The linear charge density on the wire is approximately 0.002 N/Cm.
Explanation:The linear charge density on the wire can be calculated using the formula:
λ = E × r / 2πk
where λ is the linear charge density, E is the electric field, r is the distance from the wire, and k is the Coulomb's constant. Plugging in the known values of E = 20.0 N/C and r = 2.00 cm = 0.02 m, we can solve for λ:
λ = (20.0 N/C) × (0.02 m) / (2π × 8.99 × 10^9 Nm^2/C^2) ≈ 0.002 N/Cm
Therefore, the linear charge density on the wire is approximately 0.002 N/Cm.
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Rank from largest to smallest the values of the magnetic field at the following distances from the axis of the conducting cylinder: Ra = 7.75 cm , Rb = 4.95 cm , r = 5.40 cm , and r>Ra.
The ranking from largest to smallest the values of the magnetic field should be like
Rb = 4.95 cm
r= 5.40 cm
Ra =7.75 cm
r> Ra
What is the magnetic field?It is a vector field that explained the magnetic impact on moving electric charges, electric currents, and magnetic materials.
In this, the force should be perpendicular to the velocity and the magnetic field. Also, it should be inversely proportional with respect to the distance with the axis of the cylinder.
So,
The largest magnetic field =Rb =4.95 cm
And,
Smallest magnetic field = r>Ra
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The magnetic field inside a conducting cylinder is maximized at certain points before dropping off outside. The ranked values from largest to smallest are: r = 5.40 cm, Rb = 4.95 cm, Ra = 7.75 cm, and r > Ra.
To determine the magnetic field at different distances from the axis of a conducting cylinder, we need to understand how the magnetic field varies inside and outside a cylinder carrying a current.
Given the distances Ra = 7.75 cm, Rb = 4.95 cm, r = 5.40 cm, and r > Ra, we will rank the magnetic field intensities:
1.) r = 5.40 cm:
Since r is within the cylinder where current is distributed uniformly, the magnetic field at this point can be calculated using Ampère's Law.
2.) Rb = 4.95 cm:
Still within the cylinder but closer to the center, thus the field here will not yet be maximized.
3.) Ra = 7.75 cm:
Outside the cylinder where the current configuration influences how the field drops off with increasing distance.
4.) r > Ra:
This point is furthest from the axis, hence the magnetic field will be the smallest.
The magnetic field reaches a maximum inside the conducting cylinder before decreasing outside of it.
A 7.7 kg sphere makes a perfectly inelastic collision with a second sphere initially at rest. The composite system moves with a speed equal to one third the original speed of the 7.7 kg sphere. What is the mass of the second sphere
Answer:
15.4 kg.
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = V(m+m').................... Equation 1
Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.
Given: m = 7.7 kg, u' = 0 m/s (at rest)
Let: u = x m/s, and V = 1/3x m/s
Substitute into equation 1
7.7(x)+m'(0) = 1/3x(7.7+m')
7.7x = 1/3x(7.7+m')
7.7 = 1/3(7.7+m')
23.1 = 7.7+m'
m' = 23.1-7.7
m' = 15.4 kg.
Hence the mass of the second sphere = 15.4 kg
Answer:
The mass of the second sphere is 15.4 kg
Explanation:
Given;
mass of the first sphere, m₁ = 7.7 kg
initial velocity of the second sphere, u₂ = 0
let mass of the second sphere = m₂
let the initial velocity of the first sphere = u₁
final velocity of the composite system, v = ¹/₃ x u₁ = [tex]\frac{u_1}{3}[/tex]
From the principle of conservation of linear momentum;
Total momentum before collision = Total momentum after collision
m₁u₁ + m₂u₂ = v(m₁ + m₂)
Substitute the given values;
[tex]7.7u_1 + 0=\frac{u_1}{3} (7.7+m_2)[/tex]
Divide through by u₁
7.7 = ¹/₃(7.7 + m₂)
multiply both sides by 3
23.1 = 7.7 + m₂
m₂ = 23.1 - 7.7
m₂ = 15.4 kg
Therefore, the mass of the second sphere is 15.4 kg
A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50km/s in the +x-direction experiences a force of 2.06�10-16N in the +y-direction, and an electron moving at 4.40km/s in the -z-direction experiences a force of 8.40�10-16N in the +y-direction.
Part A
What is the magnitude of the magnetic field?
B = T
Part B
What is the direction of the magnetic field? (in the xz-plane)
theta = from the -z-direction
Part C
What is the magnitude of the magnetic force on an electron moving in the -y-direction at 3.70km/s ?
F = N
Part D
What is the direction of this the magnetic force? (in the xz-plane)
theta = from the -x-direction
Answer:
Explanation:
velocity of proton v = 1.5 x 10³ i m /s
charge on proton e = 1.6 x 10⁻¹⁹ C
Let the magnetic field be B = Bx i + Bz k
force on charged particle ( proton )
F = e ( v x B )
2.06 x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ [ 1.5 x 10³ i x ( Bx i + Bz k) ]
2.06 x10⁻¹⁶ j = - 1.6 x 10⁻¹⁹ x 1.5 x 10³ Bz j) ]
2.06 x10⁻¹⁶ = - 1.6 x 10⁻¹⁹ x 1.5 x 10³ Bz
Bz = - .8583
force on charged particle ( electron )
F = e ( v x B )
8.40 x10⁻¹⁶ j = -1.6 x 10⁻¹⁹ [ - 4.4 x 10³ k x ( Bx i + Bz k) ]
8.4 x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ x 4.4 x 10³ Bx j ]
- 8.4 x10⁻¹⁶ = 1.6 x 10⁻¹⁹ x 4.4 x 10³ Bx
Bx = - 1.19
Magnetic field = - 1.19 i - .8583 k
magnitude = √ (1.19² + .8583²)
= 1.467 T
If it is making angle θ with x - axis in x -z plane
Tanθ = (.8583 / 1.19 )
36⁰ .
C )
v = - 3.7 x 10³j m /s
e = - 1.6 x 10⁻¹⁶ C
Force = F = e ( v x B )
= -1.6 x 10⁻¹⁹ [ -3.7 x 10³ j x ( Bx i + Bz k) ]
= - 1.6 x 10⁻¹⁹ x 3.7 x 10³ Bx k -1.6 x 10⁻¹⁹ x 3.7 x 10³Bzi ]
= 5.08 i - 7.04 k
Tanθ = 54 ° .
Jet fighter planes are launched from aircraft carriers with the aid of their own engines and a catapult. If in the process of being launched from rest, the thrust of a jet's engines is 2.30 105 N and after moving through a distance of 90 m the plane lifts off with a kinetic energy of 3.60 107 J, what is the work done (in J) on the jet by the catapult?
Answer:
The work done on the jet by the catapult is 1.53 x 10⁷ J
Explanation:
Given;
force of thrust engines, F = 2.3 x 10⁵ N
distance moved by the thrust engines, d = 90 m
kinetic energy of the plane, K.E = 3.60 x 10⁷ J
Based on work-energy theorem;
the net work done on the plane by catapult and the thrust engine is given as;
[tex]W_n = K.E_f = 3.6 *10^7 \ J[/tex]
work done by the thrust engine on the jet;
[tex]W_T_._e_n = fd\\\\W_T_._e_n = 2.3*10^5 * 90\\\\W_T_._e_n = 2.07*10^7 \ J[/tex]
Work done on the jet by the catapult;
[tex]W_c_p= W_n - W_T_._e_g\\\\W_c_p= (3.6*10^7 - 2.07*10^7) J\\\\W_c_p= 1.53*10^7 \ J[/tex]
Therefore, the work done on the jet by the catapult is 1.53 x 10⁷ J
A person is diving in a lake in the depth of h = 15 m. The density of the water is rho = 1.0 x103 kg/m3. The pressure of the atmosphere is P0 = 1.0 x 105 Pa. The surface area of the top of the person's head is A = 0.036 m2.
(a) Express the absolute pressure at the depth of h , Po, Q in terms of Poe, and h
Answer:
Pabs = 247150 [Pa]
Explanation:
The pressure in the depth h can be calculated by the following expression.
Pabs = Po + (rho * g * h)
Where:
g = gravity = 9.81[m/s^2]
rho = density = 1000 [kg/m^3]
h = depth = 15 [m]
Po = 100000 [Pa]
Pabs = 100000 + (1000*9.81*15)
Pabs = 247150 [Pa]
The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the eyes of air-dwelling mammals. In mammalian eyes, the shape (and therefore the focal length) of the lens changes to enable the eye to focus at different distances. In amphibian eyes, the shape of the lens doesn't change. Amphibians focus on objects at different distances by using specialized muscles to move the lens closer to or farther from the retina, like the focusing mechanism of a camera. In air, most frogs are nearsighted; correcting the distance vision of a typical frog in air would require contact lenses with a power of about −6.0D .A frog can see an insect clearly at a distance of 10cm . At that point the effective distance from the lens to the retina is 8mm .If the insect moves 5cm farther from the frog, by how much and in which direction does the lens of the frog's eye have to move to keep the insect in focus?0.02cm, toward the retina.0.02cm, away from the retina.0.06cm, toward the retina.0.06cm, away from the retina.
Answer:
0.2cm towards the retina.
Explanation:
the focal length of the frog eye is
(1/f) = (1/10) + (1/0.8)
f = 0.74cm
Since the distance of the object is 15cm Hence
(1/0.74) = (1/15) + (1/V)
V = 0.78cm
Therefore the distance the retina is to move is
0.78cm - 0.8cm = 0.02cm towards the retina.
A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-47 m/s) and a second piece, also of mass m, moves with velocity (-47 m/s) . The third piece has mass 3m. Just after the explosion, what are the (a) magnitude and (b) direction (as an angle relative to the +x axis) of the velocity of the third piece?
Answer:
a) 22.06m/s
b)45°
Explanation:
Let 'm' be mass of 1st and 2nd pieces
mass of 3rd piece=3m
[tex]v_{1}[/tex]=velocity of 1st piece = -47iˆ m/s
[tex]v_{2}[/tex]=velocity of 2nd piece = -47jˆ m/s
[tex]v_{3}[/tex]=velocity of 3rd piece=?
By considering conservation of linear momentum, we have
m[tex]v_{1}[/tex] + m[tex]v_{2}[/tex] + 3m[tex]v_{3}[/tex]=0
[tex]v_{1}[/tex] + [tex]v_{2}[/tex] + 3[tex]v_{3}[/tex]=0
3[tex]v_{3}[/tex]= - ([tex]v_{1}[/tex] + [tex]v_{2}[/tex] )
[tex]v_{3}[/tex] = -[tex]\frac{1}{3}[/tex] ([tex]v_{1}[/tex] + [tex]v_{2}[/tex] )
Substituting the values of [tex]v_{1}[/tex] and [tex]v_{2}[/tex] in above equation
[tex]v_{3}[/tex] = -[tex]\frac{1}{3}[/tex] (-47iˆ -47jˆ ) => 15.6iˆ + 15.6jˆ
(a)Magnitude of the velocity of the third piece is given by
|[tex]v_{3}[/tex]| =√15.6²+15.6² => 22.06m/s
(b) its direction (as an angle relative to the +x axis) 'θ'
θ= [tex]tan^{-1} (\frac{15.6}{15.6} )[/tex] => [tex]tan^{-1} (1 )[/tex] =>45°
A step-up transformer is connected to a generator that is delivering 141 V and 110 A. The ratio of the turns on the secondary to the turns on the primary is 1110 to 5. What voltage is across the secondary
Final answer:
To find the voltage across the secondary winding of a step-up transformer with a primary voltage of 141 V and a turns ratio of 1110 to 5, the secondary voltage is calculated to be 31302 V.
Explanation:
Calculating the Voltage Across the Secondary in a Step-Up Transformer
To calculate the voltage across the secondary winding of a step-up transformer, we use the transformer equation that relates the primary and secondary voltages with the number of turns on the primary (Np) and secondary (Ns) windings:
Vs / Vp = Ns / Np
Given that the primary voltage (Vp) is 141 V, and the ratio of the number of turns is 1110 to 5 (Ns / Np), we can rearrange the equation to solve for the secondary voltage (Vs) as follows:
Vs = Vp × (Ns / Np)
Vs = 141 V × (1110 / 5)
Vs = 141 V × 222
Vs = 31302 V
Thus, the voltage across the secondary winding is 31302 V.
A wire of length L is wound into a square coil with 117 turns and used in a generator that operates at 60.0 Hz and 120 V rms value in a 0.023-T magnetic field. What is the length L of the wire used to construct the coil?
Answer: The length L of the wire used to construct the coil is 191.4m
Explanation: Please see the attachments below
A bat, flying at 5.1 m/s, pursues an insect that is flying at 1.1 m/s in the same direction. The bat emits a 47000-Hz sonar pulse. Take the speed of sound to be 343 m/s. show answer No Attempt At what frequency, in hertz, does the bat hear the pulse reflected back from the insect? f = | sin() cos() tan() cotan() asin() acos() atan() acotan() sinh() cosh() tanh() cotanh() Degrees Radians π ( ) 7 8 9 HOME E ↑^ ^↓ 4 5 6 ← / * 1 2 3 → + - 0 . END √() BACKSPACE DEL CLEAR Grade Summary Deductions 0% Potential 100% Submissions Attempts remaining: 3 (4% per attempt) detailed view Hints: 4% deduction per hint. Hints remaining: 2 Feedback: 5% deduction per feedback.
Answer:
The frequency of the sona-pulse reflected back is [tex]f_b = 48109.22Hz[/tex]
Explanation:
From the question we are told that
The speed of the bat is [tex]v = 5.1 m/s[/tex]
The speed of the insect is [tex]v_i = 1.1 m/s[/tex]
The frequency emitted by the bat is [tex]f = 47000 \ Hz[/tex]
The speed of sound is [tex]v_s = 343 m/s[/tex]
Let look at this question in this manner
At the first instant the that the bat emits the sonar pulse
Let the bat be the source of sound
Let the insect be the observer
This implies that the frequency of sound the the insect would receive is mathematically represented as
[tex]f_a = [\frac{v_s - v_i}{v_s - v}] f[/tex]
Substituting values
[tex]f_a = [\frac{343 - 1.1}{345 -5.1} ] * 47000[/tex]
[tex]f_a = 47556.4 Hz[/tex]
Now at the instant the sonar pules reaches the insect
Let the bat be the observer
Let the insect be the source of the sound
Here the sound wave is reflected back to the bat
This implies that the frequency of sound the the bat would receive is mathematically represented as
[tex]f_b =[ \frac{x}{y} ] * f_a[/tex]
[tex]f_b =[ \frac{v_s + v}{v_s + v_i} ] * f_a[/tex]
[tex]f_b =[ \frac{343 + 5.1}{343 + 1.1} ] * 47556.4[/tex]
[tex]f_b = 48109.22Hz[/tex]
Final answer:
To calculate the frequency a bat hears from a sonar pulse reflected back by an insect, the Doppler effect formula is used. By substituting the given values into the formula, the result is approximately 47,544 Hz, which is the frequency at which the bat detects the echo from the insect.
Explanation:
To calculate the frequency at which a bat hears the sonar pulse reflected back from an insect, we can use the Doppler effect formula for sound in the same direction as the source of the sound is moving:
[tex]f' =\frac{f (v + v_d)}{(v + v_s)}[/tex]
Where:
f' is the frequency heard by the bat.
f is the emitted frequency of the bat's sonar pulse (47,000 Hz).
v is the speed of sound (343 m/s).
[tex]v_d[/tex] is the speed of the detector (bat), which is 5.1 m/s.
[tex]v_s[/tex] is the speed of the source (insect), which is 1.1 m/s.
Substituting the values into the equation:
[tex]f' = \frac{47000 Hz \times (343 m/s + 5.1 m/s)}{(343 m/s + 1.1 m/s)}[/tex]
[tex]f' = \frac{47000 Hz * 348.1 m/s}{344.1 m/s}[/tex]
[tex]f' = 47000 Hz \times 1.0116[/tex]
f' = 47544.2 Hz
Therefore, the bat hears the pulse reflected back from the insect at approximately 47,544 Hz.
A battery with emf e m f is connected in series with an inductance L and a resistance R. (a) Assuming the current has reached steady state when it is at 99% of its maximum value, how long does it take to reach steady state, assuming the initial current is zero? (Use any variable or symbol stated above as necessary. To represent e m f, use E.) t99% = (b) If an emergency power circuit needs to reach steady state within 1.2 ms of turning on and the circuit has a total resistance of 72 Ω, what values of the total inductance of the circuit are needed to satisfy the requirement? (Give the maximum value.) H
Answer:
a) [tex]t = 4.6\tau[/tex]
b) [tex]L = 0.0187 \: H[/tex]
Explanation:
The current flowing in a R-L series circuit is given by
[tex]I = I_{0} (1 - e^{\frac{-t}{\tau} })[/tex]
Where τ is the time constant and is given by
[tex]\tau = \frac{L}{R}[/tex]
Where L is the inductance and R is the resistance
Assuming the current has reached steady state when it is at 99% of its maximum value,
[tex]0.99I_{0} = I_{0} (1 - e^{\frac{-t}{\tau} })\\0.99 = (1 - e^{\frac{-t}{\tau} })\\1 - 0.99 = e^{\frac{-t}{\tau}}\\ln(0.01) = ln(e^{\frac{-t}{\tau}})\\-4.6 = \frac{-t}{\tau}\\t = 4.6\tau\\[/tex]
Therefore, it would take t = 4.6τ to reach the steady state.
(b) If an emergency power circuit needs to reach steady state within 1.2 ms of turning on and the circuit has a total resistance of 72 Ω, what values of the total inductance of the circuit are needed to satisfy the requirement?
[tex]t = 4.6\frac{L}{R}\\t = 4.6\frac{L}{R}\\0.0012 = 4.6\frac{L}{72}\\0.0864 = 4.6 L\\L = 0.0864/4.6\\L = 0.0187 \: H[/tex]
Therefore, an inductance of 0.0187 H is needed to satisfy the requirement.
A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is attached to a spring that has a spring constant k k of 205 N/m. 205 N/m. The block and bullet continue to move, compressing the spring by 35.0 cm 35.0 cm before the whole system momentarily comes to a stop. Assuming that the surface on which the block is resting is frictionless, determine the mass m w mw of the wooden block.
Answer:
0.372 kg
Explanation:
The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:
[tex]mu=(M+m)v[/tex] (1)
where
[tex]m=11.9 g = 11.9\cdot 10^{-3}kg[/tex] is the mass of the bullet
[tex]u=261 m/s[/tex] is the initial velocity of the bullet
[tex]M[/tex] is the mass of the block
[tex]v[/tex] is the velocity at which the bullet and the block travels after the collision
We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write
[tex]\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2[/tex] (2)
where
k = 205 N/m is the spring constant
x = 35.0 cm = 0.35 m is the compression of the spring
From eq(1) we get
[tex]v=\frac{mu}{M+m}[/tex]
And substituting into eq(2), we can solve to find the mass of the block:
[tex](M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg[/tex]
The mass of the wooden block is approximately 0.404 kg.
The initial kinetic energy of the bullet is given by:
KE_initial = 0.5 * mb * vb^2
where:
`m_b` is the mass of the bullet (11.9 g = 0.0119 kg)
`v_b` is the velocity of the bullet (261 m/s)
The potential energy stored in the compressed spring is given by:
PE_spring = 0.5 * k * x^2
where:
`k` is the spring constant (205 N/m)
`x` is the compression of the spring (0.35 m)
Since the system comes to a stop, the initial kinetic energy is converted to potential energy stored in the spring:
KE_initial = PE_spring
Solving for the mass of the wooden block, we get:
m_w = (2 * KE_initial) / (v_b^2 - (k * x) / m_b)
Plugging in the values, we get,
m_w = (2 * 0.5 * 0.0119 kg * (261 m/s)^2) / ((261 m/s)^2 - (205 N/m * 0.35 m) / 0.0119 kg)
m_w ≈ 0.404 kg
Therefore, the mass of the wooden block is approximately 0.404 kg.
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if two objects each have a mass of 10 kg , then the force of gravity between them...
A) is constant
B) is 100 kg
C)is greater when they are closer together
D) depends only on their masses
if two objects each have a mass of 10 kg, then the force of gravity between them C) is greater when they are closer together.
We have two objects, each with a mass of 10 kg. We can calculate the force of gravity (F) between them using Newton's law of universal gravitation.
[tex]F = G \frac{m_1m_2}{r^{2} }[/tex]
where,
G: gravitational constantm₁, m₂: masses of both objectsr: distance between the objectsAs we can see from this expression. the gravitational force between the objects depends on their masses and the distance between them. The closer they are, the stronger the gravity force.
if two objects each have a mass of 10 kg, then the force of gravity between them...
A) is constant. No. It varies with the distance.
B) is 100 kg. No. kg is not a unit of force.
C) is greater when they are closer together. Yes.
D) depends only on their masses. No. It also depends on the distance between them.
if two objects each have a mass of 10 kg, then the force of gravity between them C) is greater when they are closer together.
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The symbol, "I" represents-
a. voltage
b. power
c. current
d. resistance
Answer:
The answer is C.
Explanation:
Current can be represented in the symbol of I.
The symbol for voltage is V.
The symbol for power is W.
The symbol for resistance is R.
Determine the absolute pressure on the bottom of a swimming pool 30.0 m by 8.9 m whose uniform depth is 1.9 m . Express your answer using two significant figures. P = nothing N/m2 Request Answer Part B Determine the total force on the bottom of a swimming pool. Express your answer using two significant figures. F = nothing N Request Answer Part C What will be the pressure against the side of the pool near the bottom? Express your answer using two significant figures. P = nothing N/m2 Request Answer Provide Feedback
Answer:
A)P = 1.2 × 10⁵Pa
B)F = 3.2 × 10⁷N
C) P = 1.2 × 10⁵Pa
Explanation:
Part A)
The relative pressure at the bottom of a column of fluid is given by
[tex]p_r = \rho g h[/tex]
where
[tex]\rho[/tex] is the fluid density
g is the gravitational acceleration
h is the height of the column of fluid
At the bottom of the swimming pool, h=1.9 m, and the water density is
[tex]\rho[/tex] = 1000 kg/m^3, therefore the relative pressure is
[tex]p_r = (1000 kg/m^3)(9.81 m/s^2)(1.9 m)=1.86 \cdot 10^4 Pa[/tex]
To find the absolute pressure, we must add to this the atmospheric pressure, [tex]p_a[/tex] :
[tex]p= p_r + p_a\\= 1.86 \cdot 10^4 Pa + 1.01 \cdot 10^5 Pa \\=1.2 \times 10^5 Pa[/tex]
part B
Total force acting on the bottom
force = pressure * area
area of pool = 30.0 m × 8.9 m
= 267m²
Force F =
1.2 × 10⁵ * 267m² N
= 32040000 N
F = 3.2 × 10⁷N
Part C
The pressure acting on the side wall will be
now the pressure at the side of the pool at the bottom is simply equal to absolute pressure as they are at same level
P = 1.2 × 10⁵
A long straight wire carries a current of 40 A to the right. An electron, traveling at 2.7 10 7 m/s, is 5.3 cm from the wire. What force, magnitude and direction, acts on the electron if the electron velocity is directed toward the wire
Answer:
[tex]6.53\times10^-^1^7N[/tex]
Explanation:
The magnet of the magnetic field is 53 cm = 0.53m from wire is
[tex]B = \frac{\mu_0 I}{2\pi d}[/tex]
[tex]= \frac{(4\pi \times 10^-^7)(40)}{2 \pi (0.53)} \\\\= \frac{5.0265\times 10^-^5}{3.33} \\\\= 1.5095 \times 10^-^5[/tex]
the magnetic force exerted by the wire on the electron is
[tex]F = Bqv \sin \theta\\\\= 1.5095 \times 10^-^5 \times1.602\times10^-^1^9\times2.7\times10^7\\\\= 6.53\times10^-^1^7N[/tex]
From the right hand rule the direction of the force is parallel to the current (since the particle is electron)
Answer: f = 6.52*10^-16 N
Explanation:
if we assume that the force is directed at the y positive direction, then
B = μi / 2πr, where
μ = 4π*10^-7
B= (4π*10^-7 * 40) / 2 * π * 5.3*10^-2
B = 5.027*10^-5 / 0.333
B = 1.51*10^-4 T
Since v and B are perpendicular, then,
F = qvB
F = 1.6*10^-19 * 2.7*10^7 * 1.51*10^-4
F = 2.416*10^-23 * 2.7*10^7
F = 6.52*10^-16 N
Therefore, the magnitude of the force is, F = 6.52*10^-16 N and it moves in the i negative direction
Derivation of Eq. (3): o Basic physics principles: Justify equations (1) and (2) in your own words. . Doing the algebra: From equations (1) and (2), show that equation (3) holds. mu? =eu B This would be nice if we knew the velocity. Fortunately, we know the voltage through which the electrons are accelerated. Setting the change in electrical potential energy equal to the final kinetic energy of the electrons, we find: eV = From this point, some algebra lets us eliminate the velocity variable entirely, giving us: e 2V m (B_T)
Answer:
About eq (1)
[tex]mv^2/r = eVB[/tex]
when a charged particle (electron) enters into the magnetic field which is perpendicular to direction of motion than there will be magnetic force on particle and particle will travel in circular path in with constant speed.
So using force balance on charged particle:
[tex]F_{net} = Fc - Fm[/tex]
Since particle is traveling at constant speed, So acceleration is zero, and
[tex]F_{net} = 0[/tex]
[tex]Fc - Fm=0[/tex]
[tex]Fc = Fm[/tex]
Fc = centripetal force on particle [tex]= m*v^2/r[/tex]
Fm = magnetic force on electron = [tex]q*VxB = q*V*B*sin \theta[/tex]
q = charge on electron = e
since magnetic field is perpendicular to the velocity of particle, So theta = 90 deg
sin 90 deg = 1
So,
[tex]m*v^2/r = e*v*B[/tex]
About equation 2:
When this charged particle is released from rest in a potential difference V, and then it enters into above magnetic field, then using energy conservation on charge particle
[tex]KEi + PEi = KEf + PEf[/tex]
KEi = 0, since charged particle started from rest
[tex]PEi - PEf = q*dV[/tex]
[tex]PEi - PEf = eV[/tex]
KEf = final kinetic energy of particle when it leaves [tex]= (1/2)*m*v^2[/tex]
So,
[tex]0 + eV = (1/2)*m*v^2[/tex]
[tex]eV = (1/2)*m*v^2[/tex]
From above equation (1) and (2)
[tex]m*v^2/r = evB[/tex]
[tex]e/m = v/(r*B)[/tex]
Now
[tex]eV = (1/2)*m*v^2[/tex]
[tex]v = \sqrt{(2*e*V/m)}[/tex]
[tex]e/m = \sqrt{ (2*e*V/m)/(r*B)}[/tex]
[tex]\frac{e^2}{m^2} = \frac{2*e*V}{(m*r^2*B^2)}[/tex]
[tex]e/m = 2*V/(r^2*B^2)[/tex]
[tex]e/m = 2V/(Br)^2[/tex]
A copper rod is sliding on two conducting rails that make an angle of 19o with respect to each other, as in the drawing. The rod is moving to the right with a constant speed of 0.60 m/s. A 0.63-T uniform magnetic field is perpendicular to the plane of the paper. Determine the magnitude of the average emf induced in the triangle ABC during the 7.5-s period after the rod has passed point A.
Answer:
0.2923 V
Explanation:
Given that
Angle between the rails, θ = 19°
Speed of the rod, v = 0.6 m/s
Magnetic field present, B = 0.63 T
Time used, t = 7.5 s
E = -ΔΦ/Δt
where, Φ = BA, so
E = -BΔA / Δt
To get the area, if we assume the rails are joined in a triangular fashion(see attachment)
E = -B(1/2 * AC * BC) / Δt
E = -B(vΔt * vΔt tanθ) / 2Δt
E = -(B * v² * Δt² * tanθ) / 2Δt
E = -Bv²Δt.tanθ/2
E = -(0.63 * 0.6² * 7.5 * tan 19) / 2
E = -0.5857 / 2
E = -0.2923
Thus, the magnitude of average emf induced if 0.2923 V
A man pushes a shopping cart at a constant speed. What happens if the man increases the force on the cart?
Answer:
The cart would speed up.
Explanation:
According to Newton's 1st law, object subjected to no force, or net force 0, would have a constant speed. In our case the cart is initially at constant speed, meaning the man exerts a force that is equal to friction force. If he increases the force on the cart, the net force would no longer be 0. The cart would gain an acceleration and increases its speed.
Answer:
A. The cart accelerates
Explanation:
20 POINTS! TRUE OR FALSE
Friction is a force that speeds up motion between two surfaces that are in contact with each other.
A) TRUE
B) FALSE
Answer:
False
Explanation:
Friction is the resisting force of motion that converts kinetic energy into heat energy. Kinetic energy is the energy of movement. Removing kinetic energy makes the object slower.
Block A weighs 12 N and has an apparent weight of 8 N when completely submerged in an ideal fluid. Block B weighs 20 N and has an apparent weight of 12 N when completely submerged in the same ideal fluid.
The ratio of the densities A/ B equals:
Final answer:
The calculated density ratio is 0.5.
The ratio of the densities of two blocks, A and B, is found by comparing the apparent weight reductions when they are submerged in a fluid. According to Archimedes' principle, this reduction is due to the buoyant force, which is equal to the weight of the fluid displaced.
Explanation:
The question revolves around calculating the ratio of the densities of two blocks, A and B, using the concept of buoyant force according to Archimedes' principle. The apparent weight of each block when submerged in a fluid gives us the buoyant force acting upon it, which is equal to the weight of the fluid displaced. Since both blocks are submerged in the same fluid, the difference in weights in air versus submerged gives us a measure of the volume of fluid displaced due to the respective densities of the blocks.
To determine the ratio of densities (A/B), we use the formula:
Ratio of densities = (Weight of A in air - Apparent weight of A) / (Weight of B in air - Apparent weight of B)
Substituting the given weights:
Ratio of densities = (12 N - 8 N) / (20 N - 12 N) = 4 N / 8 N = 0.5
Thus, the ratio of the densities of block A to block B is 0.5.
Octane (C8H18) is burned with dry air. The volumetric analysis of the products on a dry basis is as below. (Fig. 15–12) Determine (a) the air–fuel ratio, (b) the percentage of theoretical air used, and (c) the amount of H2O that condenses as the products are cooled to 25°C at 100 kPa
Answer:
a. Air fuel Ratio = 19.76 kg air/kg fuel
b. % Theoretical air used = 131%
c. Amount of H2O that condenses as the products are cooled to 25°C at 100kPa = 6.59 kmol
Explanation:
A particle with charge 1.60×10−19 C is placed on the x axis in a region where the electric potential due to other charges increases in the +x direction but does not change in the y or z direction.The particle, initially at rest is acted on by the electricforce and moves from point a to point b along the x axis increasingits kinetic energy by 8.00 x 10 -19J In what direction and through what potential difference does the particlemove
If the particle moves from point b to point c in the ydirection what is the change in its potential energy
Answer:
Explanation:
According to energy conservation principle;
[tex]\delta \ K.E = \delta \ P.E[/tex]
[tex]8.00*10^{-19} \ J = (1.6*10^{-19} \ C ) |V_b-V_a|[/tex]
[tex]|V_b-V_a|= \frac {8.00*10^{-19} }{(1.6*10^{-19} ) }[/tex]
[tex]|V_b-V_a|= 5 \ V[/tex]
The potential is positive hence the electric fielf=d is negative along the x-axis.
We can then say that the movement of the particle goes to the left through a potential difference of [tex]|V_b-V_a|= 5 \ V[/tex].
There will be no significant change in the y-direction of the potential energy when the particle moves from point b to point c in the y-direction.
The human ear is sometimes called a Fourier analyzer. What does this mean and why is it an apt description? 1. Our ears can sort out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone. 2. Our ears measure the intensity of sound, which is just what a Fourier analyzer does. 3. Our ears can measure the speed of sound. 4. Our ears have nothing to do with a Fourier analyzer
Answer:
1. Our ears can sort out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone.
Explanation:
A complex tone is a sound wave that consist of two or more forms of audible sound frequencies. Sound wave is a mechanical wave that is longitudinal, and could be represented by a sine wave because of it sinusoidal manner of propagation.
A Fourier analyzer can be used to differentiate individual sine waves from a combination of two or more of it; which is as the same function performed by human ear. To the human ear, a sound wave that consist of more than one sine wave will have perceptible harmonics which would be distorted and turn to a noise.
Thus, the human ear makes it possible to hear the pure tones that make up a complex tone.
Answer:
1. A Fourier analyzer sorts out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone.
Explanation:
Fourier analysis is a technique that is used to determine which sine waves constitute a given signal, i.e. to deconstruct the signal into its individual sine waves. It is the process of decomposing a periodic function into its constituent sine or cosine waves.
What goes on inside our ears is a mathematical process called a Fourier transform. In the ear, there's a combination of different waves, Fourier analysis identifies contributions at different frequencies, allowing us to reconstruct the individual signals that go into it.
A complex tone perceived by the air is is an individual sine wave that the ear, by acting as a Fourier analyzer, decomposes to serious of sine waves that we hear as pure tones.
Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 103 kg, and the second a mass of 7.50 ✕ 103kg. If the two satellites collide elastically rather than dock, what is their final relative velocity? Adopt the reference frame in which the second satellite is initially at rest and assume that the positive direction is directed from the second satellite towards the first satellite.
...............m/s
Answer: Their final relative velocity is -0.412 m/s.
Explanation:
According to the law of conservation,
[tex]m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v[/tex]
Putting the given values into the above formula as follows.
[tex]m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v[/tex]
[tex]2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v[/tex]
[tex]-4.12 \times 10^{3} kg m/s = (10^{4} kg) v[/tex]
v = [tex]\frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}[/tex]
= -0.412 m/s
Thus, we can conclude that their final relative velocity is -0.412 m/s.
A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of 0.390.
How far along the rough ice does she go before stopping?
Answer:
18.8m
Explanation:
Let the distance traveled before stopping be 'd' m.
Given:
Mass of the skater (m) = 90 kg
Initial velocity of the skater (u) = 12.0 m/s
Final velocity of the skater (v) = 0 m/s (Stops finally)
Coefficient of kinetic friction (μ) = 0.390
Acceleration due to gravity (g) = 9.8 m/s²
Now, we know from work-energy theorem, that the work done by the net force on a body is equal to the change in its kinetic energy.
Hence, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.
Frictional force is given as:
[tex]f=\mu N[/tex]
Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg
[tex]f=\mu mg\\=0.390\times 90\times 9.8\\=343.98\ N[/tex]
work done by friction is a negative work as friction and displacement are in opposite direction and is given as
[tex]W=-fd=-343.98d[/tex]
Now, change in kinetic energy is given as:
[tex]\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J[/tex]
Therefore, from work-energy theorem,
[tex]W=\Delta K\\\\-343.98d=6480\\\\d=\frac{6480}{343.98}\\\\d=18.8m[/tex]
Hence, the skater covers a distance of 18.8 m before stopping.
Answer:
18.84m
Explanation:
We are given;
Mass; m = 90 Kg
Initial velocity; u = 12 m/s
Coefficient of friction; μ = 0.390
Let,the combined mass of the ice skater and the skate be M
Thus, So,if the coefficient of frictional force is μ,then frictional force acting is; μN. N= Mg. Thus F_friction = μMg
Now, the deceleration due to friction will be, F_friction/M
Thus, deceleration = μMg/M
M will cancel out and we have; μg
Now, deceleration means negative acceleration. Thus acceleration (a) =
-μg
Now, to find the distance, let's use equation of motion which is;
v² = u² + 2as
Putting -μg for a, we have;
v² = u² + 2(-μg)s
v² = u² — 2μgs
We want to know the distance covered before coming to rest. Thus, v = 0m/s
So,
0² = 12² - (2 x 0.39 x 9.8 x s)
0 = 144 - 7.644s
Thus,
7.644s = 144
Thus, s = 144/7.644 = 18.84m
A block of plastic in the shape of a rectangular solid that has height 8.00 cm and area A for its top and bottom surfaces is floating in water. You place coins on the top surface of the block (at the center, so the top surface of the block remains horizontal). By measuring the height of the block above the surface of the water, you can determine the height hh below the surface. You measure hh for various values of the total mass mm of the coins that you have placed on the block. You plot h versus m and find that your data lie close to a straight line that has slope 0.0890 m/kg and y-intercept 0.0312 mm.
What is the mass of the block?
Answer:
m_{p} = 0.3506 kg
Explanation:
For this exercise we use Newton's equilibrium equation
B - Wc-Wp = 0
where B is the thrust of the water, Wc is the weight of the coins and Wp is the weight of the plastic block
B = Wc + Wp
the state push for the Archimeas equation
B = ρ_water g V
the volume of the water is the area of the block times the submerged height h, which is
h´ = 8 - h
where h is the height out of the water
ρ_water g A h´ = [tex]m_{c}[/tex] g + [tex]m_{p}[/tex] g
ρ_water A h´ = m_{c} + m_{p}
write this equation to make the graph
h´= 1 /ρ_water A (m_{c} +m_{p})
h´ = 1 /ρ_waterA (m_{c} + m_{p})
if we graph this expression, we get an equation of the line
y = m x + b
where
y = h´
m = 1 /ρ_water A
b = mp /ρ_water A
whereby
m_{p} = b ρ_water A
ρ_water = 1000 kg / m³
b = 0.0312 m
m = 0.0890 m / kg
we substitute the slope equation
b = m_{p} / m
calculate
m_{p}= 0.0312 / 0.0890
m_{p} = 0.3506 kg
A negative charge of - 0.0005 C exerts an attractive force of 19.0 N on a second charge that is 25 m away. What is the magnitude of the second charge?
Answer:
0.002638 C or 2.6388*10^-3 C
Explanation:
Given that
Quantity of the charge, q = -0.0005 C
Force on the charge of magnitude, F1 = 19 N
Distance from the second charge, r = 25 m
Magnitude of force of the second charge, q2 = ? N
F = (kq1q2) / r², where
k = 9*10^9
19 = (9*10^9 * 0.0005 * q2) / 25²
19 * 625 = 4.5*10^6 * q2
q2 = 11875 / 4.5*10^6
q2 = 0.002638 C or 2.6388*10^-3 C
Thus, the magnitude of the second charge is 0.002638 C or 2.6388*10^-3 C
Most days in the year are weekdays (as opposed to weekends). There will come a day when a huge asteroid strikes the earth. Thus, the day a huge asteroid strikes the earth will likely be a weekday. – The previous argument is Group of answer choices
Answer:
True.
The day a huge asteroid strikes the earth will likely be a weekday.
Explanation:
Mathematically, there are 5 weekdays in the week and there are only 2 weekend days.
Total number of days in a week = 7
Since, the asteroids can hit on any day,
The probability of the asteroid hitting on a weekend day = (2/7) = 0.2857
The probability that an asteroid hitting on a weekday = (5/7) = 0.7143.
So, by the virtue of the probability of the asteroid hitting on a weekday or on a weekend day, the asteroid is more likely to hit on a weekday as
P(weekday) > P(weekend)
0.7143 > 0.2857.
Hope this Helps!!!
A police officer's radar gun emits radio waves at a frequency of 10.7 GHz10.7 GHz . What is the difference in the frequencies reflected back to the radar gun by a car traveling at the 75 mph75 mph speed limit and a car travelling 9 mph9 mph faster? Assume that the cars are both headed directly away from the stationary officer.
Answer:
Explanation:
A police car's radar gun emits microwaves with frequency f. 1. =36 GHz. The beam reflects from a car that speeds away from the cruiser with 43 m/s. The receiver ...