Answer:
av=0.333m/s, U=3.3466J
b.
[tex]v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s[/tex]
Explanation:
a. let [tex]m_A[/tex] be the mass of block A, and[tex]m_B=10.0kg[/tex] be the mass of block B. The initial velocity of A,[tex]\rightarrow v_A_1=2.0m/s[/tex]
-The initial momentum =Final momentum since there's no external net forces.
[tex]pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}[/tex]
Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):
[tex]v_A_1-v_B_1=v_{B2}-v_{A2}[/tex]
-Applying the conservation of momentum. The blocks have the same velocity after collision:
[tex]v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s[/tex]
#Total Mechanical energy before and after the elastic collision is equal:
[tex]K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J[/tex]
Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s
b. Taking the end collision:
From a above, [tex]m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0[/tex]
We plug these values in the equation:
[tex]m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}[/tex]
[tex]2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s[/tex]
(a) The maximum energy stored in the spring bumpers during the collision is 3.00 J, and the velocity of both the blocks is 0.50 m/s.
(b) After they move apart, block A has a velocity of -1.00 m/s and block B has a velocity of 1.00 m/s.
You can follow these simple steps to find the required solution -
(a) Maximum Energy Stored in the Spring Bumpers
To find the maximum energy stored in the spring bumpers, we will use the conservation of momentum and energy principles.
Initially, block A (mass 2.00 kg) is moving at 2.00 m/s, and block B (mass 6.00 kg) is at rest. The total initial momentum ([tex]p_{initial[/tex]) is:
[tex]p_{\text{initial}} = m_A \cdot v_A + m_B \cdot v_B = 2.00 \, \text{kg} \cdot 2.00 \, \text{m/s} + 6.00 \, \text{kg} \cdot 0 \, \text{m/s} = 4.00 \, \text{kg} \cdot \text{m/s} \\[/tex]At the point of maximum compression, both blocks momentarily move with the same velocity ([tex]v_{common[/tex]). Using the conservation of momentum:
[tex]p_{\text{final}} = (m_A + m_B) \cdot v_{\text{common}} = 4.00 \, \text{kg} \cdot \text{m/s} \\[/tex]Solving for [tex]v_{common[/tex]:
[tex]v_{\text{common}} = \frac{4.00 \, \text{kg} \cdot \text{m/s}}{2.00 \, \text{kg} + 6.00 \, \text{kg}} = 0.50 \, \text{m/s} \\[/tex]Next, determine the initial kinetic energy ([tex]KE_{initial[/tex]):
[tex]KE_{\text{initial}} = 0.5 \cdot m_A \cdot (v_A)^2 + 0.5 \cdot m_B \cdot (v_B)^2 = 0.5 \cdot 2.00 \, \text{kg} \cdot (2.00 \, \text{m/s})^2 + 0.5 \cdot 6.00 \, \text{kg} \cdot (0 \, \text{m/s})^2 = 4.00 \, \text{J} \\[/tex]The kinetic energy at the point of maximum compression ([tex]KE_{final[/tex]) is:
[tex]KE_{\text{final}} = 0.5 \cdot (m_A + m_B) \cdot (v_{\text{common}})^2 = 0.5 \cdot 8.00 \, \text{kg} \cdot (0.50 \, \text{m/s})^2 = 1.00 \, \text{J} \\[/tex]The maximum energy stored in the spring bumpers ([tex]E_{spring[/tex]) is the difference between [tex]KE_{initial[/tex] and [tex]KE_{final[/tex]:
[tex]E_{\text{spring}} = KE_{\text{initial}} - KE_{\text{final}} = 4.00 \, \text{J} - 1.00 \, \text{J} = 3.00 \, \text{J} \\[/tex](b) Velocity of Each Block After Collision
After they have moved apart, assuming an elastic collision where both kinetic energy and momentum are conserved, the final velocities ([tex]v_A_{final[/tex] and [tex]v_B_{final[/tex]) can be found using the respective equations:
For block A:
[tex]v_{A_{\text{final}}} = \frac{(m_A - m_B)}{(m_A + m_B)} \cdot v_{A_{\text{initial}}} + \frac{(2 \cdot m_B)}{(m_A + m_B)} \cdot v_{B_{\text{initial}}} \\[/tex][tex]v_{A_{\text{final}}} = \frac{(2.00 \, \text{kg} - 6.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 2.00 \, \text{m/s} + \frac{(2 \cdot 6.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 0 \, \text{m/s} = -1.00 \, \text{m/s} \\[/tex]For block B:
[tex]v_{B_{\text{final}}} = \frac{(2 \cdot m_A)}{(m_A + m_B)} \cdot v_{A_{\text{initial}}} + \frac{(m_B - m_A)}{(m_A + m_B)} \cdot v_{B_{\text{initial}}} \\[/tex][tex]v_{B_{\text{final}}} = \frac{(2 \cdot 2.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 2.00 \, \text{m/s} + \frac{(6.00 \, \text{kg} - 2.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 0 \, \text{m/s} = 1.00 \, \text{m/s} \\[/tex]Therefore, the final velocities are:
[tex]v_{A_{\text{final}}} = -1.00 \, \text{m/s} \quad \text{and} \quad v_{B_{\text{final}}} = 1.00 \, \text{m/s}[/tex]Two stones are launched from the top of a tall building. One stone is thrown in a direction 15.0 ∘∘ above the horizontal with a speed of 20.0 m/sm/s ; the other is thrown in a direction 15.0 ∘∘ below the horizontal with the same speed.
Which stone spends more time in the air? (Neglet air resistance)
a. The stone thrown upward spends more time in the air.
b. The stone thrown downward spends more time in the air.
c. Both stones spend the same amount of time in the air.
Answer:a
Explanation:
Given
First stone is thrown [tex]15^{\circ}[/tex] above the horizontal with some speed let say u
Second stone is thrown at [tex]15^{\circ}[/tex] below the horizontal with speed u
For a height h of building
For first stone (motion in vertical direction)
using
[tex]v^2-u^2=2ah [/tex]
where v=final velocity
u=initial velocity
a=acceleration
h=displacement
[tex]h=u\sin 15(t_1)-\frac{1}{2}gt_1^2---1[/tex]
For second stone
[tex]h=(-u\sin 15)(t_2)-\frac{1}{2}gt_2^2----2[/tex]
Equating 1 and 2
[tex]u\sin 15(t_1+t_2)-\frac{1}{2}g(t_1-t_2)(t_1+t_2)=0[/tex]
[tex](t_1+t_2)[u\sin 15-4.9(t_1-t_2)]=0[/tex]
as [tex]t_1+t_2[/tex] cannot be zero
so [tex]t_1-t_2=1.05\ s[/tex]
[tex]t_1=t_2+1.056[/tex]
therefore time taken by first stone(thrown upward) will be more.
Answer:
a. The stone thrown upward spends more time in the air.
Explanation:
Given:
projection of first stone, [tex]\theta_1=15^{\circ}[/tex] above the horizontal
initial velocity of projectiles, [tex]u_1=u_2=20\ m.s^{-1}[/tex]
projection of second stone,[tex]\theta_2=15^{\circ}[/tex] below the horizontal
The stone thrown upward will spend more time in the air because it travels more distance than the one thrown downwards.
The stone thrown upwards faces deceleration due to the gravity because it goes opposite to the gravity initially, then reaches a velocity zero for a moment and then falls freely from a greater height.
While the second stone posses an initial velocity downward in the direction of the gravity and which further increases its velocity and it travels a short distance.
KATZPSEF1 30.P.041. My Notes Ask Your Teacher A charged particle enters a region of space with a uniform magnetic field B = 1.98 i T. At a particular instant in time, it has velocity v = (1.46 ✕ 106 i + 2.42 ✕ 106 j) m/s. Based on the observed acceleration, you determine that the force acting on the particle at this instant is F = 1.60 k N. What are the following?(a) the sign of the charged particle
(b) the magnitude of the charged particle
Answer: The charge on the particle is positive
While the magnitude = 0.00028C
Explanation:
Please find the attached file for the solution
a straight, 2.5-m wire carries a typical household current of 1.5 a (in one direction) at location where the earth's magnetic field i 0.55 gauss from south to north at a location where the earth's magnetic field is 0.55 gauss from south to north. find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) is the magnetic force ever large enough to cause significant effects under normal household current of 1.5 A
Answer:
a) When the current is from west to east and the magnetic field is from south to north the magnitude of the force is 2.1x10⁻⁴N and the direction is upwards.
b) The current is moving vertically upward, the magnitude of the force is 2.1x10⁻⁴N and the direction is west.
c) The force is 0 because the magnetic field and the direction of the current are in parallel.
d) No, the force is less.
Explanation:
Given:
L = length of the wire = 2.5 m
i = current in wire = 1.5 A
B = magnetic field = 0.55x10⁻⁴T
a) The magnitude of magnetic force is equal to:
[tex]F=BiL=0.55x10^{-5} *1.5*2.5=2.1x10^{-4} N[/tex]
b) The same way to a):
F = 2.1x10⁻⁴N
c) F = 0
The magnetic field and the direction of the current are in parallel.
d) The answer is no, the force is less
A 1200 kg car reaches the top of a 100 m high hill at A with a speed vA. What is the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0? Neglect friction.
Answer:
the value of vA that will allow the car to coast in neutral so as to just reach the top of the 150 m high hill at B with vB = 0 is 31.3 m/s
Explanation:
given information
car's mass, m = 1200 kg
[tex]h_{A}[/tex] = 100 m
[tex]v_{A}[/tex] = [tex]v_{A}[/tex]
[tex]h_{B}[/tex] = 150 m
[tex]v_{B}[/tex] = 0
according to conservative energy
the distance from point A to B, h = 150 m - 100 m = 50 m
the initial speed [tex]v_{A}[/tex]
final speed [tex]v_{B}[/tex] = 0
thus,
[tex]v_{B}[/tex]² = [tex]v_{A}[/tex]² - 2 g h
0 = [tex]v_{A}[/tex]² - 2 g h
[tex]v_{A}[/tex]² = 2 g h
[tex]v_{A}[/tex] = √2 g h
= √2 (9.8) (50)
= 31.3 m/s
A ball of mass I .5 kg falls vertically downward. Just before striking the floor, its speed is 14 m/s. Just after rebounding upward, its speed is 10 m/s.
If this change of velocity took place in 0.20 seconds, what is the average force of the ball on the floor?
Answer:
180 N
Explanation:
We know that acceleration is the rate of change of speed per unit time hence
[tex]a=\frac {v_f-v_i}{t}[/tex] where v and t are velocity and time respectively, f and i represent final and initial.
Also, from Newton's law of motion, F=ma and replacing a with the above then
[tex]F=m\frac {v_f-v_i}{t}[/tex]
Substituting 1.5 Kg for mass, m -14 m/s for i and 10 m/s for for v then
[tex]F=1.5\times \frac {10--14}{0.2}=180 N[/tex]
Therefore, the force is 180 N
The average force exerted by a 1.5 kg ball on the floor, falling with a speed of 14 m/s and rebounding with a speed of 10 m/s over 0.20 seconds, is 180 N.
Explanation:The subject of this question is Physics and from the concept of impulse and momentum. The change in momentum equals the product of force and the time over which the force is applied. So, we can calculate the force using this formula: Force = Change in momentum / Time.
In this case, the ball's momentum changes by the difference in velocity multiplied by the mass of the ball, which is (14 m/s + 10 m/s) * 1.5 kg. The reason the velocities are added is because the direction of velocity changes, making the speed of ball before and after striking the floor of equal magnitude but opposite in direction. So, the change in momentum becomes 36 kg*m/s. Given that the time is 0.20 seconds, the force would be 36 kg*m/s divided by 0.20 s, or 180 N.
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a solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls without slipping a distance L = 6.0 m down a roof that is inclined at the angle theta = 30degree.
(a) What is the angular speed of the cylinder about its center as it leaves the roof?
(b) The roofs edge is at height H = 5.0 m. How far horizontally from the roof's edge does the cylinder hit the level ground?
Answer:
Explanation:
Acceleration of cylinder
a = g sin 30 / 1+ k² / r² where k is radius of gyration and r is radius of cylinder.
For cylinder k² = (1 / 2) r²
acceleration
= gsin30 / 1.5
= g / 3
= 3.27
v² = u² + 2as
= 2 x 3.27 x 6
v = 6.26 m /s
v = angular velocity x radius
6.26 = angular velocity x .10
angular velocity = 62.6 rad / s
b ) vertical component of velocity
= 6.26 sin 30
= 3.13 m /s
h = ut + 1/2 g t²
5 = 3.13 t + .5 t²
.5 t²+ 3.13 t- 5 = 0
t = 1.32 s
horizontal distance covered
= 6.26 cos 30 x 1.32
= 7.15 m
The conservation of energy and kinematics allows to find the results for the questions about the movement of the cylinder on the ceiling and when falling are:
a) The angular velocity is w = 6.26 rad / s
b) the distance to the ground is: x = 7,476 m
Given parameters
Cylinder radius r = 10 cm = 0.10 m Mass m = 12 kg Distance L = 6.0 m Roof angle θ = 30º Ceiling height H = 5.0 mTo find
a) The angular velocity.
b) Horizontal distance.
Mechanical energy is the sum of kinetic energy and potential energies. If there is no friction, it remains constant at all points.
Linear and rotational kinematics study the motion of bodies with linear and rotational motions.
a) Let's write the mechanical energy at the points of interest.
Starting point. When it comes out of the top
Em₀ = U = m g h
Final point. On the edge of the roof.
[tex]Em_f[/tex] = K = ½ mv² + ½ I w²
Since the cylinder does not slide, friction is zero and energy is conserved.
Em₀ = [tex]Em_f[/tex]
mg h = ½ m v² + ½ I w²
The moment of inertia of the cylinder is;
I = ½ m r²
Linear and angular variables are related.
v = w r
let's substitute.
m g h = ½ m (wr) ² + ½ (½ m r²) w²
gh = ½ w² r² (1 + ½) = ½ w² r² [tex]\frac{3}{2}[/tex]
w² = [tex]\frac{4}{3 } \ \frac{gh}{r^2}[/tex]
Let's use trigonometry to find the height of the ceiling.
sin θ = h / L
h = L sin θ
We substitute.
w= [tex]\sqrt{ \frac{4}{3} \ \frac{g \ L sin \theta }{r^2} }[/tex]
Let's calculate.
w = [tex]\sqrt{\frac{4}{3} \frac{9.8 \ 6.0 sin 30}{0.10^2} }[/tex]
Let's calculate
w = Ra 4/3 9.8 6.0 sin 30 / 0.10²
w = 62.6 rad / s
b) For this part we can use the projectile launch expressions.
Let's find the time it takes to get to the floor.
y = y₀ + go t - ½ g t²
The initial height is y₀ = H, when it reaches the ground its height is y = 0 and let's use trigonometry for the vertical initial velocity.
sin 30 = [tex]\frac{v_{oy}}{v_o}[/tex]I go / v
[tex]v_{oy}[/tex] = v sin 30 = wr sin 30
[tex]v_{oy}[/tex] = 62.6 0.1 sin 30
[tex]v_{oy}[/tex] = 3.13 m / s
0 = H + voy t - ½ g t²
0 = 5 + 3.13 t - ½ 9.8 t³
t² - 0.6387 t - 1.02 = 0
We solve the quadratic equation.
t =[tex]\frac{0.6387 \pm \sqrt{0.6387^2 - 4 \ 1.02} }{2}[/tex]
t = [tex]\frac{0.6378 \pm 2.118}{2}[/tex]
t₁ = 1.379 s
t₂ = -0, 7 s
The time o must be a positive quantity, therefore the correct answer is:
t = 1.379 s
We look for the horizontal distance.
x = v₀ₓ t
vₓ = v cos θ
v = wr
Let's substitute.
x = wr cos t
Let's calculate.
x = 62.6 0.1 cos 30 1.379
x = 7.476 m
In conclusion using the conservation of energy and kinematics we can find the results for the questions about the movement of the cylinder on the ceiling and when falling are:
a) The angular velocity is w = 6.26 rad / s
b) the distance to the ground is: x = 7,476 m
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Using the first definition of coefficient of elasticity given in the lab (based on velocity), if a ball strikes a surface with a speed of 10 m/s and rebounds just after the collsion with a speed of 3 m/s, what is the coefficient of elasticity
Final answer:
The coefficient of elasticity is a measure of the elasticity of a collision, defined as the ratio of speeds after and before the collision. In this case, the coefficient of elasticity is 0.3.
Explanation:
The coefficient of elasticity, also known as the coefficient of restitution (c), is a measure of the elasticity of a collision between a ball and an object. It is defined as the ratio of the speeds after and before the collision.
In this case, the ball strikes a surface with a speed of 10 m/s and rebounds with a speed of 3 m/s. To find the coefficient of elasticity, we can use the formula c = (v_final / v_initial), where v_final is the final velocity and v_initial is the initial velocity.
Therefore, the coefficient of elasticity in this case would be c = (3 m/s / 10 m/s) = 0.3.
A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 562 Hz tone, what is the wavelength of that tone in air at standard conditions?
To solve this problem we will apply the concept of wavelength, which warns that this is equivalent to the relationship between the speed of the air (in this case in through the air) and the frequency of that wave. The air is in standard conditions so we have the relation,
Frequency [tex]= f = 562Hz[/tex]
Speed of sound in air [tex]= v = 331m/s[/tex]
The definition of wavelength is,
[tex]\lambda = \frac{v}{f}[/tex]
Here,
v = Velocity
f = Frequency
Replacing,
[tex]\lambda = \frac{331m/s}{562Hz}[/tex]
[tex]\lambda = 0.589m[/tex]
Therefore the wavelength of that tone in air at standard conditions is 0.589m
The wavelength of the tone in air 0.59 Hz
The trombone can produce pitches wavelength ranging from 85 Hz to 660 Hz
The trombone produces a tone of 562 Hz
The tone of air is at standard conditions, hence the velocity of the sound in air is 331 m/s
velocity= frequency/wavelength
331= 562/wavelength
wavelength= 331/562
= 0.59 HZ
Hence the wavelength of the tone is 0.59 Hz
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he deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1000 ohm resistor in series with the deflection plates. How long do
Answer:
Incomplete question
This is the complete question
The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 V potential difference is suddenly applied to the initially uncharged plates through a 1000 Ω resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 95 V?
Explanation:
Given that,
The dimension of 10cm by 2cm
0.1m by 0.02m
Then, the area is Lenght × breadth
Area=0.1×0.02=0.002m²
The distance between the plate is d=1mm=0.001m
Then,
The capacitance of a capacitor is given as
C=εoA/d
Where
εo is constant and has a value of
εo= 8.854 × 10−12 C²/Nm²
C= 8.854E-12×0.002/0.001
C=17.7×10^-12
C=17.7 pF
Value of resistor resistance is 1000ohms
Voltage applied is V = 100V
This Is a series resistor and capacitor (RC ) circuit
In an RC circuit, voltage is given as
Charging system
V=Vo[1 - exp(-t/RC)]
At, t=0, V=100V
Therefore, Vo=100V
We want to know the time, the voltage will deflect 95V.
Then applying our parameters
V=Vo[1 - exp(-t/RC)]
95=100[1-exp(-t/1000×17.7×10^-12)]
95/100=1-exp(-t/17.7×10^-9)
0.95=1-exp(-t/17.7×10^-9)
0.95 - 1 = -exp(-t/17.7×10^-9)
-0.05=-exp(-t/17.7×10^-9)
Divide both side by -1
0.05=exp(-t/17.7×10^-9)
Take In of both sides
In(0.05)=-t/17.7×10^-9
-2.996=-t/17.7×10^-9
-2.996×17.7×10^-9=-t
-t=-53.02×10^-9
Divide both side by -1
t= 53.02×10^-9s
t=53.02 ns
The time to deflect 95V is 53.02nanoseconds
Calculate the range of wavelengths for the frequencies found on the FM band. Take a look at the "whip" radio antennas on cars, comment on the size of the wavelength and the antenna. {2.8 – 3.4 m}
Answer:
2.8m-3.4m
Explanation:
Radio waves are electromagnetic waves and they all travel with a speed of
[tex]3*10^8m/s[/tex] in air.
The range of frequencies in the FM band is from 88MHz to 108MHz.
Generally, the relationship between velocity, frequency and wavelength for electromagnetic waves is given by equation (1);
[tex]v=\lambda f..............(1)[/tex]
From equation (1), we can write,
[tex]\lambda=\frac{v}{f}...............(2)[/tex]
For the upper frequency of 108MHz, the wavelength is given by;
[tex]\lambda_1=\frac{3*10^8}{108*10^6}\\\lambda_1=0.02778*10^2\\\lambda_1=2.78m[/tex]
Similarly, for the lower frequency of 88MHz, the wavelength is given by;
[tex]\lambda_2=\frac{3*10^8}{88*10^6}\\\lambda_2=0.0341*10^2\\\lambda_2=3.41m[/tex]
The range of wavelength therefore is [tex]\lambda_1-\lambda_2=2.8m-3.4m[/tex] approximately.
Please note the following:
[tex]108MHz=108*10^6Hz\\88MHz=88*10^6Hz[/tex]
A rectangular coil 20 cm by 33 cm has 110 turns. This coil produces a maximum emf of 72 V when it rotates with an angular speed of 200 rad/s in a magnetic field of strength B.Find the value of B.
Answer:
0.05T
Explanation:
Data given,
area, A=20cm*33cm=0.2m*0.33m=0.066m^2
Number of turns, N=110 turns,
Emf= 72v,
angular speed, W= 200rad/s
magnetic field strength, B= ??
from the expression showing the relationship between induced emf and magnetic field is shown below
[tex]E=NBAW[/tex]
Where N is the number of turns,
E=is the emf,
Bis the magnetic field strength
if we substitute values, we arrive at
[tex]E=NABW\\72=110*0.066*B*200\\B=\frac{72}{1452}\\ B=0.05T[/tex]
Explanation:
Below is an attachment containing the solution.
The insulating solid sphere in the previous Example 24.5 has the same total charge Q and radius a as the thin shell in the example here. Consider the electric fields due to the sphere (E_sphere), the shell (E_shell), and a point charge Q (E_Q). Rank the strength of the electric fields due to these three charged objects at the same three points ('1' represents the object with the strongest, '2' for the next in strength and so on. For example, you can have the triad {3, 1, 2} as your answer, or {2, 1, 2} if the shell is strongest while the sphere and the point charge have equal strengths.): r greaterthan a a. {1, 1, 2} b. {2, 1, 1} c. {1, 1, 1} d. {2, 2, 1} e. {1, 2, 3} r lesserthan a a. {2,2, 1} b. {1,1,1} c. {2, 3, 1} d. {2,1,1} e.{1, 2, 1}
Answer and Explanation:
Using Gauss's law,
If r>a
then charge enclosed in all the three cases is same as Q.
So Electric field for all three is same.
So {1,1,1}.
(b) r<a,
Charge enclosed in case of shell is zero since all charge is present on the surface. So E = 0.
Charge enclosed by incase of point charge is Q.
Charge enclosed in case of sphere is Qr3/a3 which is less than Q.
So ranking {2,3,1}
a 1-kg discus is thrown with a velocity of 19 m/s at an angle of 35 degrees from the vertical direction. calculate the vertical and horizontal velocity components.
Answer:
Vx = 10.9 m/s , Vy = 15.6 m/s
Explanation:
Given velocity V= 19 m/s
the angle 35 ° is taken from Y-axis so the angle with x-axis will be 90°-35° = 55°
θ = 55°
to Find Vx = ? and Vy= ?
Vx = V cos θ
Vx = 19 m/s × cos 55°
Vx = 10.9 m/s
Vx = V sin θ
Vy = 19 m/s × sin 55°
Vy = 15.6 m/s
15.56m/s and 10.90m/s respectively
Explanation:The vertical and horizontal components of a given vector, say A, are given by
[tex]A_{Y}[/tex] = A sin θ ----------------(i)
[tex]A_{X}[/tex] = A cos θ ----------------(ii)
Where;
[tex]A_{Y}[/tex] is the vertical component of the vector A
[tex]A_{X}[/tex] is the horizontal component of the vector A
A is the magnitude of the vector A
θ is the angle the vector makes with the positive x-axis (horizontal direction).
Now, from the question;
The vector is the velocity of the 1-kg discus. Lets call it vector V
The magnitude of the velocity vector V = V = 19m/s
The angle that the vector makes with the positive x-axis = θ
To calculate θ;
Notice that the velocity vector makes an angle of 35° from the vertical direction rather than the horizontal direction.
Therefore, to get the horizontal direction of the velocity vector, we subtract 35° from 90° as follows;
θ = 90° - 35° = 55°
Now, the vertical and horizontal components of the velocity vector, V, are given by
[tex]V_{Y}[/tex] = V sin θ --------------------(iii)
[tex]V_{X}[/tex] = V cos θ ------------------------(iv)
Substitute all the necessary values into equations(iii) and (iv) as follows;
[tex]V_{Y}[/tex] = 19 sin 55° = 19m/s x 0.8192 = 15.56m/s
[tex]V_{X}[/tex] = 19 cos 55° = 19m/s x 0.5736 = 10.90m/s
Therefore, the vertical and horizontal velocity components are respectively 15.56m/s and 10.90m/s.
A proton moves through a magnetic field at 26.7 % 26.7% of the speed of light. At a location where the field has a magnitude of 0.00687 T 0.00687 T and the proton's velocity makes an angle of 101 ∘ 101∘ with the field, what is the magnitude of the magnetic force acting on the proton?
Answer:
[tex]8.64283\times 10^{-14}\ N[/tex]
Explanation:
q = Charge of proton = [tex]1.6\times 10^{-19}\ C[/tex]
v = Velocity of proton = [tex]0.267\times c[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
B = Magnetic field = 0.00687 T
[tex]\theta[/tex] = Angle = [tex]101^{\circ}[/tex]
Magnetic force is given by
[tex]F=qvBsin\theta\\\Rightarrow F=1.6\times 10^{-19}\times (0.267\times 3\times 10^8)\times 0.00687\times sin101\\\Rightarrow F=8.64283\times 10^{-14}\ N[/tex]
The magnetic force acting on the proton is [tex]8.64283\times 10^{-14}\ N[/tex]
A constant voltage of 8.00 V has been observed over a certain time interval across a 2.30 H inductor. The current through the inductor, measured as 1.00 A at the beginning of the time interval, was observed to increase at a constant rate to a value of 8.00 A at the end of the time interval. How long was this time interval
Answer:2.01 s
Explanation:
Given
Applied voltage [tex]V=8\ V[/tex]
Inductance [tex]L=2.3\ H[/tex]
Change in Current [tex]\Delta i=8-1=7\ A[/tex]
Induced EMF is given by
[tex]V=L\times \dfrac{\Delta A}{\Delta t}[/tex]
[tex]8=2.3\times \dfrac{7}{\Delta t}[/tex]
[tex]\Delta t=\dfrac{2.3\times 7}{8}[/tex]
[tex]\Delta t=2.0125\ s[/tex]
1.!(1)!A!hiker!determines!the!length!of!a!lake!by!listening for!the!echo!of!her!shout!reflected!by!a! cliff!at!the!far!end of!the!lake.!She!hears!the!echo!2.0!s!after!shouting.!Estimate the!length!of!the! lake.
Answer:
The length of the lake is 340 meters.
Explanation:
It is given that, a hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake. She hears the echo 2 s after shouting. We need to find the length of the lake.
The distance covered by the person in 2 s is :
[tex]d=vt[/tex]
v is the speed of sound
[tex]d=340\ m/s\times 2\ s[/tex]
[tex]d=680\ m[/tex]
The length of the lake is given by :
[tex]l=\dfrac{d}{2}[/tex]
[tex]l=\dfrac{680\ m}{2}[/tex]
l = 340 meters
So, the length of the lake is 340 meters. Hence, this is the required solution.
A 0.453 kg pendulum bob passes through the lowest part of its path at a speed of 2.58 m/s. What is the tension in the pendulum cable at this point if the pendulum is 75.1 cm long? Submit Answer Tries 0/12 When the pendulum reaches its highest point, what angle does the cable make with the vertical? Submit Answer Tries 0/12 What is the tension in the pendulum cable when the pendulum reaches its highest point?
Answer with Explanation:
Mass of pendulum bob, m=0.453 kg
Speed, [tex]v_1=[/tex]2.58 m/s
a.r=75.1 cm=[tex]75.1\times 10^{-2}m[/tex]=0.751 m
[tex] 1cm=10^{-2} m[/tex]
Tension in the pendulum cable is given by
Tension=Centripetal force+force due to gravity
[tex]T=\frac{mv^2}{r}+mg[/tex]
Where [tex]g=9.8 m/s^2[/tex]
Substitute the values
[tex]T=\frac{0.453(2.58)^2}{75.1\times 10^{-2}}+0.453\times 9.8[/tex]
[tex]T=8.45 N[/tex]
b.When the pendulum reaches its highest point,then
Final velocity, [tex]v_2=0[/tex]
According to law of conservation of energy
[tex]mgh_1+\frac{1}{2}mv^2_1=mgh_2+\frac{1}{2}mv^2_2[/tex]
[tex]gh_1+\frac{1}{2}v^2_1=gh_2+\frac{1}{2}v^2_2[/tex]
[tex]h_1=0[/tex]
Substitute the values
[tex]9.8\times 0+\frac{1}{2}(2.58)^2=9.8\times h_2+\frac{1}{2}(0)^2[/tex]
[tex]3.3282=9.8h_2[/tex]
[tex]h_2=\frac{3.3282}{9.8}=0.34 m[/tex]
The angle mad by cable with the vertical=[tex]cos\theta=\frac{0.751-0.34}{0.751}=0.55[/tex]
[tex]\theta=cos^{-1}(0.55)=56.6^{\circ}[/tex]
c.When the pendulum reaches at highest point then
Acceleration, a=0
Therefore, the tension in the pendulum cable
[tex]T=mgcos\theta[/tex]
Substitute the values
[tex]T=0.453\times 9.8cos56.6[/tex]
[tex]T=2.4 N[/tex]
A block with mass m is pulled horizontally with a force F_pull leading to an acceleration a along a rough, flat surface.
Find the coefficient of kinetic friction between the block and the surface.
Answer:
[tex]\mu_k=\frac{a}{g}[/tex]
Explanation:
The force of kinetic friction on the block is defined as:
[tex]F_k=\mu_kN[/tex]
Where [tex]\mu_k[/tex] is the coefficient of kinetic friction between the block and the surface and N is the normal force, which is always perpendicular to the surface that the object contacts. So, according to the free body diagram of the block, we have:
[tex]N=mg\\F_k=F=ma[/tex]
Replacing this in the first equation and solving for [tex]\mu_k[/tex]:
[tex]ma=\mu_k(mg)\\\mu_k=\frac{a}{g}[/tex]
The emission spectrum of an incandescent light bulb is continuous rather than discrete. This is true even through the filament is made of tungsten, an element. Why do you think this is
Answer:
This is the property of metals like W (Tungsten) to produce a continuous spectrum.
Explanation:
The molecules of a gas are highly differentiated and single atoms absorb photons instead of bulk like metals and only give line spectrum for example line spectrum given by hydrogen etc. On the other hand, metals like Tungsten gets very hot and emission of every possible wavelength of light is observed. This emission is continuous because the wavelengths of light in the spectrum has no breakages.
g A current loop, carrying a current of 5.6 A, is in the shape of a right triangle with sides 30, 40, and 50 cm. The loop is in a uniform magnetic field of magnitude 62 mT whose direction is parallel to the current in the 50 cm side of the loop. Find the magnitude of (a) the magnetic dipole moment of the loop in amperes-square meters and (b) the torque on the loop.
Answer:
(a) 0.336 A m²
(b) 0 Nm
Explanation:
(a) Magnetic dipole moment, μ, is given by
[tex]\mu = IA[/tex]
I is the current in the loop and A is the area of the loop.
The loop is a triangle. To find its area, observe that the dimensions form a Pythagorean triple, making it a right-angled triangle with base and height of 30 cm and 40 cm.
[tex]A = \frac{1}{2}\times(0.3\text{ m})\times(0.4\text{ m})=0.06\text{ m}^2[/tex]
[tex]\mu = (5.6\text{ A})(0.06\text{ m}^2) = 0.336\text{ A}\,\text{m}^2[/tex]
(b) Torque is given by
[tex]\tau = \mu B\sin\theta[/tex]
where B is the magnetic field and [tex]\theta[/tex] is the angle between the loop and the magnetic field. Since the field is parallel, [tex]\theta[/tex] is 0.
[tex]\tau = \mu B\sin0 = 0\text{ Nm}[/tex]
Delicate measurements indicate that the Earth has an electric field surrounding it, similar to that around a positively charged sphere. Its magnitude at the surface of the Earth is about 100 N/C. What charge would an oil drop of mass 2.0 x 10 15 kg have to have, in order to remain suspended by the Earth’s electric field? Give your answer in Coulombs ?
Answer:
q = 1.96 10⁴ C
Explanation:
The elective force is given by
[tex]F_{e}[/tex] = q E
Where E is the electric field and q the charge.
Let's use Newton's law of equilibrium for the case of the suspended drop
F_{e} –W = 0
F_{e} = W
q E = m g
q = m g / E
Let's calculate
q = 2.0 10⁵ 9.8 / 100
q = 1.96 10⁴ C
To maintain suspension in the Earth’s electric field, an oil drop with a mass of 2.0 × 10^-15 kg requires a charge of 1.96 × 10^-16 Coulombs, calculated by equating the electric force with the gravitational force.
Explanation:To determine the charge needed for an oil drop of mass 2.0 × 10-15 kg to remain suspended by the Earth’s electric field of 100 N/C, we can apply the equilibrium condition between the electric force and the gravitational force. The electric force (Felectric) is equal to the charge (q) multiplied by the electric field (E), so Felectric = q × E. The gravitational force (Fgravity) is the mass (m) multiplied by the acceleration due to gravity (g), which is approximately 9.8 m/s2.
For the oil drop to remain suspended, these two forces must be equal: q × E = m × g, which gives us q = (m × g) / E. Using the provided values, the charge q is calculated as follows:
q = (2.0 × 10-15 kg × 9.8 m/s2) / 100 N/C
q = (2.0 × 10-15 × 9.8) / 100
q = 1.96 × 10-16 C
Therefore, the oil drop must have a charge of 1.96 × 10-16 Coulombs to remain suspended in the Earth’s electric field.
A piston raises a weight, then lowers it again to its original height. If the process the system follows as the piston rises is isothermal, and as it falls is isobaric, then is the work done by the gas positive, negative, or zero? Explain
Answer:
Negative
Explanation:
First law of thermodynamic also known as the law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.
The first law relates relates changes in internal energy to heat added to a system and the work done by a system by the conservation of energy.
The first law is mathematically given as ΔU = [tex]U_{F}[/tex] - [tex]U_{0}[/tex] = Q - W
Where Q = Quantity of heat
W = Work done
From the first law The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system.
Analyzing the pistol when it raises in isothermal and when it falls in isobaric state.The following can be said:
In the Isothermal compression of a gas there is work done on the system to decrease the volume and increase the pressure. For work to be done on the system it is a negative work done then.
In the Isobaric State An isobaric process occurs at constant pressure. Since the pressure is constant, the force exerted is constant and the work done is given as PΔV.If a gas is to expand at a constant pressure, heat should be transferred into the system at a certain rate.Isobaric is a fuction of heat which is Isothermal Provided the pressure is kept constant.
In Isobaric definition above it can be seen that " Heat should be transferred into the system ata certain rate. For heat to be transferred into the system work is deinitely been done on the system thereby favouring the negative work done.
You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought gravity was just a lot weaker up there. Convince them and yourself that it isn't so by calculating how much weaker gravity is 400 km above the Earth's surface.
Answer:
It's only 1.11 m/s2 weaker at 400 km above surface of Earth
Explanation:
Let Earth radius be 6371 km, or 6371000 m. At 400km above the Earth surface would be 6371 + 400 = 6771 km, or 6771000 m
We can use Newton's gravitational law to calculate difference in gravitational acceleration between point A (Earth surface) and point B (400km above Earth surface):
[tex]g = G\frac{M}{r^2}[/tex]
where G is the gravitational constant, M is the mass of Earth and r is the distance form the center of Earth to the object
[tex]\frac{g_B}{g_A} = \frac{GM/r^2_B}{GM/r^2_A}[/tex]
[tex]\frac{g_B}{g_A} = \left(\frac{r_A}{r_B}\right)^2 [/tex]
[tex]\frac{g_B}{g_A} = \left(\frac{6371000}{6771000}\right)^2 [/tex]
[tex]\frac{g_B}{g_A} = 0.94^2 = 0.885[/tex]
[tex]g_B = 0.885 g_A[/tex]
So the gravitational acceleration at 400km above surface is only 0.885 the gravitational energy at the surface, or 0.885*9.81 = 8.7 m/s2, a difference of (9.81 - 8.7) = 1.11 m/s2.
A small toy car draws a 0.50-mA current from a 3.0-V NiCd battery. In 10 min of operation, (a) how much charge flows through the toy car, and (b) how much energy is lost by the battery? 4. (Resistance and Ohm’s law, Prob. 17.16, 1.0 point) How
Answer:
(a) 0.3 C
(b) 0.9 J
Explanation:
(a)
Given:
Current drawn (I) = 0.50 mA = 0.50 × 10⁻³ A
Terminal voltage (V) = 3.0 V
Time of operation (t) = 10 min = 10 × 60 = 600 s
Charge flowing through the toy car 'Q' is given as:
[tex]Q=It[/tex]
Plug in the given values and solve for 'Q'. This gives,
[tex]Q=(0.50\times 10^{-3}\ A)(600\ s)\\\\Q=0.3\ C[/tex]
Therefore, 0.3 C charge flows the toy car.
(b)
Energy lost by the battery is equal to the product of power consumed by the battery and time of operation.
Power consumed by the battery is given as:
[tex]P=VI[/tex]
Plug in the given values and solve for 'P'. This gives,
[tex]P=(3.0\ V)(0.50\times 10^{-3}\ A)\\\\P=1.5\times 10^{-3}\ W[/tex]
Therefore, the energy lost by the battery is given as:
[tex]E=P\times t\\\\E=1.5\times 10^{-3}\ W\times 600\ s\\\\E=0.9\ J[/tex]
Therefore, the energy lost by the battery is 0.9 J
A polarized Light of intensity I0 is incident on an analyzer. What should the angle between the axis of polarization of the light and the transmission axis of the analyzer be to allow 44% of the total intensity to be transmitted?
Answer:
You could use Malus's Law. Malus's Law tells us that if you have a polarized wave (of intensity I 0 0 ) passing through a polarizer the emerging intensity ( Y OR ) will be proportional to the square cosine of the angle between the polarization direction of the incoming wave and the axis of the polarizer.
Explanation:
OR: I = I 00 ⋅ cos two ( e )
Daisy walks across a force platform, and forces exerted by her foot during a step are recorded. The peak velocity reaction force is 1200 (this force acts upward on Daisy). At the same instant, the frictional force is 200N(this force acts forward on Daisy).(a) how large is the resultant of these forces(b) What is the direction of the resultant forces?
Answer:
(a). The resultant of these forces is 1216.55 N.
(b). The direction of the resultant forces is 80.53°.
Explanation:
Given that,
First force = 1200 N
Second force = 200 N
(a). We need to calculate the resultant of these forces
Using cosine law
[tex]F=\sqrt{F_{1}^2+F_{2}^2+2F_{1}F_{2}\cos\theta}[/tex]
Put the value into the formula
[tex]F=\sqrt{1200^2+200^2+2\times1200\times200\cos90}[/tex]
[tex]F=\sqrt{1200^2+200^2}[/tex]
[tex]F= 1216.55\ N[/tex]
The resultant of these forces is 1216.55 N.
(b). We need to calculate the direction of the resultant forces
Using formula of direction
[tex]\tan\alpha=\dfrac{F_{1}}{F_{2}}[/tex]
Put the value into the formula
[tex]\alpha=\tan^{-1}(\dfrac{1200}{200})[/tex]
[tex]\alpha=80.53^{\circ}[/tex]
Hence, (a). The resultant of these forces is 1216.55 N.
(b). The direction of the resultant forces is 80.53°.
Answer:
a) [tex]F_r=1216.55\ N[/tex]
b) [tex]\theta=80.54^{\circ}[/tex]
Explanation:
Given:
force acting upward on the, [tex]F_y=1200\ N[/tex]force acting forward on daisy, [tex]F_x=200\ N[/tex]a)
Now the resultant of these forces:
Since the forces are mutually perpendicular,
[tex]F_r=\sqrt{F_x^2+F_y^2}[/tex]
[tex]F_r=\sqrt{200^2+1200^2}[/tex]
[tex]F_r=1216.55\ N[/tex]
b)
The direction of this force from the positive x-direction:
[tex]\tan\theta=\frac{F_y}{F_x}[/tex]
[tex]\tan\theta=\frac{1200}{200}[/tex]
[tex]\theta=80.54^{\circ}[/tex]
In another experiment, the intensity of the incident light and the temperature of the metal are held constant. Assuming that the initial light incident on the metal surface causes electrons to be ejected from the metal, what happens if the frequency of the incident light is increased?
Answer:
Explanation:
Answer:
Explanation:
The intensity and the temperature of metal is constant so the number of photo electrons remains constant. As the number of photo electrons remains same so the photo electric remains constant.
As the frequency is increased, the kinetic energy of the photo electrons increases and thus, the speed of photo electrons increases.
When the frequency of the incident light is increased above a certain threshold, electrons will start being ejected from the metal's surface, with their maximum velocity increasing linearly with the frequency.
As the frequency of the incident light is increased beyond a certain threshold frequency, electrons will begin to be ejected from the surface of the metal. However, their kinetic energy will not increase proportionally with the frequency, rather the maximum velocity of the ejected electrons will increase linearly with the frequency of the incident light.
An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.60 A in the -c-direction. At an instant when the electron is at point (0, 0.200 m, 0) and the electron's velocity is v(5.00 What is the force that the wire exerts on the electron? Enter the z, y, and z components of the force separated by commas. 104 m/s)^-(3.00 x 104 m/s)3.
Answer:
The force that the wire exerts on the electron is [tex]-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
Explanation:
Given that,
Current = 8.60 A
Velocity of electron [tex]v= (5.00\times10^{4})i-(3.00\times10^{4})j\ m/s[/tex]
Position of electron = (0,0.200,0)
We need to calculate the magnetic field
Using formula of magnetic field
[tex]B=\dfrac{\mu I}{2\pi d}(-k)[/tex]
Put the value into the formula
[tex]B=\dfrac{4\pi\times10^{-7}\times8.60}{2\pi\times0.200}[/tex]
[tex]B=0.0000086\ T[/tex]
[tex]B=-8.6\times10^{-6}k\ T[/tex]
We need to calculate the force that the wire exerts on the electron
Using formula of force
[tex]F=q(\vec{v}\times\vec{B}[/tex]
[tex]F=1.6\times10^{-6}((5.00\times10^{4})i-(3.00\times10^{4})j\times(-8.6\times10^{-6}) )[/tex]
[tex]F=(1.6\times10^{-19}\times3.00\times10^{4}\times(-8.6\times10^{-6}))i+(1.6\times10^{-19}\times5.00\times10^{4}\times(-8.6\times10^{-6}))j+0k[/tex]
[tex]F=-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
Hence, The force that the wire exerts on the electron is [tex]-4.128\times10^{-20}i-6.88\times10^{-20}j+0k[/tex]
Part A) Components of the Force The force components on the electron are: [tex]\[F_x = -8.26 \times 10^{-20} \, \text{N}, \quad F_y = -1.38 \times 10^{-19} \, \text{N}, \quad F_z = 0 \, \text{N}\][/tex]
Part B) Magnitude of the Force The magnitude of the force is:[tex]\[F \approx 1.60 \times 10^{-19} \, \text{N}\][/tex]
Part A: Calculate the force components
The force on a moving charge in a magnetic field is given by the Lorentz force equation:
[tex]\[\vec{F} = q \vec{v} \times \vec{B}\][/tex]
First, we need to find the magnetic field [tex]\(\vec{B}\)[/tex] produced by the wire at the position of the electron. The magnetic field due to a long, straight current-carrying wire is given by:
[tex]\[B = \frac{\mu_0 I}{2 \pi r}\][/tex]
where:
- [tex]\(\mu_0 = 4 \pi \times 10^{-7} \, \text{T} \cdot \text{m/A}\)[/tex] (the permeability of free space)
- [tex]\(I = 8.60 \, \text{A}\)[/tex] (the current through the wire)
- [tex]\(r = 0.200 \, \text{m}\)[/tex] (the distance from the wire to the electron)
Calculating [tex]\(B\)[/tex]:
[tex]\[B = \frac{4 \pi \times 10^{-7} \times 8.60}{2 \pi \times 0.200} = \frac{4 \times 10^{-7} \times 8.60}{0.200} = \frac{3.44 \times 10^{-6}}{0.200} = 1.72 \times 10^{-5} \, \text{T}\][/tex]
The direction of [tex]\(\vec{B}\)[/tex] follows the right-hand rule. Since the current flows in the [tex]\(-x\)[/tex]-direction, at the point [tex]\((0, 0.200, 0)\)[/tex], the magnetic field [tex]\(\vec{B}\)[/tex] is directed into the page (negative [tex]\(z\)[/tex]-direction):
[tex]\[\vec{B} = -1.72 \times 10^{-5} \hat{k} \, \text{T}\][/tex]
Now we use the Lorentz force equation with:
[tex]\[q = -1.60 \times 10^{-19} \, \text{C} \quad (\text{charge of an electron})\][/tex]
[tex]\[\vec{v} = (5.00 \times 10^4 \hat{i} - 3.00 \times 10^4 \hat{j}) \, \text{m/s}\][/tex]
[tex]\[\vec{B} = -1.72 \times 10^{-5} \hat{k} \, \text{T}\][/tex]
The cross product [tex]\(\vec{v} \times \vec{B}\)[/tex]:
[tex]\[\vec{v} \times \vec{B} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\5.00 \times 10^4 & -3.00 \times 10^4 & 0 \\0 & 0 & -1.72 \times 10^{-5}\end{vmatrix}= \hat{i}( (-3.00 \times 10^4)(-1.72 \times 10^{-5}) - 0) - \hat{j}( (5.00 \times 10^4)(-1.72 \times 10^{-5}) - 0)\][/tex]
[tex]\[= \hat{i}( 5.16 \times 10^{-1}) - \hat{j}( -8.60 \times 10^{-1})\][/tex]
[tex]\[= 0.516 \hat{i} + 0.860 \hat{j} \, \text{N/C}\][/tex]
Now, multiply by the charge of the electron:
[tex]\[\vec{F} = q \vec{v} \times \vec{B} = -1.60 \times 10^{-19} (0.516 \hat{i} + 0.860 \hat{j})\][/tex]
[tex]\[\vec{F} = -0.516 \times 1.60 \times 10^{-19} \hat{i} - 0.860 \times 1.60 \times 10^{-19} \hat{j}\][/tex]
[tex]\[\vec{F} = -8.26 \times 10^{-20} \hat{i} - 1.376 \times 10^{-19} \hat{j} \, \text{N}\][/tex]
So, the components of the force are:
[tex]\[F_x = -8.26 \times 10^{-20} \, \text{N}, \quad F_y = -1.376 \times 10^{-19} \, \text{N}, \quad F_z = 0 \, \text{N}\][/tex]
Part B: Calculate the magnitude of the force
The magnitude of the force is given by:
[tex]\[F = \sqrt{F_x^2 + F_y^2 + F_z^2}\][/tex]
[tex]\[F = \sqrt{(-8.26 \times 10^{-20})^2 + (-1.376 \times 10^{-19})^2}\][/tex]
[tex]\[F = \sqrt{(6.82 \times 10^{-39}) + (1.89 \times 10^{-38})}\][/tex]
[tex]\[F = \sqrt{2.57 \times 10^{-38}}\][/tex]
[tex]\[F \approx 1.60 \times 10^{-19} \, \text{N}\][/tex]
So, the magnitude of the force is approximately [tex]\(1.60 \times 10^{-19} \, \text{N}\).[/tex]
The complete question is attached here:
An electron is moving in the vicinity of a long, straight wire that lies along the z-axis. The wire has a constant current of 8.60 A in the -z-direction. At an instant when the electron is at point (0, 0.200 m, 0) and the electron's velocity is (5.00 x 104 m/s) -(3.00 x 104 m/s).
Part A:What is the force that the wire exerts on the electron?
Part B:Calculate the magnitude of this force.
We would like to use the relation V(t)=I(t)R to find the voltage and current in the circuit as functions of time. To do so, we use the fact that current can be expressed in terms of the voltage. This will produce a differential equation relating the voltage V(t) to its derivative. Rewrite the right-hand side of this relation, replacing I(t) with an expression involving the time derivative of the voltage. Express your answer in terms of dV(t)/dt and quantities given in the problem introduction.
Answer:
Explanation:
The explanation or solution is given in the attach document
Final answer:
To find the voltage and current as functions of time in a circuit, replace current I(t) with C*dV(t)/dt, and substitute this into Ohm's law V(t)=I(t)R to obtain a differential equation V(t) = (C*dV(t)/dt)*R.
Explanation:
To rewrite the relation V(t)=I(t)R by replacing I(t) with an expression involving the time derivative of the voltage, first understand that the current I(t) is the rate of change of charge Q with respect to time, that is I(t) = dQ/dt. However, since Q is related to the voltage across a capacitor by the equation Q=CV(t), where C is the capacitance, the current can also be expressed in terms of the voltage as I(t) = C*(dV(t)/dt). Replacing this into the original equation gives us V(t) = (C * dV(t)/dt) * R, which is a first-order differential equation relating the voltage to its time derivative.
Jeff was riding in a car. Which change demonstrates the GREATEST effect of inertia?
Answer: The car makes a sudden stop.
Explanation:
Manuel is coasting on his bike. Because he is not pedaling, his bike will come to a stop. Which of these will cause Manuel's bike to stop? - the force of friction.
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