constructive interference and the regions of destructive interference.
Answer:
Option E (greater in regions of constructive interference than in regions of destructive interference)Explanation:
Constructive interference occurs when the maxima of two waves add together (the two waves are in phase), so that the amplitude of the resulting wave is equal to the sum of the individual amplitudes. Whereas in destructive interference opposite is true.
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A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 254 kcal of heat is added to the gas, the volume is observed to increase slowly from 12.0 m3 to 16.2 m3 . Part A Calculate the work done by the gas. Express your answer with the appropriate units. W = nothing nothing Request Answer Part B Calculate the change in internal energy of the gas. Express your answer with the appropriate units. ΔU = nothing nothing
The work done by the gas when the volume increases from 12.0 m³ to 16.2 m³ at atmospheric pressure is 425.565 kJ. The change in internal energy of the gas when 254 kcal of heat is added is 636.931 kJ.
Explanation:To calculate the work done by the gas during a quasi-static expansion, we can use the formula W = P ΔV, where W is work, P is pressure, and ΔV is the change in volume. Given the atmospheric pressure is 1 atm or 101,325 Pa, and the volume change is from 12.0 m³ to 16.2 m³, the work done by the gas can be calculated as:
W = P ΔV = 101,325 Pa × (16.2 m³ - 12.0 m³)
W = 101,325 Pa × 4.2 m³
W = 425,565 J or 425.565 kJ
To calculate the change in internal energy (ΔU) of the gas, we can use the first law of thermodynamics, which states that ΔU = Q - W, where Q is the heat added to the system. The heat is given as 254 kcal, which needs to be converted to joules (1 kcal = 4.184 kJ).
ΔU = Q - W = (254 kcal × 4.184 kJ/kcal) - 425.565 kJ
ΔU = 1062.496 kJ - 425.565 kJ
ΔU = 636.931 kJ
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Rank from largest to smallest the values of the magnetic field at the following distances from the axis of the conducting cylinder: Ra = 7.75 cm , Rb = 4.95 cm , r = 5.40 cm , and r>Ra.
The ranking from largest to smallest the values of the magnetic field should be like
Rb = 4.95 cm
r= 5.40 cm
Ra =7.75 cm
r> Ra
What is the magnetic field?It is a vector field that explained the magnetic impact on moving electric charges, electric currents, and magnetic materials.
In this, the force should be perpendicular to the velocity and the magnetic field. Also, it should be inversely proportional with respect to the distance with the axis of the cylinder.
So,
The largest magnetic field =Rb =4.95 cm
And,
Smallest magnetic field = r>Ra
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The magnetic field inside a conducting cylinder is maximized at certain points before dropping off outside. The ranked values from largest to smallest are: r = 5.40 cm, Rb = 4.95 cm, Ra = 7.75 cm, and r > Ra.
To determine the magnetic field at different distances from the axis of a conducting cylinder, we need to understand how the magnetic field varies inside and outside a cylinder carrying a current.
Given the distances Ra = 7.75 cm, Rb = 4.95 cm, r = 5.40 cm, and r > Ra, we will rank the magnetic field intensities:
1.) r = 5.40 cm:
Since r is within the cylinder where current is distributed uniformly, the magnetic field at this point can be calculated using Ampère's Law.
2.) Rb = 4.95 cm:
Still within the cylinder but closer to the center, thus the field here will not yet be maximized.
3.) Ra = 7.75 cm:
Outside the cylinder where the current configuration influences how the field drops off with increasing distance.
4.) r > Ra:
This point is furthest from the axis, hence the magnetic field will be the smallest.
The magnetic field reaches a maximum inside the conducting cylinder before decreasing outside of it.
Suppose that the resistive force of the air on a skydiver can be approximated by f = −bv2. If the terminal velocity of an 82.0 kg skydiver is 33.4 m/s, what is the value of b (in kg/m)?
Answer:
The value of b 0.7351 kg/m
Explanation:
Given that;
Mass of sky diver = 82 kg
Velocity = 33.4 m/s
f = −bv²
It is a resistance force, therefore the negative sign is ignored.
since; f = mg
∴ mg = bv²
b = mg / v ² ........ (1)
At terminal velocity a = 0
Put parameters in (1)
b = (82 × 10) / (33.4)²
b = 820 / 1,115.56
b = 0.7351