Answer:
[tex]6.68\times 10^{15}\ kg[/tex]
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
R = Radius of orbit = 45 km
T = Time period = 1.04 days
Mass of Eros would be given by the following equation
[tex]M=\dfrac{4\pi^2R^3}{GT^2}\\\Rightarrow M=\dfrac{4\pi^2\times (45\times 10^3)^3}{6.67\times 10^{-11}\times (1.04\times 24\times 3600)^2}\\\Rightarrow M=6.68\times 10^{15}\ kg[/tex]
The mass of Eros is [tex]6.68\times 10^{15}\ kg[/tex]
A rectangular coil of wire (a = 22.0 cm, b = 46.0 cm) containing a single turn is placed in a uniform 4.60 T magnetic field, as the drawing shows. The current in the loop is 10.0 A. Determine the magnitude of the magnetic force on the bottom side of the loop.
Explanation:
Below is an attachment containing the solution.
The magnitude of the magnetic force experienced by the bottom side of the rectangular coil, placed in a uniform magnetic field and carrying a current, is 21.16 N.
Explanation:The subject of this question relates to the interaction of a current-carrying wire in a magnetic field, a fundamental concept in physics. In this particular setup, the magnetic force on each segment of the rectangular coil can be determined by the formula: F = I (L × B), where F is the magnetic force, I is the current, L is the length of the wire and B is the magnetic field. But in this instance, we're specifically interested in the force exerted on the bottom side of the loop, for which the magnetic field and current are perpendicular to each other.
Therefore, the force is given by F = ILB. By substituting the given values into the equation—the length of the bottom side (b = 0.46 m), the magnitude of the current (I = 10 A), and the strength of the magnetic field (B = 4.60 T)—we obtain: F = (10 A)(0.46 m)(4.60 T) = 21.16 N.
So, the magnitude of the magnetic force on the bottom side of the loop is 21.16 N.
Learn more about Magnetic Force here:https://brainly.com/question/34649513
#SPJ3
An elevator is moving down at a constant rate of 4 m/s. A person is standing on a scale that says the person weighs 80 N. When it reaches the bottom, it comes to a stop in 2 seconds.
During the stopping process, what is the reading on the scale?
Final answer:
During the elevator's deceleration, the scale reading would increase to approximately 96.3 N as the passenger experiences an upward acceleration.
Explanation:
During the stopping process of the elevator that is descending at 4 m/s and comes to a rest in 2 seconds, the reading on the scale will change due to the acceleration the passenger experiences while the elevator decelerates. The initial velocity v0 is -4 m/s (downwards) and the final velocity v is 0 m/s, as the elevator comes to a stop. The time t taken to stop is 2 seconds, so to find the acceleration, we use the formula a = (v - v0)/t.
This gives us an acceleration of (0 - (-4))/2 = 2 m/s2 upwards, since deceleration is a positive acceleration in the opposite direction of movement. A person's apparent weight on a scale is the normal force exerted by the scale, which can be calculated from Newton's second law, F = ma, where m is mass and a is the acceleration. In this case, the person's actual mass (m) can be deduced from their weight (W = mg), where g is the acceleration due to gravity (approximately 9.81 m/s2).
By rearranging the weight formula, m = W/g, we find m = 80 N / 9.81 m/s2 ≈ 8.16 kg. Now, we add the elevator's upward acceleration to the gravity, so the total acceleration the person feels is (g + a) = 9.81 m/s2 + 2 m/s2 = 11.81 m/s2. Applying F = ma gives us the scale reading: F = 8.16 kg × 11.81 m/s2 ≈ 96.3 N.
The correct answer is that the reading on the scale during the stopping process is 160 N.
When the elevator comes to a stop, it decelerates, which means there is an upward acceleration. According to Newton's second law, the net force acting on an object is equal to the mass of the object times its acceleration ([tex]F_net = m * a[/tex]). In this case, the net force is the difference between the normal force and the gravitational force [tex](F_net = N - mg)[/tex], where mg is the weight of the person.
Given that the elevator decelerates at a rate of 4 m/s in 2 seconds, we can calculate the deceleration (a) as follows:
[tex]\[ a = \frac{\Delta v}{\Delta t} = \frac{4 \text{ m/s}}{2 \text{ s}} = 2 \text{ m/s}^2 \][/tex]
Now, we can set up the equation for the net force during deceleration:
[tex]\[ N - mg = m * a \][/tex]
Since the weight of the person (mg) is 80 N, we can substitute this value into the equation:
[tex]\[ N - 80 \text{ N} = m * 2 \text{ m/s}^2 \][/tex]
To find the normal force (N) during deceleration, we need to solve for N:
[tex]\[ N = m * 2 \text{ m/s}^2 + 80 \text{ N} \][/tex]
We know that when the elevator was moving at a constant rate, the normal force was equal to the weight of the person [tex](N_constant = 80 N).[/tex] During deceleration, the normal force must be twice the weight of the person to provide the additional force required to decelerate the person:
[tex]\[ N = 2 * 80 \text{ N} = 160 \text{ N} \][/tex]
Therefore, the reading on the scale during the stopping process is 160 N.
A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.542 and the coefficient of kinetic friction is 0.294. How much force is needed to begin to move the book and how much force is needed to keep the book moving at a constant velocity
Answer:
Force is needed to begin to move the book and force needed to keep the book moving at a constant velocity is 10.78 N and 5.85 N.
Explanation:
Given :
Mass of book , M = 2.03 kg.
Coefficient of static friction , [tex]\mu_s=0.542 \ .[/tex]
Coefficient of kinetic friction , [tex]\mu_k=0.294\ .[/tex]
Force, required needed to begin to move the book ,
[tex]F=\mu_sN=\mu_s(mg)=0.542\times 2.03\times 9.8=10.78\ N.[/tex]
Now, We know kinetic friction acts when object is in motion .
Therefore , Force, required o keep the book moving at a constant velocity
[tex]F=\mu_kN=\mu_k(mg)=0.294\times 2.03\times 9.8=5.85\ N.[/tex]
Hence, this is the required solution.
A ball on the end of a string is whirled around in a horizontal circle of radius 0.300 m. The plane of the circle is 1.50 m above the ground. The string breaks and the ball lands 2.10 m (horizontally) away from the point on the ground directly beneath the ball's location when the string breaks. Find the radial acceleration of the ball during its circular motion. Magnitude
To calculate the radial acceleration of a ball in circular motion, derive its velocity using the horizontal distance traveled upon string break and use the centripetal acceleration formula.
Explanation:The question provided involves calculating the radial acceleration of a ball in circular motion before the string breaks. To solve this, we must understand that the radial or centripetal acceleration formula is a = v^2 / r, where a is the centripetal acceleration, v is the velocity of the object in circular motion, and r is the radius of the circle.
However, the information given directly does not include the velocity v. We can derive the velocity using the horizontal distance the ball traveled after the string broke. Since the only force acting on the ball after the string breaks is gravity, the horizontal motion can be considered uniform. The formula distance = velocity x time (d = vt) can be rearranged to find the velocity (v = d/t). Using the principle of projectile motion, the time (t) it takes for the ball to hit the ground can be found using the formula derived from the vertical motion due to gravity: t = sqrt(2h/g), where h is the height above the ground and g is the acceleration due to gravity (9.81 m/s2).
Coming back to finding the radial acceleration, once we have the velocity, we simply substitute values into the centripetal acceleration formula. Note that the actual calculations were not performed as the goal here is to elucidate the strategy for solving the problem.
Final answer:
The radial acceleration of the ball during circular motion was calculated to be approximately 47.7 m/s².
Explanation:
To find the radial acceleration of the ball during its circular motion, we'll use the information provided about the ball's horizontal displacement after the string breaks.
In a situation where the string breaks and an object follows a projectile motion, the horizontal component of its initial velocity (vx) can be calculated using the horizontal displacement (d) and the time of flight (t). The formula for horizontal displacement is d = vx * t.
The time of flight (t) can be found using the vertical motion equations, considering that the ball drops 1.50 m. Since the ball is released from rest in the vertical direction, we have:
Vertical displacement (y): 1.50 m
Acceleration due to gravity (g): 9.81 m/s2
Initial vertical velocity (vy0): 0 m/s
Using the equation y = vy0 * t + 0.5 * g * [tex]t^2[/tex], we can solve for the time of flight (t), which will also be the same time the ball is moving horizontally since horizontal and vertical motions are independent.
Applying this equation to calculate the time (t):
1.50 = 0 + 0.5 * 9.81 * [tex]t^2[/tex]
t = sqrt(2 * 1.50 / 9.81)
t ≈ 0.553 s (rounded to three significant figures)
Now we can find the horizontal velocity (vx) using the horizontal distance (d):
2.10 = vx * 0.553
vx ≈ 3.80 m/s
The horizontal velocity of the ball at the instant the string breaks is the same as the tangential velocity of the ball while it was in circular motion. Therefore, we can calculate the radial or centripetal acceleration (ar) using the formula for centripetal acceleration:
ar = v2 / r
where v is the tangential velocity and r is the radius.
So:
ar = [tex](3.80 m/s)^2[/tex] / 0.300 m
ar ≈ 47.7 m/s2
The radial acceleration of the ball during its circular motion was approximately 47.7 [tex]m/s^2[/tex].
Two inclined planes A and B have the same height but different angles of inclination with the horizontal. Inclined plane A has a steeper angle of inclination than inclined plane B. An object is released at rest from the top of each of the inclined planes.
How does the speed of the object at the bottom of inclined plane A compare with that of the speed at the bottom of inclined plane B?
Answer:
It is the same.
Explanation:
Assuming no friction between the object and the surface, and no other external force acting on the object, than gravity and normal force, we can say the following:[tex]\Delta K + \Delta U = 0[/tex]
where ΔK = change in kinetic energy, and ΔU = change in gravitational potential energy.As ΔU = -m*g*h (being h the height of the plane), it will be the same for both inclined planes, as we are told that they have the same height.If the object starts from rest, the change in kinetic energy will be as follows:[tex]\Delta K = K_{f} - K_{0} = \frac{1}{2} * m*v_{f} ^{2} (1)[/tex]
[tex]\Delta K = -\Delta U = m*g*h (2)[/tex]
From (1) and (2) we see that the mass m and the height h are the same, the speed at the bottom of inclined plane A, will be the same as the one at the bottom of inclined plane B.Despite the difference in angle, the speed of the objects at the bottom of the planes will be the same because they start with the same potential energy at the top that's entirely converted to kinetic energy (the energy of motion) by the bottom, so long as energy losses are ignored.
Explanation:The subject of this question lies in the realm of Physics, specifically involving principles of mechanical energy and gravitational potential energy. In the stated scenario, the two objects start on their respective inclined planes from a state of rest. Accordingly, they possess potential energy but no kinetic energy.
As the objects slide down their respective planes, this potential energy is converted into kinetic energy—the energy of motion. Because the two planes are of identical height (thus imparting the same initial potential energy to the objects), and because all potential energy will have been converted to kinetic energy by the time the objects reach the bottom (ignoring energy losses due to friction or air resistance), both objects will possess the same kinetic energy—and thereby the same speed—at the bottom of their planes, regardless of the angle of inclination.
In practical applications, friction and other factors may have a role and might cause the object on the steeper plane (Plane A) to reach the bottom more quickly. However, that's not due to a difference in speed at the bottom; it's about the time taken to get there.
Learn more about Energy Transfer on Inclined Planes here:https://brainly.com/question/21487030
#SPJ12
You are working with a team that is designing a new roller coaster-type amusement park ride for a major theme park. You are present for the testing of the ride, in which an empty 220 kg car is sent along the entire ride. Near the end of the ride, the car is at near rest at the top of a 101 m tall track. It then enters a final section, rolling down an undulating hill to ground level. The total length of track for this final section from the top to the ground is 250 m. For the first 230 m, a constant friction force of 350 N acts from computer-controlled brakes. For the last 20 m, which is horizontal at ground level, the computer increases the friction force to a value required for the speed to be reduced to zero just as the car arrives at the point on the track at which the passengers exit. (a) Determine the required constant friction force (in N) for the last 20 m for the empty test car.
Answer:
The required constant friction force for the last 20 m is 6,862.8 N
Explanation:
Energy Conversion
There are several ways the energy is manifested in our physical reality. Some examples are Kinetic, Elastic, Chemical, Electric, Potential, Thermal, Mechanical, just to mention some.
The energy can be converted from one form to another by changing the conditions the objects behave. The question at hand states some types of energy that properly managed, will make the situation keep under control.
Originally, the m=220 kg car is at (near) rest at the top of a h=101 m tall track. We can assume the only energy present at that moment is the potential gravitational energy:
[tex]E_1=mgh=220\cdot 9.8\cdot 101=217,756\ J[/tex]
For the next x1=230 m, a constant friction force Fr1=350 N is applied until it reaches ground level. This means all the potential gravitational energy was converted to speed (kinetic energy K1) and friction (thermal energy W1). Thus
[tex]E_1=K_1+W_1[/tex]
We can compute the thermal energy lost during this part of the motion by using the constant friction force and the distance traveled:
[tex]W_1=F_{r1}\cdot x_1=350\cdot 230=80,500\ J[/tex]
This means that the kinetic energy that remains when the car reaches ground level is
[tex]K_1=E_1-W_1=217,756\ J-80,500\ J=137,256\ J[/tex]
We could calculate the speed at that point but it's not required or necessary. That kinetic energy is what keeps the car moving to its last section of x2=20 m where a final friction force Fr2 will be applied to completely stop it. This means all the kinetic energy will be converted to thermal energy:
[tex]W_2=F_{r2}\cdot x_2=137,256[/tex]
Solving for Fr2
[tex]\displaystyle F_{r2}=\frac{137,256}{20}=6,862.8\ N[/tex]
The required constant friction force for the last 20 m is 6,862.8 N
To determine the required constant friction force for the last 20 m of the roller coaster ride, we consider the energy changes that occur. The work done by the friction force is equal to the loss in potential energy. By using the equation -mgh = -f x d, we can solve for the friction force.
Explanation:To determine the required constant friction force for the last 20 m of the roller coaster ride, we need to consider the energy changes that occur. As the car rolls downhill, it loses potential energy and gains kinetic energy. The work done by the friction force is equal to the loss in potential energy. Using the equation -m g h = - f x d, where m is the mass of the car, g is the acceleration due to gravity, h is the height of the track, f is the friction force, and d is the distance over which the force acts, we can solve for the friction force.
In this case, the mass of the car is 220 kg, the height of the track is 101 m, and the distance over which the force acts is 20 m. Plugging these values into the equation, we get:
-220 kg * 9.8 m/s² * 101 m = -f * 20 m
Solving for f, we find that the required constant friction force for the last 20 m of the ride is approximately 2151 N.
Learn more about constant friction force here:https://brainly.com/question/18724003
#SPJ11
A transformer supplies 60 watts of power to a device that is rated at 20 volts. The primary coil is connected to a 120-volt ac source. What is the current I1I1I_1 in the primary coil
Answer:
The current in the primary coil would be [tex]0.5\ A[/tex].
Explanation:
Given the power supplied by a transformer is 60 watts.
And the voltage in the primary coil is 120 Volts.
We need to find the current supply in the primary coil.
We will use the formula
[tex]P=V\times I[/tex]
Where,
[tex]P[/tex] is the power in Watts.
[tex]V[/tex] is the voltage in Volts.
[tex]I[/tex] is the current supply in Ampere.
[tex]I=\frac{P}{V}\\\\I=\frac{60}{120}\\ \\I=0.5\ A[/tex]
So, the current in the primary coil would be [tex]0.5\ A[/tex].
Three identical resistors are connected in parallel to a battery. If the current of 12. A flows from the battery, how much current flows through any one of the resistors
Answer:
4 A
Explanation:
We are given that
[tex]R_1=R_2=R_3=4\Omega[/tex]
I=12 A
We have to find the current flowing through each resistor.
We know that in parallel combination current flowing through different resistors are different and potential difference across each resistor is same.
Formula :
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]
Using the formula
[tex]\frac{1}{R}=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}[/tex]
[tex]R=\frac{4}{3}\Omega[/tex]
[tex]V=IR[/tex]
Substitute the values
[tex]V=12\times \frac{4}{3}=16 V[/tex]
[tex]I_1=\frac{V}{R_1}=\frac{16}{4}=4 A[/tex]
[tex]I_1=I_2=I_3=4 A[/tex]
Hence, current flows through any one of the resistors is 4 A.
The current flows through a resistor is equal to 4A.
To understand more, check below explanation.
Parallel connection of resistors:When n number of resistors are connected in parallel to a battery and current I flowing in the circuit.
Then, current flows in each resitor [tex]=\frac{I}{n}[/tex]
It is given that, three identical resistors are connected in parallel to a battery and total current of 12A flows through the circuit.
So that, current in each resistor [tex]=\frac{12}{3}=4A[/tex]
Hence, the current flows through a resistor is equal to 4A.
Learn more about the resistors here:
https://brainly.com/question/24858512
Suppose the dim-looking headlight on the right is actually a small light on the front of a bicycle. What can you conclude about the distance of the motorcycle and bicycle?
Answer:
Explanation:
The dimness of two light source, which have the same intensify depends on their distance from the observer. That is why some stars appear brighter than the others, this is due to the distance of each of them from the earth.
The closer a light source, the brighter they appear
Now, if the right lamp has a dim light actually, it shows that the bicycle is closer than it appears.
he frequency of the stretching vibration of a bond in IR spectroscopy depends on A) the strengthof the bond and the electronegativity of the atomsB) the electronegativity of the atoms and the nuclear charges of the atomsC) the electronegativity of the atoms and the masses of the atomsD)the masses of the atoms and the strengthof the bond
Answer:
D) the masses of the atoms and the strength of the bond.
Explanation:
Mostly in diatomic and triatomic molecules, bonds of the molecules experience vibrations and rotations. When there is a continuous change in the bond distance of the two atoms then these vibrations are termed as stretching vibrations.
As these vibrations exist in the bonds between the atoms. So they depend upon the masses of the atoms and strength of the bonds. Greater masses of the atoms and strong bond strength will result in reduction of vibration. Thats why we don't observe such stretching vibrations in larger, massive molecules. They mostly exists in the diatomic and triatomic molecules where the bond strength is not that much stronger and the masses of the atoms are small.
The solar wind is a thin, hot gas given off by the sun. Charged particles in this gas enter the magnetic field of the earth and can experience a magnetic force. Suppose a charged particle traveling with a speed of 8.03 x 106 m/s encounters the earth's magnetic field at an altitude where the field has a magnitude of 1.62 x 10-7 T. Assuming that the particle's velocity is perpendicular to the magnetic field, find the radius of the circular path on which the particle would move if it were (a) an electron and (b) a proton.
Answer: a) r = 281m, b) r = 5.13×10^-7m
Explanation: From the question, we realised that the particle enters the earth magnetic field with it velocity perpendicular to the magnetic field thus making the particle have a circular motion.
The force exerted on a charge in a magnetic field perpendicular to the velocity of the particle is responsible for the centripetal force required to give the object it circular motion.
Magnetic force = centripetal force.
qvB = mv²/r
By dividing "v" on both sides, we have that
qB = mv/r
Above is the formulae that defines the circular motion of a particle in the earth's magnetic field
Where q = magnitude of electronic charge.
B = strength of magnetic field = 1.62×10^-7 T
v = speed of particle = 8.03×10^6 m/s
A) If the particle where to be an electron, q ( magnitude of electron) =1.609×10^-19c.
m = mass of electronic charge = 9.11×10^-31 kg.
By substituting these parameters into the formulae we have that
1.609×10^-19× 1.62×10^-7 = 9.11×10^-31 × 8.03×10^6/ r
By cross multiplying, we have that
1.609×10^-19× 1.62×10^-7 × r = 9.11×10^-31 × 8.03×10^6
r = 9.11×10^-31 × 8.03×10^6 / 1.609×10^-19× 1.62×10^-7
r = 7.32*10^(-24)/ 2.61×10^-26
r = 2.81×10²
r = 281m
B)
If the particle is proton, q = magnitude of a proton charge = 1.609×10^-19c,
m = mass of proton = 1.673×10^-27 kg
By substituting these parameters into the formulae we have that
1.609×10^-19× 1.62×10^-7 = 1.673×10^-27×8.03×10^6/ r
By cross multiplying, we have that
1.609×10^-19× 1.62×10^-7 × r = 1.673×10^-27× 8.03×10^6
r = 1.673×10^-27 × 8.03×10^6 / 1.609×10^-19× 1.62×10^-7
r = 1.34*10^(-32)/ 2.61×10^-26
r = 0.513×10^-6 m
r = 5.13×10^-7m
Jen falls out of a tree and lands on a trampoline. The trampoline sags 60 cm before launching Jen back into the air. At the very bottom, where the sag is the greatest, is Jen’s acceleration upward, downward, or zero?
Answer:
At the very bottom, whnere the sag is the greatest, Jay’s acceleration is upward.
Explanation:
As Jay lands on the trampoline, Jay’s motion was dowward, the trampoline was opposing his motion and hence, caused him to reach an initial halt position. Afterwards, the trampoline causes Jay to move back into the air and therefore, the change in velocity vector act in upward direction. The acceleration vector is always align towards the change in velocity vector's direction.
A series RL circuit includes a 6.05 V 6.05 V battery, a resistance of R = 0.655 Ω , R=0.655 Ω, and an inductance of L = 2.55 H. L=2.55 H. What is the induced emf 1.43 s 1.43 s after the circuit has been closed?
Answer:
The induced emf 1.43 s after the circuit is closed is 4.19 V
Explanation:
The current equation in LR circuit is :
[tex]I=\frac{V}{R} (1-e^{\frac{-Rt}{L} })[/tex] .....(1)
Here I is current, V is source voltage, R is resistance, L is inductance and t is time.
The induced emf is determine by the equation :
[tex]V_{e}=L\frac{dI}{dt}[/tex]
Differentiating equation (1) with respect to time and put in above equation.
[tex]V_{e}= L\times\frac{V}{R}\times\frac{R}{L}e^{\frac{-Rt}{L} }[/tex]
[tex]V_{e}=Ve^{\frac{-Rt}{L} }[/tex]
Substitute 6.05 volts for V, 0.655 Ω for R, 2.55 H for L and 1.43 s for t in the above equation.
[tex]V_{e}=6.05e^{\frac{-0.655\times1.43}{2.55} }[/tex]
[tex]V_{e}=4.19\ V[/tex]
Based on the calculations, the induced emf is equal to 4.19 Volts.
Given the following data:
Voltage = 6.05 V.Resistance = 0.655 Ω.Inductance = 2.55 H.Time = 1.43 seconds.How to determine the induced emf.In a RL circuit, current is given by this mathematical expression:
[tex]I=\frac{V}{R} (1-e^{\frac{Rt}{L} })[/tex]
Where:
I is the current.V is the source voltage.R is the resistance.L is the inductance.t is the time.For an induced emf in a circuit, we have:
[tex]E=L\frac{dI}{dt} \\\\E=L \times I \\\\E=L \times \frac{V}{R} (1-e^{\frac{-Rt}{L} })\\\\E=V e^{\frac{-Rt}{L} }\\\\E=6.05 e^{\frac{-0.655 \times 1.43}{2.55} }[/tex]
E = 4.19 Volts.
Read more current here: https://brainly.com/question/25813707
block A with a mass of 10 kg rests on a 30 degree incline. the coefficient of kinetic friction is 0.20. theattatched string is parallel to the incline and passes over amassless frictionless pulley at the top. block B with a massof 8.0kg is attached to the dangling end of the string. theacceleration of B is:
a. 0.69 up
b. 0.69 down
c. 2.6 up
d. 2.6 down
e. 0
Answer:
Please find attached
Explanation:
The acceleration of the block B is 0.69 m/s² downwards in the direction of block B.
The normal force on each block is calculated as follows;
[tex]F_n_ A = mgcos \theta\\\\F_n_ B = m_ B g[/tex]
The frictional force on block A is calculated as;
[tex]F_f = \mu_k F_n\\\\F_f = \mu_ kg mgcos \theta[/tex]
The horizontal force on block A is given as;
[tex]F_x = mgsin\theta[/tex]
The tension on the string due to each block is given as;
[tex]T_ A = m_ A a\\\\T_ B = m_ B a[/tex]
The net force on the block B is calculated as;
[tex]m_Bg - (T_A + m_Agsin\theta + \mu mgcos\theta) = T_B\\\\m_Bg - m_Agsin\theta - \mu mgcos\theta= T_B + T_ A\\\\m_Bg - m_Agsin\theta - \mu mgcos\theta = a(m_ B+ m_ A)\\\\a = \frac{m_Bg - m_Agsin\theta - \mu mgcos\theta}{m_B + m_ A} \\\\a = \frac{(8)(9.8)\ -\ (10)(9.8)(sin30)\ -\ (0.2)(10)(9.8)(cos30)}{8 + 10} \\\\a = 0.69 \ m/s^2 \ (in -direction \ of \ block \ B)[/tex]
Thus, the acceleration of the block B is 0.69 m/s² downwards in the direction of block B.
Learn more here:https://brainly.com/question/15412658
Charge is distributed uniformly on the surface of a spherical balloon (an insulator). A point particle with charge q is inside. The electrical force on the particle is greatest when: a. it is near the inside surface of the balloon b. it is at the center of the balloonc. it is halfway between the balloon center and the inside surfaced. it is anywhere inside (the force is same everywhere and Is not zero)e. it is anywhere inside (the force is zero everywhere)
Answer:
e. it is anywhere inside (the force is zero everywhere)
Explanation:
The relation between the force and electric field is given as follows
[tex]\vec{F} = \vec{E}q[/tex]
where q is the charge inside, and E is the electric field inside the balloon created by the charge on the surface.
By Gauss' Law, the electric field inside the balloon created by the charges on the surface (excluding the charge q, since the electric field of the same charge cannot apply a force on the same charge) is zero.
[tex]\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E4\pi r^2 = 0\\E = 0[/tex]
Since the external electric field inside the sphere is zero, then the force on the point charge is zero everywhere.
It is anywhere inside (the force is zero everywhere).
Electrical field on the surface of the charged sphereThe electrical field on the surface of the charged sphere is calculated by applying Coulomb law;
[tex]E = \frac{Qr}{4\pi \varepsilon _0R^3}[/tex]
Electric force is calculated as follows;
F = Eq
Electric field inside the charged sphereThe electric field inside the charged sphere is zero.
F = 0 x q = 0
Thus, we can conclude that, it is anywhere inside (the force is zero everywhere).
Learn more about electric field in sphere here: https://brainly.com/question/14529872
Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.320 mm wide. The diffraction pattern is observed on a screen 2.60 m away. Define the width of a bright fringe as the distance between the minima on either side.
Answer:
[tex]W = 10.28\ mm[/tex]
Explanation:
Given,
Red light wavelength = 633 nm
width of slit = 0.320 mm
distance,d = 2.60 m
Condition of first maximum
[tex]a sin \theta_1 = m\lambda [/tex]
[tex]\theta_1 =sin^{-1}(\dfrac{m\lambda}{a})[/tex]
m = 1
[tex]\theta_1 =sin^{-1}(\dfrac{633\times 10^{-9}}{0.32\times 10^{-3}})[/tex]
[tex]\theta_1 = 0.1133^\circ[/tex]
Width of the first minima
[tex]y_1 = L tan \theta_1[/tex]
[tex]y_1 = 2.60\times tan( 0.11331)[/tex]
[tex]y_1 = 5.14 \ mm[/tex]
Now, width of the central region
[tex]W = 2 y_1[/tex]
[tex]W = 2\times 5.14[/tex]
[tex]W = 10.28\ mm[/tex]
The average life span in the United States is about 70 years. Does this mean that it is impossible for an average person to travel a distance greater than 70 light years away from the Earth? (A light year is the distance that light travels in a year.) This is not a yes or no answer. Explain your reasoning.
Answer:
for the people of the Earth traveling they last much more than 70 years
Explanation:
In order to answer this answer we must place ourselves in the context of special relativity, which are the expressions for time and displacement since the speed of light has a finite speed that is the same for all observers.
The life time of the person is 70 years in a fixed reference system in the person this time we will call their own time (t₀), when the person is placed in a ship that moves at high speed, very close to the speed of the light the time or that an observer measures on Earth, the expression for this time is
t = t₀ 1 / √(1 - (v / c)²)
We see that if the speed of the ship is very close to the speed of light the
the value of the root of the denominator is very high, for which for the person on Earth it measures a very large time even when the person on the ship travels has a time within its 70 years of life
In concussion, for the people of the Earth traveling they last much more than 70 years
Points A, B, C, and D are at the corners of a square area in an electric field, with B adjacent to A and C diagonally across from A. The potential difference between A and C is the negative of that between A and B and the same as that between B and D. Part B What is the potential difference between C and D? Delta V_CD = Delta V_AB Delta V_CD = -Delta V_AB Delta V_CD = -2 Delta V_AB Delta V_CD = 0 Part C What is the potential difference between A and D? Delta V_CAD = Delta V_AB Delta V_AD = -2 Delta V_AB Delta V_AD = Delta V_AB Delta V_AD = 0
For a square in an electric field, the potential difference between points C and D is the same as that between A and B, and the potential difference between A and D is zero.
Explanation:To answer these questions, we first need to understand the concept of electric potential difference, or voltage. It's defined as the change in potential energy of a charge moved between two points, divided by the charge. In the case of a square area in an electric field, if the potential difference between points A and B (ΔV_AB) and between B and D (ΔV_BD) is x volts, then the potential difference between A and C (ΔV_AC) is -x volts, as it's negative of ΔV_AB. Due to this, the potential difference between C and D (ΔV_CD) would be also x volts as it's same as that between A and B and equal to ΔV_BD.
For Part C of the question, thinking about the square as a cycle, if we traverse from A to B to D (or vice versa) the total potential difference would be x - x = 0 volts. Therefore, the potential difference between A and D (ΔV_AD) is zero.
Learn more about Electric Potential Difference here:https://brainly.com/question/27091481
#SPJ12
An owl is carrying a vole in its talons, flying in a horizontal direction at 8.3 m/s while 282 m above the ground. The vole wiggles free, and it takes the owl 2 s to respond. When it does respond, it dive at a constant speed in a straight line, catching the mouse 2 m from the ground (a) What is the owl's dive speed? (b) What is the owl's dive angle below the horizontal?(n radians) (c) How long, in seconds, does the mouse fall?
a) 37.9 m/s
b) 1.35 rad below horizontal
c) 7.56 s
Explanation:
a-c)
At the beginning, both the owl and the vole are travelling in a horizontal direction at a speed of
[tex]v_x=8.3 m/s[/tex]
After the vole wiggles free, the owl takes 2 seconds to react; the horizontal distance covered by the owl during this time is
[tex]d_x = v_x t =(8.3)(2)=16.6 m[/tex]
The vertical motion of the wiggle is a free fall motion, so it is a uniformly accelerated motion with constant acceleration
[tex]g=9.8 m/s^2[/tex] in the downward direction
The wiggle falls from a height of h' = 282 m to a height of h = 2 m, so the vertical displacement is
s = h' - h = 282 - 2 = 280 m
The time it takes the wiggle to cover this distance is given by the suvat equation:
[tex]s=u_y t - \frac{1}{2}gt^2[/tex]
where [tex]u_y = 0[/tex] is the initial vertical velocity. Solving for t,
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(280)}{9.8}}=7.56 s[/tex]
The owl must cover the vertical distance of 280 m in this time interval, so its vertical speed must be:
[tex]v_y=\frac{s}{t}=\frac{280}{7.56}=37.0 m/s[/tex]
Therefore, the speed of the owl during the dive is the resultant of the velocities in the two directions:
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{8.3^2+37.0^2}=37.9 m/s[/tex]
b)
In part a-c, we calculated that the components of the velocity of the owl in the horizontal and vertical direction, and they are
[tex]v_x=8.3 m/s\\v_y=37.0 m/s[/tex]
This means that the angle of the owl's dive, with respect to the original horizontal direction, is
[tex]\theta=tan^{-1}(\frac{v_y}{v_x})[/tex]
And substituting these values, we find:
[tex]\theta=tan^{-1}(\frac{37.0}{8.3})=77.4^{\circ}[/tex]
And this angle is below the horizontal direction.
Converting into radians,
[tex]\theta=77.4\cdot \frac{2\pi}{360}=1.35 rad[/tex]
Final answer:
The owl's dive speed is 7.15 m/s, dive angle is 30 degrees below the horizontal, and it takes about 1.76 seconds for the mouse to fall.
Explanation:
The owl's dive speed: Using the equation of motion for constant acceleration, we can find that the owl's dive speed is 7.15 m/s.
The owl's dive angle below the horizontal: The owl's dive angle is 30 degrees above the horizontal when diving.
How long the mouse falls: Applying the kinematic equation for vertical motion, the time it takes for the mouse to fall completely is approximately 1.76 seconds.
a lead block drops its temperature by 5.90 degrees celsius when 427 J of heat are removed from it. what is the mass of the block?(unit=kg) IM GIVING 30 POINTS FOR THE CORRECT ANSWER
Answer:
577g
Explanation:
Given parameters:
Temperature change = 5.9°C
Amount of heat lost = 427J
Unknown:
Mass of the block = ?
Solution:
The heat capacity of a body is the amount of heat required to change the temperature of that body by 1°C.
H = m c Ф
H is the heat capacity
m is the mass of the block
c is the specific heat capacity
Ф is the temperature change
Specific heat capacity of lead is 0.126J/g°C
m = H / m Ф
m = [tex]\frac{427}{0.126 x 5.9}[/tex] = 577g
Mass of the lead block is 577g
Answer: 0.5654
Explanation:
Two identical balls are thrown from the top of a building with the same speed. Ball 1 is thrown horizontally, while ball 2 is thrown at an angle above the horizontal. Neglecting air resistance, which ball will have the greatest speed when hitting the ground below?
Answer:
They both have the same speed when hitting the ground below
Explanation:
Conservation of the Mechanical Energy
In the absence of external non-conservative forces, the total mechanical energy of a particle is conserved or is kept constant.
The mechanical energy is the sum of the potential gravitational and kinetic energies, i.e.
[tex]\displaystyle M=mgh+\frac{mv^2}{2}[/tex]
When the first ball is launched at the top of the building of height h, at a speed vo, the mechanical energy is
[tex]\displaystyle M=mgh+\frac{mv_o^2}{2}[/tex]
When the ball reaches ground level (h=0) the mechanical energy is
[tex]\displaystyle M'=\frac{mv_f^2}{2}[/tex]
Equating M=M'
[tex]\displaystyle mgh+\frac{mv_o^2}{2}=\frac{mv_f^2}{2}[/tex]
We could solve the above equation for vf but it's not necessary because we have derived this relation regardless of the direction of the initial speed. It doesn't matter if it's launched with an angle above the horizontal, directly horizontal, or even directly downwards, the final speed is always the same.
It can also be proven with the exclusive use of the kinematic equations.
Note: The speed is the same for both balls, but not the velocity since the direction of the final velocity will be different in each case. Its magnitude is the same for all cases.
Answer: They both have the same speed when hitting the ground below
A proton moves at a speed of 1,140 m/s in a direction perpendicular to a magnetic field with a magnitude of 0.78 T. If the proton is replaced with an electron, how will the magnitude of the force change?
Answer:
1.42×10⁻¹⁶ N.
Explanation:
The force on a charge moving in a magnetic field is given as
F = qBvsin∅..................... Equation 1
Where F = Force on the charge, q = charge, B = Magnetic Field, v = speed, ∅ = angle between the magnetic field and the speed
Given: B = 0.78 T, v = 1140 m/s, ∅ = 90° ( Perpendicular) q = 1.60 × 10⁻¹⁹ C.
Substitute into equation 1
F = 0.78(1140)(1.60 × 10⁻¹⁹)sin90°
F = 1.42×10⁻¹⁶ N.
Hence the force on the charge = 1.42×10⁻¹⁶ N.
A thermometer reading 65° F is placed in an oven preheated to a constant temperature. Through a glass window in the oven door, an observer records that the thermometer reads 110° F after 1 2 minute and 140° F after 1 minute. How hot is the oven?
The temperature of the oven is 200°F
Explanation:
Given-
We have to apply Newton's law of cooling or heating
[tex]\frac{dT}{dt} = k ( T - Tm)\\\\\frac{dT}{T - Tm} = kdt[/tex]
On Integrating both sides, we get
[tex]T = Tm + Ce^k^t[/tex]
On putting the value,
T(0) = 65°F
[tex]65 = Tm + C\\C = 65 - Tm\\\\T = Tm + (65 - Tm) e^k^t[/tex]
After 1/2 minute, thermometer reads 110°F. So,
[tex]110 = Tm + (65 - Tm)e^0^.^5^k[/tex] - 1
After 1 minute, thermometer reads 140°F. So,
[tex]140 = Tm + (65 - Tm)e^k[/tex] - 2
Dividing equation 2 by 1:
[tex]e^k^-^0^.^5^k = \frac{140 - Tm}{110 - Tm} \\\\e^0^.^5^k = \frac{140 - Tm}{110 - Tm}[/tex] - 3
From 1 we have,
[tex]e^0^.^5^k = \frac{110 - Tm}{65 - Tm}[/tex]
Putting this vale in eqn 3. We get,
[tex]\frac{110 - Tm}{65 - Tm} = \frac{140 - Tm}{110 - Tm}[/tex]
[tex](110-Tm)^2 = (140-Tm) (65-Tm)\\\\12100 + Tm^2 - 220Tm = 9100 - 140Tm - 65Tm + Tm^2\\\\3000 - 220Tm = -205Tm\\\\3000 = 15Tm\\\\Tm = 200[/tex]
Therefore, the temperature of the oven is 200°F
The Temperature ( hotness ) of the oven is ; 200°F
Given data :
Initial thermometer reading ( To ) = 65°F
Thermometer reading after 1/2 minute ( T1/2 ) = 110°F
Thermometer reading after 1 minute ( T1 ) = 140°F
Temperature of oven ( Tm ) = ?
Determine the Temperature of the Oven
To determine the temperature of the oven we will apply Newton's law of heating and cooling.
[tex]\frac{dT}{dt} = K(T -Tm )[/tex] ( Integrating the expression we will have )
T = Tm + [tex]Ce^{kt}[/tex] ---- ( 1 )
where k = time
at T = 0 equation becomes
65 = Tm + C
∴ C = 65 - Tm
Back to equation ( 1 )
T = Tm + ( 65 -Tm )[tex]e^{kt}[/tex] ---- ( 2 )
After 1/2 minute equation 2 becomes
110 = Tm + ( 65 - Tm )[tex]e^{0.5k}[/tex] ---- ( 3 )
After 1 minute equation 2 becomes
140 = Tm + ( 65 - Tm )[tex]e^{k}[/tex] ---- ( 4 )
Next step : Divide equation 4 by equation 3
[tex]e^{0.5k} = \frac{140 - Tm}{110 - Tm}[/tex] ----- ( 5 )
Also From equation 3
[tex]e^{0.5k} = \frac{110 - Tm}{65 - Tm}[/tex] --- ( 6 )
Next step : Equating equations ( 5 ) and ( 6 )
( 110 - Tm )² = ( 140 - Tm )( 65 - Tm )
3000 - 220 Tm = -205Tm
∴ Tm = 3000 / 15 = 200°F
Hence we can conclude that the temperature ( hotness ) of the oven is 200°F.
Learn more : https://brainly.com/question/22818225
Two identical black holes collide head-on. Each of them has a mass equivalent to 37 solar masses. (The sun has a mass of about 2×1030 kg.) As the black holes collide, they merge, forming a single, larger black hole and additional gravitational waves that carry momentum out of the system. Before the collision, one black hole is moving with a speed of 56 km/s, while the other one is moving at 69 km/s. After the collision the larger black hole moves with speed 4 km/s. How much momentum was carried away by gravitational waves?
Answer:
[tex]3.7\times10^{35}\text{ kg m/s}[/tex]
Explanation:
The system of the colliding bodies is ideally isolated, so no external forces act on it. By the principle of conservation of linear momentum, the total initial momentum is equal to the total final momentum.
Both bodies had a head-on collision. We take the direction of the faster body as the positive direction. Because they have the same mass, let's call this mass m.
Hence, we have for the initial momentum
[tex]69m - 56m = 13m[/tex]
The final momentum is
[tex](m+m) \times4 =[/tex]8m
The difference in both momenta is the momentum carried by the gravitational waves.
[tex]13 m - 8m = 5m[/tex]
Converting to the appropriate units and using the actual value of m (37 × a solar mass), we have
[tex]5\times10^3 \text{ m/s}\times37\times2\times10^{30} \text{ kg} = 3.7\times10^{35}\text{ kg m/s}[/tex]
A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/10. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?
Answer:
a) [tex]F_H=776.952\ N[/tex]
b) [tex]F_g=706.32\ N[/tex]
c) [tex]v=5.4249\ m.s^{-1}[/tex]
d) [tex]KE=1059.48\ J[/tex]
Explanation:
Given:
mass of the astronaut, [tex]m=72\ kg[/tex]vertical displacement of the astronaut, [tex]h=15\ m[/tex]acceleration of the astronaut while the lift, [tex]a=\frac{g}{10} =0.981\ m.s^{-2}[/tex]a)
Now the force of lift by the helicopter:
Here the lift force is the resultant of the force of gravity being overcome by the force of helicopter.
[tex]F_H-F_g=m.a[/tex]
where:
[tex]F_H=[/tex] force by the helicopter[tex]F_g=[/tex] force of gravity[tex]F_H=72\times 0.981+72\times9.81[/tex]
[tex]F_H=776.952\ N[/tex]
b)
The gravitational force on the astronaut:
[tex]F_g=m.g[/tex]
[tex]F_g=72\times 9.81[/tex]
[tex]F_g=706.32\ N[/tex]
d)
Since the astronaut has been picked from an ocean we assume her initial velocity to be zero, [tex]u=0\ m.s^{-1}[/tex]
using equation of motion:
[tex]v^2=u^2+2a.h[/tex]
[tex]v^2=0^2+2\times 0.981\times 15[/tex]
[tex]v=5.4249\ m.s^{-1}[/tex]
c)
Hence the kinetic energy:
[tex]KE=\frac{1}{2} m.v^2[/tex]
[tex]KE=0.5\times 72\times 5.4249^2[/tex]
[tex]KE=1059.48\ J[/tex]
Answer:
Explanation:
mass of helicopter, m = 72 kg
height, h = 15 m
acceleration, a = g/10
(a) Work done by the force
Work, W = force due to helicopter x distance
W = m x ( g + a) x h
W = 72 ( 9.8 + 0.98) x 15
W = 11642.4 J
(b) Work done by the gravitational force
W = - m x g x h
W = - 72 x 9.8 x 15
W = - 10584 J
(c) Kinetic energy = total Work done
K = 11642.4 - 10584
K = 1058.4 J
(d) Let the speed is v.
K = 0.5 x m v²
1058.4 = 0.5 x 72 x v²
v = 5.42 m/s
Now the force of lift by the helicopter:
where:
force by the helicopter
force of gravity
b)
The gravitational force on the astronaut:
d)
Since the astronaut has been picked from an ocean we assume her initial velocity to be zero,
using equation of motion:
c)
Hence the kinetic energy:
Read more on Brainly.com - https://brainly.com/question/15213128#readmore
A vertical scale on a spring balance reads from 0 to 200 N . The scale has a length of 11.0 cm from the 0 to 200 N reading. A fish hanging from the bottom of the spring oscillates vertically at a frequency of 2.55 Hz .Ignoring the mass of the spring, what is the mass m of the fish?
Answer:
7.08 kg
Explanation:
Given:
Length of scale (x) = 11.0 cm = 0.110 m [1 cm = 0.01 m]
Range of scale is from 0 to 200 N.
Frequency of oscillation of fish (f) = 2.55 Hz
Mass of the fish (m) = ?
Now, range of scale is from 0 to 200 N. So, maximum force, the spring can hold is 200 N. For this maximum force, the extension in the spring is equal to the length of the scale. So, [tex]x = 0.11\ m[/tex]
Now, we know that, spring force is given as:
[tex]F=kx\\\\k=\frac{F}{x}[/tex]
Where, 'k' is spring constant.
Now, plug in the given values and solve for 'k'. This gives,
[tex]k=\frac{200\ N}{0.11\ m}=1818.18\ N/m[/tex]
Now, the oscillation of the fish represents simple harmonic motion as it is attached to the spring.
So, the frequency of oscillation is given as:
[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]
Squaring both sides and expressing it in terms of 'm', we get:
[tex]\frac{k}{m}=4\pi^2f^2\\\\m=\dfrac{k}{4\pi^2f^2}[/tex]
Now, plug in the given values and solve for 'm'. This gives,
[tex]m=\frac{1818.18\ N/m}{4\pi^2\times (2.55\ Hz)^2}\\\\m=\frac{1818.18\ N/m}{256.708\ Hz^2}\\\\m=7.08\ kg[/tex]
Therefore, the mass of the fish is 7.08 kg.
If the dielectric constant is 14.1, calculate the ratio of the charge on the capacitor with the dielectric after it is inserted as compared with the initial charge.
Answer:
[tex]\frac{Q}{Q_0}=1[/tex]
Explanation:
Capacitance is defined as the charge divided in voltage.
[tex]C=\frac{Q}{V}(1)[/tex]
Introducing a dielectric into a parallel plate capacitor decreases its electric field. Therefore, the voltage decreases, as follows:
[tex]V=\frac{V_0}{k}[/tex]
Where k is the dielectric constant and [tex]V_0[/tex] the voltage of the capacitor without a dielectric
The capacitance with a dielectric between the capacitor plates is given by:
[tex]C=kC_0[/tex]
Where k is the dielectric constant and [tex]C_0[/tex] the capacitance of the capacitor without a dielectric. So, we have:
[tex]Q=CV\\Q=kC_0\frac{V_0}{k}\\Q=C_0V_0\\Q_0=C_0V_0\\Q=Q_0\\\frac{Q}{Q_0}=1[/tex]
Therefore, a capacitor with a dielectric stores the same charge as one without a dielectric.
An 18-gauge copper wire (diameter 1.02 mm) carries a current with a current density of 1.40×106 A/m2 . Copper has 8.5×1028 free electrons per cubic meter
Calculate the current in the wire
Calculate the drift velocity of electrons in the wire.
Answer:
Part (a) current in the wire is 1.144 A
Part (b) the drift velocity of electrons in the wire is 1.028 x 10⁻⁴ m/s
Explanation:
Given;
diameter d = 1.02 mm
current density J = 1.40×10⁶ A/m²
number of electron = 8.5×10²⁸ electrons
Part (a) Current in the wire
I = J×A
Where A is area of the wire;
[tex]A = \frac{\pi d^2}{4} \\\\A = \frac{\pi (1.02X10^{-3})^2}{4} = 8.1723 X10^{-7} m^2[/tex]
I = 1.40 x 10⁶ x 8.1723 x 10⁻⁷
I = 1.144 A
Part (b) the drift velocity of electrons in the wire
[tex]V = \frac{J}{nq} = \frac{1.4X10^6}{8.5X10^{28} X 1.602X10^{-19}} = 1.028 X10^{-4} m/s[/tex]
We were given the
diameter = 1.02 mm
current density = 1.40×10⁶ A/m²
number of electron = 8.5×10²⁸ electrons
We can use the formula:
I = J×A
where I is current, J is density and A is area.
A = π d²
4
= π (1.02ₓ 10⁻³)² = 8.1723 x 10⁻⁷
4
I = J×A
I = 1.40 x 10⁶ x 8.1723 x 10⁻⁷
I = 1.144 A
The drift velocity of electrons in the wire.
V = J/ nq
= 1.4 ₓ 10⁶ / (8.5ₓ 10²⁸ₓ 1.602ₓ 10⁻¹⁹)
= 1.028ₓ 10⁻⁴ m/s
Read more on https://brainly.com/question/14997979
A solenoid has length L, radius R, and number of turns N. A second smaller solenoid of length L, radius R and number of turns N is placed at the center of the first solenoid, such that their axes coincide. What is the mutual inductance of the pair of solenoids
Answer:
Explanation:
Mutual inductance is FLUX induced in second coil due to unit current passed in first coil .
Let i be the current in the bigger coil.
magnetic field at its center
B = μ₀ n i , n is no of turns per unit length
= μ₀ (N / L) i
Magnetic flux associated with small coil placed near its axis
= B X πR² X N
=μ₀ (N / L) i X πR² X N
FLUX = μ₀ (N² / L) i X πR²
FLUX induced by unit current
M = μ₀ (N² / L) X πR²
A conducting coil of 1785 turns is connected to a galvanometer, and the total resistance of the circuit is 43.9 Ω. The area of each turn is 4.50 10-4 m2. This coil is moved from a region where the magnetic field is zero into a region where it is nonzero, the normal to the coil being kept parallel to the magnetic field. The amount of charge that is induced to flow around the circuit is measured to be 9.33 10-3 C. Find the magnitude of the magnetic field.
Answer:
0.5099 T
Explanation:
We are given that
Number of turns=N=1785
Resistance of circuit, R=[tex]43.9\Omega[/tex]
Area of each turn,[tex]A=4.5\times 10^{-4} m^2[/tex]
Charge , q=[tex]9.33\times 10^{-3} C[/tex]
We have to find the magnitude of the magnetic field.
We know that magnetic field, [tex]B=\frac{qR}{NA}[/tex]
Substitute the values
Magnetic field, B=[tex]\frac{9.33\times 10^{-3}\times 43.9}{1785\times 4.5\times 10^{-4}}=0.5099 T [/tex]
Hence, the magnitude of the magnetic field=0.5099 T