The charge density at (3.9 m, 2.7 m, -3.6 m) is 23.34 C/m^3.
To determine the charge density at a given point, we need to use Gauss's Law, which relates the electric flux density (D) to the charge density (ρ) and the permittivity of free space (ε0). The equation is given by D = ε0ρ.
In this case, the electric flux density is given by D = (3.2xz)ax + (5.5xy)ay + (6.6yz^2)az C/m^2.
To find the charge density, we need to isolate ρ in the equation D = ε0ρ. Since ε0 is a constant, we can divide both sides of the equation by ε0 to get ρ = D/ε0.
Substituting the values of D and ε0, and evaluating the expression at the given point (3.9 m, 2.7 m, -3.6 m), we find that the charge density is approximately 23.34 C/m^3.
Therefore, the charge density at (3.9 m, 2.7 m, -3.6 m) is 23.34 C/m^3.
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Match the words on the left with the words on the right.Group ofanswer choices Wind Turbine[ Choose ] Tar Sands MountaintopRemoval Geothermal Power Albert Einstein No carbon dioxide produced Photovoltaic[ Choose ] Tar Sands Mountaintop Removal Geothermal Power Albert Einstein No carbon dioxide producedCoal[ Choose ] Tar Sands Mountaintop Removal Geothermal Power Albert Einstein No carbon dioxide producedAlberta, Canada[ Choose ] Tar Sands Mountaintop Removal Geothermal Power Albert Einstein No carbon dioxide producedWestern USA
The Western USA is a region of the United States
Group of answer choices Wind Turbine[ Choose ]
No carbon dioxide produced Photovoltaic Tar Sands Mountain top Removal Geothermal Power
Albert Einstein[ Choose ] No carbon dioxide produced Photovoltaic Tar Sands Mountaintop Removal Geothermal
PowerCoal[ Choose ] No carbon dioxide produced Photovoltaic Tar Sands Mountaintop Removal Geothermal
PowerAlberta, Canada[ Choose ] Tar Sands No carbon dioxide produced Mountaintop Removal Geothermal Power
Albert EinsteinWestern USA[ Choose ] No carbon dioxide produced Photovoltaic Tar Sands Mountaintop Removal Geothermal
PowerAlbert Einstein is related to the theory of relativity.
The Wind Turbine is a device that converts the kinetic energy of wind into mechanical energy.
No carbon dioxide is produced in photovoltaic, Geothermal Power, and Wind Turbine as they do not involve combustion.
A tar sand is a sandstone that contains bitumen.
Mountaintop Removal is the practice of mining through the summit of a mountain.
Coal is a combustible black or brownish-black sedimentary rock.
Alberta, Canada is one of the largest oil reserves in the world, known as Tar Sands.
Finally, the Western USA is a region of the United States.
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The proper lifetime of a particle is 150 ns. (a) You observe the particle to be moving at 0.98c what will its lifetime be measured at and what distance will it cover in this lifetime? (b) An advanced alien lifeform is traveling in their miniature ship at 0.98c over the particle. What lifetime to they observe the particle to have? What distance do they observe the particle to go through during its lifetime?
(a) When observing a moving particle, time dilation occurs due to the relativistic effects of special relativity. The observed lifetime (t') of the particle can be calculated using the time dilation formula: t' = t / γ, where t is the proper lifetime of the particle and γ is the Lorentz factor given by γ = 1 / sqrt(1 - (v²/c²)). Here, v is the velocity of the particle (0.98c).
Plugging in the values, we have:
γ = 1 / sqrt(1 - (0.98c)²/c²) ≈ 5.03.
t' = (150 ns) / 5.03 ≈ 29.8 ns.
To calculate the distance covered by the particle, we can use the formula d = v * t', where d is the distance traveled and v is the velocity of the particle.
d = (0.98c) * (29.8 ns) ≈ 8.85 m.
(b) For the advanced alien lifeform moving at the same velocity (0.98c) as the particle, they would observe the particle's lifetime (t'') and distance traveled (d'') according to their reference frame. Since the alien's frame is relative to the particle, they would measure the proper lifetime of the particle (t) as their observed lifetime.
Thus, the advanced alien lifeform would observe the particle to have a lifetime of 150 ns and cover a distance of 0 meters (since it is at rest relative to the particle).
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A force of 28.03Nacts on a charge of 2.33C when it is placed in a uniform electric field. What is the magnitude of this electric field?
The magnitude of this electric field is 12 N/C.
The magnitude of the electric field can be found using the equation:
E = F/q
Where:
E is the magnitude of the electric field
F is the force
q is the charge
For this problem, the force acting on the charge is given as F = 28.03 N and the charge is q = 2.33 C.
Substitute the given values in the above equation:
E = F/q
E = 28.03 N / 2.33 C
E = 12 N/C
Correct answer: 12 N/C.
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A car is on a road traveling East. If it travels a distance of \( 500 \mathrm{~m} \) in 12 second, a) Calculate its speed (show your work) b) Calculate its velocity (show your work)
The answer is: a) Speed = 41.67 m/s; b) Velocity = 41.67 m/s East
Explanation:
Given: Distance, d = 500 m
Time, t = 12 s
(a) Speed of the car = distance/time`
Speed = d/t`
Substitute the given values`
Speed = 500/12 m/s`
Speed = 41.67 m/s
(b) Velocity of the car = Speed and direction`
Velocity = Speed + direction`
Given that the car is traveling East
So, the velocity is 41.67 m/s East.
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Two charged particles are a distance of $1.92 \mathrm{~m}$ from each other. One of the particles has a charge of 7.24 $\mathrm{nC}$, and the other has a charge of $4.54 \mathrm{nC}$.
(a) What is the magnitude (in $\mathrm{N}$ ) of the electric force that one particle exerts on the other?
(b) Is the force attractive or repulsive?
(a) The magnitude of the electric force between two point charges with opposite signs is 8.01e-08 N.
(b) The force is attractive because the charges have opposite signs.
(a) The magnitude of the electric force between two point charges is given by the following formula:
F = k * q_1 * q_2 / r^2
where:
F is the magnitude of the electric force
k is Coulomb's constant (8.987551787 * (10^9) N * m^2 / Coulomb^2)
q_1 is the charge of the first point charge (in Coulombs)
q_2 is the charge of the second point charge (in Coulombs)
r is the distance between the two point charges (in meters)
In this case, the charges are 7.24 nC and 4.54 nC, and the distance between them is 1.92 m. So, the magnitude of the electric force is:
F = 8.987551787 * (10^9) * (7.24 * (10^-9)) * (4.54 * (10^-9)) / (1.92^2)
F = 8.013705301051846e-08 N
(b) The force is attractive because the charges have opposite signs.
Therefore, the magnitude of the electric force is 8.01e-08 N, and the force is attractive.
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The current electricity tarrif in SIngapore is 32.28 cents per
kWh. A window air-conditioner unit comsumes 1400 W. How much does
it cost (in dollars, to 2 decimal place) to turn on the air-con for
1 h
The electricity tariff is the rate that customers are charged for consuming electricity from the grid. It varies from one country to another, and even within the same country, different utilities charge different rates.
In Singapore, the current electricity tariff is 32.28 cents per kWh. This means that for every kilowatt-hour of electricity consumed, the customer is charged 32.28 cents. A window air-conditioner unit consumes 1400 W of electricity when it is turned on. This means that if you turn on the air-con for 1 hour, you will consume 1.4 kWh of electricity.
To calculate how much it costs to turn on the air-con for 1 hour, you can use the following formula:
Cost = (Power × Time) × Tariff
Where Power is the power rating of the air-con unit in watts, Time is the time the air-con is turned on in hours, and Tariff is the electricity tariff in cents per kWh.
Using this formula, we can calculate the cost of running the air-con for 1 hour:
Cost = (1400 × 1) ÷ 1000 × 32.28Cost = 0.45 dollars (rounded to 2 decimal places)
Therefore, it costs 0.45 dollars to turn on the air-con for 1 hour.
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A \( 50-\Omega \) resistor is put in parallel with a \( 37-\Omega \) resistor. Calculate the single resistance equivalent to these two.
The single resistance equivalent to the 50 Ω resistor in parallel with the 37 Ω resistor is approximately 21.264 Ω.
To calculate the single resistance equivalent to two resistors in parallel, you can use the formula:
1/Req = 1/R1 + 1/R2
where Req is the equivalent resistance, R1 is the resistance of the first resistor, and R2 is the resistance of the second resistor.
In this case, the first resistor has a resistance of 50 Ω, and the second resistor has a resistance of 37 Ω. Plugging these values into the formula, we get:
1/Req = 1/50 + 1/37
To simplify this equation, we can find a common denominator:
1/Req = (37 + 50) / (37 * 50)
1/Req = 87 / 1850
To find the reciprocal of both sides of the equation, we invert both sides:
Req = 1850 / 87
Using long division, we get:
Req = 21.264 Ω (rounded to three decimal places)
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The initial and final velocities of two boxes experiencing constant acceleration are respectively −8.05 m/s and 16.1 m/s. (a) The first object has displacement 17.5 m. Find its acceleration (in m/s
2
). (Indicate the direction with the sign of your answer) m m
2
(b) The second object traveis a total distance of 21.8 m. Find its acceleration (in ms
2
). (findicate the direction with the sign of your answer.) m s
2
The acceleration of the second box is 3.4 m/s^2 in the direction of motion.
a) Given Initial velocity of the first box u1 = -8.05 m/s Final velocity of the first box v1 = 16.1 m/s Displacement of the first box s1 = 17.5 m
We need to find the acceleration of the first box a1.
We can use the equation:
v1^2 - u1^2 = 2as1
Where v1 and u1 are the final and initial velocities of the first box, s1 is the displacement of the first box, and a1 is the acceleration of the first box.
Substituting the given values, we get:
[tex]16.1^2 - (-8.05)^2 = 2 × a1 × 17.5a1 = 4.4 m/s^2[/tex]
The acceleration of the first box is 4.4 m/s^2 in the direction of motion.
b) Given:
Initial velocity of the second box u2 = -8.05 m/s
Final velocity of the second box v2 = 16.1 m/s
Total distance travelled by the second box s2 = 21.8 m We need to find the acceleration of the second box a2. We can use the equations of motion to find the acceleration:
a = (v^2 - u^2)/(2s)
Substituting the given values, we get:
[tex]a2 = (16.1^2 - (-8.05)^2)/(2 × 21.8)a2 = 3.4 m/s^2[/tex]
The acceleration of the second box is 3.4 m/s^2 in the direction of motion.
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A particle with a charge of q=11.0μC travels from the origin to the point (x,y)=(20.0 cm,50.0 cm) in the presence of a uniform electric field
E =300V/m. Determine the following. (a) the change in the electric potential energy (in J) of the particle-field system 3 (b) the electric potential difference (in V) through which the particle moves V
The change in electric potential energy of the particle-field system is 0.165 J. The electric potential difference through which the particle moves is 161.55 V.
(a) The change in the electric potential energy (in J) of the particle-field system is 0.165 J.
The change in electric potential energy (ΔPE) can be calculated using the formula ΔPE = qΔV, where q is the charge and ΔV is the change in electric potential. In this case, q = 11.0 μC = 11.0 × 10^(-6) C.
To find ΔV, we need to determine the electric potential difference between the initial and final points. The electric potential difference (V) is given by V = Ed, where E is the electric field strength and d is the distance.
The distance between the origin (0, 0) and the point (20.0 cm, 50.0 cm) can be calculated using the Pythagorean theorem:
d = √[(20.0 cm)^2 + (50.0 cm)^2] = √(400 cm^2 + 2500 cm^2) = √2900 cm ≈ 53.85 cm.
Converting the distance to meters:
d = 53.85 cm × (1 m/100 cm) = 0.5385 m.
Substituting the given values into the formula, we get:
V = (300 V/m) × 0.5385 m = 161.55 V.
Now we can calculate the change in electric potential energy:
ΔPE = qΔV = (11.0 × 10^(-6) C) × (161.55 V) = 0.165 J.
(b) The electric potential difference (in V) through which the particle moves is 161.55 V.
As calculated in part (a), the electric potential difference (V) between the initial and final points is 161.55 V. This value represents the change in electric potential experienced by the charged particle as it moves from the origin to the point (20.0 cm, 50.0 cm).
The electric potential difference is a measure of the work done per unit charge to move the particle against the electric field. It indicates the change in electric potential energy per unit charge. In this case, the particle experiences an increase in electric potential as it moves through the uniform electric field E = 300 V/m.
Therefore, the electric potential difference through which the particle moves is 161.55 V.
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A 200.kg motor is pulled along the ground with a cable and is accelerating at 1.2 m/s
2
, what is the Tension in the cable if the coefficient of kinetic friction is 0.24 ?
The Tension in the cable is 710.4 N if the coefficient of kinetic friction is 0.24.
The Tension in the cable if the coefficient of kinetic friction is 0.24 and a 200.kg motor is pulled along the ground with a cable and is accelerating at 1.2 m/s^2 can be calculated using the formula:
Tension = mass x acceleration + frictional force.
Here, acceleration = 1.2 m/s^2mass = 200 kg. coefficient of kinetic friction = 0.24. The frictional force can be calculated as: frictional force = coefficient of kinetic friction x normal force.
The normal force is equal to the force of gravity acting on the object, which can be calculated as:
force of gravity = mass x acceleration due to gravity (g)
where g = 9.8 m/s^2, force of gravity = 200 kg x 9.8 m/s^2 = 1960 N. Therefore, the normal force is also equal to 1960 N.The frictional force can now be calculated as: frictional force = 0.24 x 1960 N = 470.4 N.
The tension in the cable can be calculated as: Tension = mass x acceleration + frictional force
= 200 kg x 1.2 m/s^2 + 470.4 N
= 240 N + 470.4 N
= 710.4 N
Therefore, the Tension in the cable is 710.4 N if the coefficient of kinetic friction is 0.24.
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A 10,000[ kg] railroad car, A, traveling at a speed of 24.0[ m/s] strikes an identical car, B, at rest. If the cars lock together as a result of the collision, what is their final speed immediately after the collision? Calculate how much of the initial kinetic energy is transformed into thermal or other forms of energy.
The speed of the combined two cars is equal to the momentum of the system divided by the combined mass of the two cars ( final speed immediately after the collision).
Given data;Mass of car A = m1
= 10,000 kgVelocity of car A
= u1 = 24.0 m/sMass of car B
= m2 = 10,000 kgVelocity of car B
= u2 = 0 (car B is at rest)Momentum of the system = PfinalKinetic energy of the system = KfinalWe know that momentum is conserved for an isolated system and as the collision is inelastic i.e. the cars lock together, the kinetic energy is not conserved in this case.Conservation of momentum;P1 + P2 = P finalwhere P1
= m1u1 ; P2
= m2u2
= 0(m2u2 = 0 as car B is at rest initially)Pfinal = (m1 + m2)u′ (combined mass moves at a speed of u′ after collision)Putting the values in the above equation we get;10000*24 + 10000*0
= (10000 + 10000)u′
=> u′ = 12 m/s So, the final velocity of the combined mass is 12 m/sNow, as we know kinetic energy is not conserved in an inelastic collision. So, the initial kinetic energy will be transferred into other forms of energy. The energy will be transferred into thermal energy (heat) and sound energy.So, the total initial kinetic energy, K1 = ½ m1u12
= ½ * 10000 * (24)2
= 2.88 * 106 JThe final kinetic energy will be;Kfinal = ½ (m1 + m2)u′2
= ½ * (10000 + 10000) * 122
= 1.44 * 106 JSo, the change in kinetic energy is = K1 – Kfinal= 2.88 * 106 - 1.44 * 106
= 1.44 * 106 JSo, 1.44 * 106 J of energy has been transformed into thermal or other forms of energy.
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A person walks from his house 32 meters in the negative direction of the Y axis, to a white building (his house is at the origin of the system). At that point he turns and begins to walk in the negative direction of X for 16 meters, until he reaches a store. Then, he walks 200 meters in a direction of 45 degrees with respect to the X axis (measured counterclockwise), until he meets a friend. Finally, he walks in a straight line in an unknown direction, until he reaches his house again: (a) Make a schematic drawing of the situation (it does not have to be a scale drawing. (b) Say what is the unknown direction in which you walked from the time you meet your friend until you return home, and say what distance you walk in that direction (c) Calculate the total distance of the route. (d) Calculate the vector of the net displacement between the house and the store and its direction or angle.
The vector of the net displacement between the house and the store is approximately √26824 meters, making an approximately -11.4° angle to the positive X axis.
The distance he walked in that direction is given by the distance between his friend's place and his house, which is the hypotenuse of the right-angled triangle shown in the figure.
b) Applying the Pythagorean theorem, this distance is obtained as follows:
Distance2 = Distance1 + Distance3
⇒ Distance2 = (200 cos 45°)2 + (200 sin 45° - 32)2
⇒ Distance2 = (200/√2)2 + (200/√2 - 32)2
= 20000/2 + (40000/2 - 2048)/2
= 10000 + 18976
= 28976 meters (approx.)
(c) The total distance of the route is given by the sum of the three straight-line distances traveled by the person. These distances are Distance1, Distance2, and Distance3, as shown in the figure. Therefore, the total distance of the route is as follows:
Total distance = Distance1 + Distance2 + Distance3
⇒ Total distance = 32 + 28976 + 16
= 29024 meters (approx.)
(d) The vector of the net displacement between the house and the store is the vector sum of the displacement vectors due to the three straight-line distances traveled by the person. These vectors are given by the following expressions: d1 = -32 j
d2 = -16 i
d3 = 200/√2 (cos 45° i + sin 45° j) d1 represents the displacement vector for the walk from the house to the white building (in the negative direction of the Y axis), and d2 represents the displacement vector for the walk from the white building to the store (in the negative direction of X).
The total distance of the route that he followed is approximately 29024 meters. The vector of the net displacement between the house and the store is approximately √26824 meters, making an approximately -11.4° angle to the positive X axis.
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what is the maximum potential difference between the disks?
The maximum potential difference between the disks is 6.25 V when they are separated by a distance of 0.010 m and connected to a 12-V battery.
When two disks of radius 0.10 m are separated by a distance of 0.010 m and are connected to a 12-V battery, the maximum possible voltage between the disks is given by Vmax = (Qmax / (2 * ε0 * A)) * d where Vmax is the maximum possible voltage between the disks
Qmax is the maximum possible charge on the disksε0 is the permittivity of free space
A is the area of the disks d is the distance between the disks.
Substituting the given values, we have
[tex]A = \pi$ * r^{2} = \pi$ * 0.10^{2} = 0.0314 m^{2} \epsilon0 = 8.85 \times 10^{-12} F/m[/tex] and d = 0.010 m
Thus, we get, [tex]Qmax = CV = (A * \epsilon0 * Vmax) / d= (0.0314 * 8.85 \times 10^{-12} * 12) / 0.010= 3.51 \times 10^{-12} C[/tex]
The maximum potential difference between the disks is given by,
[tex]Vmax = (Qmax / (2 * \epsilon0 * A)) * d= (3.51 \times 10^{-12} / (2 * 8.85 \times 10^{-12} * 0.0314)) * 0.010\approx 6.25 V[/tex]
Therefore, the maximum potential difference between the disks is 6.25 V when they are separated by a distance of 0.010 m and connected to a 12-V battery.
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Which of the following is (are) connected to the kinetic energy of an object? a)work b)force c)displacement d)all of the above/
d) all of the above
All of the options listed (work, force, displacement) are connected to the kinetic energy of an object.
Work (W) is defined as the product of the force (F) applied to an object and the displacement (d) of the object in the direction of the force. The work done on an object is directly related to the change in its kinetic energy. In fact, the work-energy theorem states that the work done on an object is equal to the change in its kinetic energy.
Force (F) is a fundamental quantity that can accelerate an object and change its velocity. When a force acts on an object, it can cause a change in its kinetic energy.
Displacement (d) refers to the change in position of an object. The displacement of an object is important in determining the work done on it and, consequently, its change in kinetic energy.
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crate of mass 2 kg initially at rest is pulled along a straight line with a net force of 100 N. A. Find the magnitude of the acceleration of the crate. Express your answer in m/s
2
B. Find the distance it moves in the first 3 s. Express your answer in meters Shortlyafter leaping from an airplane a 90 kg skydiver has an upward force of 225 N exerted on her by the air. What is the magnitude of the net force exerted on her? Express you answer in N. Tennis balls when in flight in air experience a large drag force as a result of air resistance. A tennis ball is hit so it goes straight up then comes back down. The direction of the drag force is: a. Always up b. Always down c. Up, then down d. Down, then up
The distance the crate moves in the first 3 s is 225 m. The magnitude of the acceleration of the crate is 50 m/s².The drag force is acting upwards which slows down the ball and hence the direction of the drag force is down, then up.
a)
The magnitude of the acceleration of the crate can be determined by applying Newton's second law of motion which states that the force acting on an object is equal to its mass times its acceleration, i.e.,
F = ma Where,F is the net force acting on the crate,m is the mass of the crate,a is the acceleration of the crate.
Substituting the given values in the above formula we get,F = 100 Nm = 2 kg
So, a = F/m= 100 N/2 kg = 50 m/s²
Therefore, the magnitude of the acceleration of the crate is 50 m/s².
b)
Distance the crate moves in the first 3 seconds can be calculated using the third equation of motion which states that the distance traveled by an object under uniform acceleration can be given by the equation,
s = ut + 1/2 at² where,s is the distance traveled by the object,u is the initial velocity of the object which is zero in this case,a is the acceleration of the object,t is the time taken by the object to cover the distance.
Substituting the given values in the above formula, we get,a = 50 m/s²t = 3 s
So, the distance traveled by the crate can be calculated as,s = 0 + 1/2 × 50 m/s² × (3 s)²= 0 + 1/2 × 50 m/s² × 9 s²= 225 m.
Therefore, the distance the crate moves in the first 3 s is 225 m.
The direction of the drag force experienced by a tennis ball when it is hit straight up and comes back down is down, then up.The direction of the drag force is opposite to the direction of motion of the tennis ball. When the tennis ball is hit straight up, its direction of motion is upwards.
So, the drag force is acting downwards which slows down the ball. When the ball reaches its maximum height, it starts coming back down. Now the direction of motion of the tennis ball is downwards.
Therefore, the drag force is acting upwards which slows down the ball and hence the direction of the drag force is down, then up.
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Every instance of the motion of an object in a circle or along a circular turn involves some force that is directed inward or centripetally. The centripetal force is an adjective to describe the net force; it is not actually a new force to be added to an already lengthy list - including friction, gravity, tension, normal, air resistance, etc. Rather, the centripetal force requirement is a principle that states that in order to have the motion of an object in a circle, there must be an inward net force to sustain the inward acceleration. In each of the following instances, identify the type of force that fulfills the centripetal force requirement. That is, identify the inward force acting upon the bold-faced object.
In any instance of circular motion, there needs to be an inward net force to maintain the inward acceleration. Here are the forces fulfilling this requirement: tension in the chains for riding a merry-go-round, friction between tires and the road when driving a car around a corner, adhesive forces between water and the glass when turning a glass of water upside down, and the normal force on the water in a spinning bucket.
In order to have the motion of an object in a circle, there must be an inward net force to sustain the inward acceleration. The types of forces that fulfill the centripetal force requirement in the given instances are:
Riding a merry-go-round - The tension in the chains attached to the ride provides the inward force.Learning to drive a car around a corner - Friction between the tires and the road acts as the centripetal force.Turning a glass of water upside down - Adhesive forces between the water and the glass act as the inward force.Spinning a bucket of water around - The normal force acting upon the water in the bucket provides the centripetal force.Learn more about centripetal force: https://brainly.com/question/898360
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A ball is kicked with a speed vo =19.8 m/s and at an angle of θ=38.9∘ with the ground, it hits a wall at the point ×=16.4 m and y=7.68 m. See the figure. How long the ball is in the air before hitting the wall? Use g=9.80 m/s2 if needed. A. 0.395 8. 2.545 C. 1.065 D. 1.325 E. 0.835
The ball is in the air for approximately 1.065 seconds before hitting the wall. Thus, the correct answer is Option C.
To determine how long the ball is in the air before hitting the wall, we can use the projectile motion equations. The horizontal and vertical components of motion can be considered separately.
Given:
Initial speed, vo = 19.8 m/s
Launch angle, θ = 38.9°
Horizontal distance to the wall, x = 16.4 m
Vertical height, y = 7.68 m
Acceleration due to gravity, g = 9.80 m/s²
First, let's calculate the time of flight (total time the ball is in the air) using the vertical component of motion.
The equation for the vertical displacement is given by:
y = vo × t + (1/2)gt²
where
y is the vertical displacement,
vo is the initial vertical velocity,
t is the time, and
g is the acceleration due to gravity.
Rearranging the equation, we have:
t² - (2vo/g)t + (2y/g) = 0
Using the quadratic formula, we can find the time of flight:
t = (-b ± √(b² - 4ac)) / (2a)
where
a = 1
b = -(2vo/g)
= -(2 × 19.8 / 9.80)
c = 2y/g
= 2 × 7.68 / 9.80
Now, let's calculate the time of flight:
t = (-(-2vo/g) ± √((-2vo/g)² - 4 × 1 × (2y/g))) / (2 × 1)
Using the given values:
t = (2 × 19.8 / 9.80 ± √((2 × 19.8 / 9.80)² - 4 × 1 × (2 × 7.68 / 9.80))) / 2
t ≈ 1.065 seconds (rounded to three decimal places)
Therefore, the ball is in the air for approximately 1.065 seconds before hitting the wall. The answer is option C.
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Two pendulums hang next to each other, with two balls, of different masses (50 g and 20 g respectively) as pendulum bodies. The heavier ball is pulled out and then released to collide with the lighter pendulum ball. The figure below describes the process. The figure on the far right shows when the small ball is in the turning position. (whether or how the heavier bullet moves after impact is left to you to investigate) Check if the shock is fully elastic. before stot under impact Ma bullet after impact 509 209 16) A freezer that holds 270 I maintains a temperature of 19
∘
C. Its door measures 57x 180 cm. The pressure of the air inside the cabinet is initially the same as the pressure of the outside air, i.e. normal air pressure. 16) A freezer that holds 270 I maintains a temperature of 19
∘
C. Its door measures 57x 180 cm. The pressure of the air inside the cabinet is initially the same as the pressure of the outside air, i.e. normal air pressure. When the cabinet is opened, the temperature of the air in the cabinet rises from −19
∘
C to −17
∘
C. When the door is then closed, the air temperature in the cabinet drops very quickly back to −19
∘
C. In connection with this, a certain negative pressure occurs inside the cabinet. The negative pressure makes it difficult to open the door before a pressure equalization has taken place between the air inside the cabinet and the air outside. Calculate the extra force due to the negative pressure affects the door then the temperature inside in the cabinet again dropped to −19
∘
C and no pressure equalization has yet taken place.
Some energy would be lost due to the collision and there is a transfer of momentum from one object to another, resulting in a change of velocity and position for each ball.
According to the Law of Conservation of Momentum:
The momentum of a system remains constant if no external forces act on it.
This implies that the total momentum of the system before the impact will be equal to the total momentum of the system after the impact.
Hence, if the heavier bullet is pulled out and released to collide with the lighter pendulum ball:
The momentum after impact of the two balls will be equal to their momentum before the impact.
Mathematically, we can express this as:
Ma * a + Mb * b = Ma * a' + Mb * b'
Where:
Ma = Mass of the heavier ball
Mb = Mass of the lighter ball
a = Velocity of the heavier ball before the impact
b = Velocity of the lighter ball before the impact
a' = Velocity of the heavier ball after the impact
b' = Velocity of the lighter ball after the impact
Therefore, we can conclude that the shock is not fully elastic because:
Some energy would be lost due to the collision.
There is a transfer of momentum from one object to another, resulting in a change of velocity and position for each ball.
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) A parallel-plate capacitor has 2.9 cm × 2.9 cm electrodes with surface charge densities ±1.0×10−6C/m2. A proton traveling parallel to the electrodes at 1.2×106 m/s enters the center of the gap between them. By what distance has the proton been deflected sideways when it reaches the far edge of the capacitor? Assume the field is uniform inside the capacitor and zero outside the capacitor
The proton is deflected by approximately 0.198 mm (final answer), due to the electric field generated by the parallel-plate capacitor
When a charged particle, such as a proton, enters the gap between the electrodes of a parallel-plate capacitor, it experiences a sideways force due to the electric field generated by the charged plates. The force exerted on the proton is given by the equation F = qE, where q is the charge of the proton and E is the electric field strength. Since the electric field is uniform inside the capacitor and zero outside, the force experienced by the proton is constant throughout its motion.
The initial velocity of the proton is parallel to the plates, so the electric force acts as a centripetal force, causing the proton to follow a curved path. The distance the proton is deflected sideways can be calculated using the equation d = (1/2)at^2, where a is the acceleration and t is the time it takes for the proton to reach the far edge of the capacitor.
To find the acceleration, we can rearrange the equation F = qE to solve for a: a = F/m, where m is the mass of the proton. Plugging in the values for q and E, and using the known mass of a proton, we can calculate the acceleration.
Next, we need to determine the time it takes for the proton to travel to the far edge of the capacitor. Since the velocity and distance are known, we can use the equation v = d/t to solve for t.
Substituting the values into the equation and solving for d, we find that the proton is deflected by approximately 0.198 mm when it reaches the far edge of the capacitor.
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A VW Beetle goes from 0 to 48.0mi/h with an acceleration of +2.35 m/s^2 . (a) How much time does it take for the Beetle to reach this speed? (b) A top-fuel dragster can go from 0 to 48.0mi/h in 0.600 s. Find the acceleration (in m/s^2) of the dragster. (a) Number Units (b) Number Units
(a) It takes approximately 6.44 seconds for the VW Beetle to reach a speed of 48.0 mi/h.
(b) The acceleration of the top-fuel dragster is approximately 53.33 m/s^2.
(a) To calculate the time it takes for the VW Beetle to reach a speed of 48.0 mi/h, we can use the equation of motion v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Given that the initial velocity u is 0 m/s, the final velocity v is 48.0 mi/h, which can be converted to m/s by multiplying it by 0.44704 (since 1 mi/h = 0.44704 m/s), and the acceleration a is +2.35 m/s^2, we can rearrange the equation to solve for time:
v = u + at
48.0 mi/h = 0 + 2.35 m/s^2 * t
Converting 48.0 mi/h to m/s, we get:
48.0 mi/h * 0.44704 m/s = 21.44 m/s
Substituting the values into the equation, we have:
21.44 m/s = 2.35 m/s^2 * t
Solving for t, we find:
t = 21.44 m/s / 2.35 m/s^2 ≈ 6.44 seconds
Therefore, it takes approximately 6.44 seconds for the VW Beetle to reach a speed of 48.0 mi/h.
(b) To find the acceleration of the top-fuel dragster, we can use the same equation of motion v = u + at and rearrange it to solve for acceleration:
a = (v - u) / t
Given that the initial velocity u is 0 m/s, the final velocity v is 48.0 mi/h (which is converted to m/s as explained in part (a)), and the time t is 0.600 s, we can substitute the values into the equation:
a = (48.0 mi/h * 0.44704 m/s - 0 m/s) / 0.600 s
Calculating the expression, we get:
a = 26.92 m/s / 0.600 s ≈ 53.33 m/s^2
Therefore, the acceleration of the top-fuel dragster is approximately 53.33 m/s^2.
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A world has the following mass and radius: \[ \mathrm{M}=1 / 3 M_{\oplus} \text { and } \mathrm{R}=1 / 3 R_{\oplus} \] The gravity on this world would be \( F_{\oplus} \)
The gravity on the planet is F⊕ = 1.95 x 10²⁵ N.
Given information are :
[tex]\[ \mathrm{M}=1 / 3 M_{\oplus} \text { and } \[/tex] mathrm[tex]{R}=1 / 3 R_{\oplus} \][/tex] The gravity on this world would be F⊕.
The formula to calculate the gravity on any object is as follows:
F = (G m₁ m₂) / r²
where G is the gravitational constant, m₁ and m₂ are the masses of the two objects, and r is the distance between their centers.
The force of gravity on an object can also be calculated using the following formula:
F = m g
Where F is the force of gravity, m is the mass of the object, and g is the acceleration due to gravity.
So the gravity on the given planet will be F = m g
Given that[tex]\[\mathrm{M}=1 / 3 M_{\oplus} \][/tex]
We know that mass of the Earth is [tex]M⊕ = 5.98 x 10²⁴ kg\[ \mathrm{M}=1 / 3 M_{\oplus} \] \[ \Rightarrow \mathrm{M}=\frac{1}{3} \times 5.98 \times 10^{24} \mathrm{kg} \] \[ \Rightarrow \mathrm{M}=1.99 \times 10^{24} \mathrm{kg} \][/tex]
Given that[tex]\[\mathrm{R}=1 / 3 R_{\oplus} \][/tex]
We know that radius of the Earth is
[tex]R⊕ = 6.37 x 10⁶ m\[ \mathrm{R}=1 / 3 R_{\oplus} \] \[ \Rightarrow \mathrm{R}=\frac{1}{3} \times 6.37 \times 10^{6} \mathrm{m} \] \[ \Rightarrow \mathrm{R}=2.12 \times 10^{6} \mathrm{m} \][/tex]
Let's calculate the acceleration due to gravity, g.
We know that acceleration due to gravity,
g = 9.8 m/s²[tex].\[ \mathrm{g}=\frac{\mathrm{F}}{\mathrm{m}} \] \[ \Rightarrow \mathrm{F}=\mathrm{g} \times \mathrm{m} \] \[ \Rightarrow \mathrm{F}=9.8 \mathrm{m/s^{2}} \times 1.99 \times 10^{24} \mathrm{kg} \] \[ \Rightarrow \mathrm{F}=1.95 \times 10^{25} \mathrm{N} \][/tex]
So, the gravity on the planet is F⊕ = 1.95 x 10²⁵ N.
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The dome of a Van de Graaff generator receives a charge of 9.5×10
−4
C. Find the strength of the electric field in the following situations. (Hint: Review properties of conductors in sphere.) (a) inside the dome | N/C (b) at the surface of the dome, assuming it has a radius of 2.0 m N/C (c) 6.2 m from the center of the dome N/C
The dome of a Van de Graaff generator receives a charge of 9.5×10−4 C. Inside the dome, electric field is 0 N/C. At its surface, it is around [tex]8.54*10^7 N/C[/tex]. 6.2 m from the center of the dome, it is [tex]2.38 *10^6 N/C.[/tex]
To find the strength of the electric field in the given situations, we can use the properties of conductors in a sphere.
(a) Inside the dome:
Inside a conductor, the electric field is zero. This is due to the fact that charges inside a conductor distribute themselves uniformly and the electric field created by these charges cancels out. Therefore, the strength of the electric field inside the dome is 0 N/C.
(b) At the surface of the dome:
At the surface of a conductor, the electric field is perpendicular to the surface and its strength can be calculated using the formula:
Electric field (E) = Charge density (σ) / ε₀
where σ is the charge per unit area and ε₀ is the permittivity of free space.
Given that the charge received by the dome is [tex]9.5*10^-^4 C[/tex] and assuming the dome has a radius of 2.0 m, we can calculate the charge density:
σ = Q / A
where Q is the charge received and A is the surface area of the dome.
Surface area of the dome (A) = 4πr²
Substituting the values:
A = 4π(2.0 m)² = 16π m²
σ = ([tex]9.5*10^-^4 C[/tex]) / (16π m²)
Now, we can calculate the electric field at the surface:
E = σ / ε₀
Substituting the value of σ and the value of ε₀ ([tex]8.854*10^-^1^2[/tex]C²/(N·m²)):
E = [([tex]9.5*10^-^4 C[/tex]) / (16π m²)] / ([tex]8.854*10^-^1^2[/tex] C²/(N·m²))
E ≈ 8.54×10^7 N/C
Therefore, the strength of the electric field at the surface of the dome is approximately [tex]8.54*10^7 N/C.[/tex]
(c) 6.2 m from the center of the dome:
To find the electric field at a distance from the center of the dome, we can use the formula for the electric field of a point charge:
E = kQ / r²
where k is the Coulomb's constant, Q is the charge received by the dome, and r is the distance from the center of the dome.
Substituting the values:
E = [tex](9*10^9 Nm^2/C^2)(9.5*10^-^4 C) / (6.2 m)^2[/tex]
E ≈ [tex]2.38 *10^6 N/C.[/tex]
Therefore, the strength of the electric field 6.2 m from the center of the dome is approximately [tex]2.38 *10^6 N/C.[/tex]
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A silver wire carries a current of 3 A for an entire day. How many electrons move past a point in the wire during this time? Ch. 20,#2 In a particular lightning strike, 15C of charge flows from cloud to ground in only 0.5 ms. a. What is the current during such a strike? b. How many electrons are involved?
(a) During an entire day, a silver wire carrying a current of 3 A would have approximately 1.62 x [tex]10^{24[/tex] electrons moving past a given point.
(b) In a lightning strike where 15 C of charge flows in 0.5 ms, the current is 30,000 A, and the number of electrons involved is approximately 9.0 x [tex]10^{19[/tex].
(a) To determine the number of electrons that move past a point in a wire, we need to use the equation relating current, charge, and time. The equation is Q = It, where Q is the charge in coulombs, I is the current in amperes, and t is the time in seconds. Given that the current is 3 A and the time is one day (24 hours or 86,400 seconds), we can calculate the total charge by multiplying the current by the time: Q = 3 A * 86,400 s = 259,200 C. Since one coulomb represents the charge of approximately 6.24 x [tex]10^{18[/tex] electrons, the total number of electrons can be found by dividing the total charge by the charge of a single electron: 259,200 C / (1.6 x [tex]10^{-19[/tex] C) ≈ 1.62 x [tex]10^{24[/tex] electrons.
(b) The current during the lightning strike can be determined by dividing the charge by the time: I = Q / t. Given that the charge is 15 C and the time is 0.5 ms (0.5 x [tex]10^{-3[/tex] seconds), we can calculate the current: I = 15 C / (0.5 x [tex]10^{-3[/tex] s) = 30,000 A. To find the number of electrons involved, we use the same conversion factor as in part (a), where one coulomb corresponds to approximately 6.24 x [tex]10^{18[/tex] electrons. Dividing the charge by the charge of a single electron gives us the number of electrons: 15 C / (1.6 x [tex]10^{-19[/tex] C) ≈ 9.0 x [tex]10^{19[/tex] electrons.
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A 1.0-µF capacitor and a 2.0-µF capacitor are connected together, and then that combination is connected across a 6.0-V potential source (a battery). What is the potential difference across the 2.0-µF capacitor if the capacitors are connected (a) in series or (b) in parallel?
(a) When the capacitors are connected in series, the potential difference across the 2.0-µF capacitor is 6.0 V.
(b) When the capacitors are connected in parallel, the potential difference across the 2.0-µF capacitor is 4.0 V.
(a) When the capacitors are connected in series, the total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of the individual capacitances:
1/C_total = 1/C1 + 1/C2
Substituting the given values, we have:
1/C_total = 1/1.0 µF + 1/2.0 µF = 2/2 µF + 1/2 µF = 3/2 µF
Therefore, C_total = 2/3 µF.
The potential difference across each capacitor connected in series is the same. Since the potential source is 6.0 V, the potential difference across the 2.0 µF capacitor is also 6.0 V.
(b) When the capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances:
C_total = C1 + C2 = 1.0 µF + 2.0 µF = 3.0 µF
In this case, the potential difference across each capacitor is determined by the ratio of its capacitance to the total capacitance. Therefore, the potential difference across the 2.0 µF capacitor connected in parallel is:
V2 = (C2 / C_total) * V_total = (2.0 µF / 3.0 µF) * 6.0 V = 4.0 V.
So, the potential difference across the 2.0 µF capacitor, when the capacitors are connected in parallel, is 4.0 V.
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The long thin cylindrical axon of a nerve carries nerve impulses. The axon can be as long as 1m. Part A Estimate the capacitance of a 1.0-m-long axon of radius 3.7 x 10 ^-6 m with a membrane thickness of 8.8 x 10 ^-9 m. The dielectric constant of the membrane material is about 6.0. Express your answer with the appropriate units. Part B Determine the magnitude of the charge on the inside (negative) and outside (positive) of the membrane wall if there is a 0.080−V potential difference across the wall. Express your answer with the appropriate units. Part C Determine the energy stored in this axon capacitor when charged. Express your answer with the appropriate units.
The capacitance of the axon is approximately 2.7 x 10^-10 Farads (F).
The magnitude of the charge on the inside (negative) and outside (positive) of the membrane wall is approximately 2.16 x 10^-11 Coulombs (C).
The energy stored in this axon capacitor when charged is approximately 8.64 x 10^-13 Joules (J).
Part A:
The capacitance of a cylindrical capacitor can be calculated using the formula:
C = (2πε₀εᵣL) / ln(b/a)
Where:
C is the capacitance
ε₀ is the vacuum permittivity (8.85 x 10^-12 F/m)
εᵣ is the dielectric constant of the membrane material (6.0)
L is the length of the axon (1.0 m)
ln is the natural logarithm
b is the outer radius of the axon (3.7 x 10^-6 m)
a is the inner radius of the axon (a = b - 2h, where h is the membrane thickness)
Given:
b = 3.7 x 10^-6 m
h = 8.8 x 10^-9 m
εᵣ = 6.0
L = 1.0 m
Calculations:
a = b - 2h = 3.7 x 10^-6 m - 2(8.8 x 10^-9 m) = 3.7 x 10^-6 m - 1.76 x 10^-8 m = 3.68 x 10^-6 m
C = (2πε₀εᵣL) / ln(b/a)
C = (2π)(8.85 x 10^-12 F/m)(6.0)(1.0 m) / ln[(3.7 x 10^-6 m) / (3.68 x 10^-6 m)]
C ≈ 2.7 x 10^-10 F
Therefore, the capacitance of the axon is approximately 2.7 x 10^-10 Farads (F).
Part B:
The charge on each side of the membrane can be calculated using the formula:
Q = CV
Where:
Q is the charge
C is the capacitance (2.7 x 10^-10 F, as calculated in Part A)
V is the potential difference across the wall (0.080 V)
Given:
C = 2.7 x 10^-10 F
V = 0.080 V
Calculations:
Q = CV
Q = (2.7 x 10^-10 F)(0.080 V)
Q ≈ 2.16 x 10^-11 C
Therefore, the magnitude of the charge on the inside (negative) and outside (positive) of the membrane wall is approximately 2.16 x 10^-11 Coulombs (C).
Part C:
The energy stored in a capacitor can be calculated using the formula:
E = (1/2)CV²
Where:
E is the energy stored
C is the capacitance (2.7 x 10^-10 F, as calculated in Part A)
V is the potential difference across the wall (0.080 V)
Given:
C = 2.7 x 10^-10 F
V = 0.080 V
Calculations:
E = (1/2)CV²
E = (1/2)(2.7 x 10^-10 F)(0.080 V)²
E ≈ 8.64 x 10^-13 J
Therefore, the energy stored in this axon capacitor when charged is approximately 8.64 x 10^-13 Joules (J).
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A 1000·kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 10·s, then the motor stops. The rocket altitude 14·s after launch is 5200·m. You can ignore any effects of air resistance. (a) What was the rocket's acceleration during the first 10·s? 128 Incorrect: Your answer is incorrect. m/s2 (b) What is the rocket's speed as it passes through a cloud 5200·m above the ground? m/s
(a) The rocket's acceleration during the first 10 seconds was 128 m/s^2.
To find the rocket's acceleration during the first 10 seconds, we need to use the formula for acceleration:
acceleration = change in velocity / time
In this case, the rocket's acceleration is constant, so we can rearrange the formula to solve for acceleration:
acceleration = change in velocity / time
We know that the time is 10 seconds, and we need to find the change in velocity. Since the rocket's acceleration is constant, we can use the following formula to find the change in velocity:
change in velocity = acceleration * time
Plugging in the values we know, we get:
change in velocity = 128 m/s^2 * 10 s
= 1280 m/s
So, the rocket's acceleration during the first 10 seconds is 128 m/s^2.
(b) To find the rocket's speed as it passes through a cloud 5200 m above the ground, we need to use the equation of motion for constant acceleration:
velocity^2 = initial velocity^2 + 2 * acceleration * displacement
In this case, the initial velocity is 0 m/s (since the motor stops), the acceleration is 128 m/s^2 (as found in part a), and the displacement is 5200 m. Plugging in the values, we get:
velocity^2 = 0^2 + 2 * 128 m/s^2 * 5200 m
= 2 * 128 * 5200 m^2/s^2
= 1331200 m^2/s^2
To find the velocity, we take the square root of both sides:
velocity = sqrt(1331200 m^2/s^2)
= 1154.7 m/s
So, the rocket's speed as it passes through a cloud 5200 m above the ground is 1154.7 m/s.
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A 0.120−kg baseball is dropped from rest. Part A If the magnitude of the baseball's momentum is 1.60 kg⋅m/s just before it lands on the ground, from what height was it dropped? Express your answer using three significant figures.
The baseball was dropped from a height of 12.5 meters.
The mechanical energy of an object consists of its kinetic energy (KE) and potential energy (PE). When the baseball is dropped from rest, it only has potential energy initially, which is given by the equation:
PE = mgh
where:
- m is the mass of the baseball (0.120 kg),
- g is the acceleration due to gravity (9.8 m/s^2), and
- h is the height from which the baseball was dropped.
The magnitude of the baseball's momentum just before it lands on the ground is equal to its mass times its velocity:
p = mv
where:
- p is the magnitude of the momentum (1.60 kg⋅m/s), and
- v is the velocity of the baseball just before it lands.
Since the baseball is dropped from rest, its initial velocity is 0 m/s, so we can calculate the final velocity using the equation:
v = sqrt(2gh)
Now, we can substitute the values into the equation for momentum and solve for h:
p = mv
1.60 kg⋅m/s = 0.120 kg * sqrt(2 * 9.8 m/s^2 * h)
Solving for h:
h = (p^2) / (2 * m * g)
Substituting the given values:
h = (1.60 kg⋅m/s)^2 / (2 * 0.120 kg * 9.8 m/s^2)
Calculating the height h:
h ≈ 0.326 m
Therefore, the baseball was dropped from a height of approximately 0.326 meters.
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A digital signal system operates at 300,000 bps. If a signal element encoded with 8 discrete levels, assume the average case (c=1/2), what is the minimum required bandwidth of the channel?
the minimum required bandwidth of the channel is 100,000 bps.The minimum required bandwidth of a channel can be calculated using the formula:
Bandwidth = Bit rate / (log2 L)
where Bit rate is the number of bits transmitted per second and L is the number of discrete signal levels.
In this case, the bit rate is given as 300,000 bps and the signal is encoded with 8 discrete levels.
Using the formula, we can calculate the minimum required bandwidth as follows:
Bandwidth = 300,000 bps / (log2 8)
First, let's calculate the log2 8:
log2 8 = log2 2^3 = 3
Now, let's substitute this value back into the formula:
Bandwidth = 300,000 bps / 3
Simplifying this expression, we get:
Bandwidth = 100,000 bps
Therefore, the minimum required bandwidth of the channel is 100,000 bps.
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The reactance of a capacitor is 60 ΩΩ at a frequency of 81 HzHz
.
What is its capacitance?.
the capacitance of the capacitor is approximately 3.16 × 10^(-5) Farads.
To calculate the capacitance of a capacitor, we can use the formula for reactance (Xc) of a capacitor in an AC circuit:
Xc = 1 / (2πfC)
Where Xc is the reactance, f is the frequency, and C is the capacitance.
In this case, we are given that the reactance is 60 Ω and the frequency is 81 Hz. We can rearrange the formula to solve for the capacitance:
C = 1 / (2πfXc)
Substituting the given values:
C = 1 / (2π * 81 Hz * 60 Ω)
Now, let's calculate the capacitance:
C = 1 / (2π * 81 * 60) Farads
C ≈ 3.16 × 10^(-5) Farads
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I don't know how to do this question. The answer is not 939mT
The magnetic field at a distance of r_1
=4.50 cm from a long straight current carrying wire, is B_1 =650mT. What is the magnetic field at a distance of r_2 =6.50 cm ? Enter your answer in mT. Your Answer: Answer units
the magnetic field at a distance of r2 = 6.50 cm from a long straight current carrying wire is 0.1725 mT (approx).
The magnetic field at a distance of r1=4.50 cm from a long straight current carrying wire is B1 =650mT.
The formula for magnetic field due to a straight current-carrying wire is given by;B = (μ₀I)/(2πr)
Where,μ₀ is the permeability of free spaceI is the current flowing in the wirer is the distance from the wire, andB is the magnetic field at the distance r.
Using this formula, we get magnetic field B1 at a distance r1 from the wire as;B1 = (μ₀I)/(2πr1)We are given that the magnetic field at a distance of r1 = 4.50 cm from a long straight current carrying wire, is B1 = 650 mT.
We need to find the magnetic field at a distance of r2 = 6.50 cm from the wire.To find the value of I, rearrange the formula as;I = (2πrB)/μ₀
Substitute the values given and get I;I = (2 x 3.14 x 4.50 x 650 x 10^(-3))/(4π x 10^(-7))I = 5.307 A
We can now use the value of I to find the magnetic field at a distance of r2 using the same formula as;B2 = (μ₀I)/(2πr2)B2 = (4π x 10^(-7) x 5.307)/(2π x 6.50 x 10^(-2))B2 = 172.48 µT
We know that 1 mT = 1000 µTTherefore, B2 = 0.1725 mT (approx)
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