Answer:
B = 9.16 10⁻² T
Explanation:
The speed selector is a configuration where the electric and magnetic force has the opposite direction, which for a specific speed cancel
q v B = q E
v = E / B
B = E / v
Let's calculate
B = 4.4 10⁵ / 4.8 10⁶
B = 9.16 10⁻² T
To select a speed of 4.8 x 10^6 m/s in a velocity selector, a magnetic field strength of 9.17 x 10^-2 T is needed.
Explanation:A velocity selector in a mass spectrometer uses a combination of electric field and magnetic field to select charged particles based on their speed. The electric field and magnetic field exert forces on the particles in opposite directions, allowing only particles with a specific speed to pass through unaffected. To determine the magnetic field strength needed to select a speed of 4.8 x 106 m/s, we can use the equation qvB = qE, where q is the charge of the particle, v is its speed, B is the magnetic field strength, and E is the electric field strength.
Given the electric field strength of 4.4 x 105 V/m, we can rearrange the equation to solve for B:
B = E/v = (4.4 x 105 V/m) / (4.8 x 106 m/s) = 9.17 x 10-2 T
Therefore, the magnetic field strength needed to select a speed of 4.8 x 106 m/s is 9.17 x 10-2 Tesla.
A current I = 20 A is directed along the positive x-axis and perpendicular to a magnetic field. A magnetic force per unit length of 0.16 N/m acts on the conductor in the negative y-direction. Calculate the magnitude and direction of the magnetic field in the region through which the current passes. magnitude T direction
Answer:
the magnitude and direction of the magnetic field in the region through which the current passes is 0.008 T and +z direction.
Explanation:
given information:
current, I = 20 A
magnetic force per unit length, F/L = 0.16 N/m
the conductor in the negative y-direction
θ = 90° (perpendicular)
as we know the formula to calculate magnetic force is
F = B I L sin θ
B = F/(I L sin θ)
= (F/L) (1/I sin θ)
= 0.16 (1/15 sin 90)
= 0.008 T
since F is in the negative y direction, based of the right hand rule the magnetic field is in positive z direction
Answer:
Explanation:
Given:
current, I = 20 A
Magnetic force per unit length, F/L
= 0.16 N/m
Conductor in the negative y-direction, therefore θ = 90° (perpendicular)
For a magnetic field,
F = B I L sin θ
B = F/(I L sin θ)
= 0.16 × (1/15 sin 90)
= 0.008 T
The field is in the +ve z - direction.
A stiff wire 35.5 cm long is bent at a right angle in the middle. One section lies along the z-axis and the other is along the line y=2x in the xy-plane. A current of 22.5 A flows in the wire-down the z-axis and out the line in the xy-plane. The wire passes through a uniform magnetic field given by B = (.318 i)T.
Determine the magnitude and the direction of the total force on the wire.
Answer:
Explanation:
Length, l = 35.5 cm = 0.355 m
y = 2 x
Slope, y / x = 2
tan θ = 2
θ = 63.4°
i = 22.5 A
B = 0.318 i Tesla
[tex]\overrightarrow{l}=0.355\widehat{k}+0.355 Cos63.4\widehat{i}+0.355 Sin63.4\widehat{j}[/tex]
[tex]\overrightarrow{l}=0.16\widehat{i} + 0.32\widehat{j}+0.355\widehat{k}[/tex]
[tex]\overrightarrow{B}=0.318\widehat{i}[/tex]
The magnetic force is given by
[tex]\overrightarrow{F}=i(\overrightarrow{l}\times \overrightarrow{B})[/tex]
[tex]\overrightarrow{F}=22.5\left ( 0.16\widehat{i}+0.32\widehat{j}+0.355\widehat{k} \right )\times 0.318\widehat{i}[/tex]
[tex]\overrightarrow{F}=2.54\widehat{j}-2.29\widehat{k}[/tex]
Magnitude of force
[tex]F=\sqrt{2.54^{2}+2.29^{2}}[/tex]
F = 3.42 N
The angle is Ф
tanФ = -2.29/2.54
Ф = 42° below y axis
The total force experienced by the wire in the magnetic field has a magnitude of 145 N and is directed -30 degrees relative to the z-axis, according to the right-hand rule and Lorentz force law.
Explanation:In order to determine the direction of the total force on the wire, you would need to use the right-hand rule. First, you know the current is flowing down the z-axis and out along the y=2x line in the xy-plane. You also know that the magnetic field, B, is given by (.318 i)T which lies along the x-axis.
According to the Lorentz force law, the force exerted on a current-carrying conductor in a magnetic field is given by F = I * (L x B), where I represents the current, L is the length vector of the wire, and x is the cross product. Because the wire is bent at a right angle, the force will have two components: one for each section of the wire.
The force along the z-axis is given by F = I*LB*sin(theta), where theta is the angle between L and B. Here, L = 35.5/2 = 17.75 cm, B = 0.318 T, I = 22.5 A, and theta = 90 degrees (since B is along x-axis and L is along z-axis). Calculating gives Fz = 22.5 A * 17.75 cm * 0.318 T * sin(90) = 126.6 N.
The force along the y-axis is similar, but with the length vector along y=2x in the xy plane and B still along the x-axis, so theta = 180 degrees - arctan(2). Substituting the values we get Fy = - 22.5 A * 17.75 cm * 0.318 T * sin(180 - arctan(2)) = -73.8 N.
Therefore, the total force is given by the vector sum of Fz and Fy, which has magnitude sqrt(Fz^2 + Fy^2) = 145 N, in the direction of atan2(Fy, Fz) = -30 degrees relative to the z-axis.
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A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the tube has magnitude f. You attach the upper end of a lightweight vertical spring of force constant k to the cap at the top of the tube, and attach the lower end of the spring to the top of the cylinder. Initially the cylinder is at rest and the spring is relaxed. You then release the cylinder. What vertical distance will the cylinder descend before it comes momentarily to rest? Express your answer in terms of the variables m, f, and constants g, k.
The cylinder will descend a distance of f/k before coming momentarily to rest.
Explanation:
When the cylinder is released, it will initially accelerate downwards due to the force of gravity. As it accelerates, the spring attached to it will begin to stretch, exerting an upward force on the cylinder. The cylinder will continue to descend until the force exerted by the spring is equal in magnitude to the force of kinetic friction between the cylinder and the walls of the tube. At this point, the net force on the cylinder will be zero, and it will come momentarily to rest.
To find the distance the cylinder descends before coming to rest, we can equate the force exerted by the spring with the force of kinetic friction:
kx = f
Where k is the force constant of the spring and x is the distance the cylinder has descended. Solving for x gives:
x = f/k
Therefore, the vertical distance the cylinder descends before it comes momentarily to rest is given by x = f/k.
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Show that one mole of any gas occupies a volume of 22.4 L at 1 atm and 0°C i.e., at standard temperature and pressure (STP). [Given: R = 0.0821 L.atm/mol.K] on T 1 . 1 (b) A tank of volume 0.3 m contains 2 moles of oxygen gas at 20°С. Find the average K.E. per atom and also find the rms speed of the atoms. [Given: k = 1.38 x 10-23 J/K; R= 8.317 J/mol, M= 16 g/mol]
Answer:
a) The statement is true, b) [tex]K_{tr.atom} = 6.073 \times 10^{-21} \frac{J}{atom}[/tex],[tex]v_{rms} \approx 477.904 \frac{m}{s}[/tex]
Explanation:
a) The standard conditions are 273.15 K and 1 atm. Let consider that gas behaves ideally, whose equation of state is:
[tex]P \cdot V = n \cdot R_{u} \cdot T[/tex]
The volume is cleared out:
[tex]V = \frac{n \cdot R_{u} \cdot T}{P}[/tex]
By replacing terms:
[tex]V = \frac{(1 mole)\cdot (0.082 \frac{L \cdot atm}{mole \cdot K} )\cdot (273.15 K)}{1 atm}\\V = 22.399 L[/tex]
Therefore, the statement is true.
b) The average kinetic energy per atom is:
[tex]K_{tr,atom} = \frac{3}{2} \cdot \frac{R_{u}\cdot T}{N_{A}}[/tex]
Where [tex]N_{A}[/tex] is the Avogadro constante and is equal to [tex]6.022 \times 10^{23} \frac{atoms}{mole}[/tex].
[tex]K_{tr,atom} = \frac{3}{2} \cdot \frac{(8.317 \frac{J}{mole\cdot K} )\cdot (293.15 K)}{6.022 \times 10^{23} \frac{atoms}{mole} } \\K_{tr,atom} = 6.073\times 10^{-21} \frac{J}{atom}[/tex]
The rms speed is determined by the following formula:
[tex]v_{rms} = \sqrt{\frac{3 \cdot k \cdot T}{n \cdot M_{O} \cdot \frac{1}{N_{A}} } }\\v_{rms} = \sqrt{\frac{3\cdot(1.38 \times 10^{-23} \frac{J}{K} )\cdot(293.15 K)}{(2 moles) \cdot (16 \frac{g}{mole} )(\frac{1 kg}{1000 g} ) \cdot (\frac{1 mole}{6.022 \times 10^{23} atoms} )}} \\v_{rms} \approx 477.904 \frac{m}{s}[/tex]
Many spacecraft have visited Mars over the years. Mars is smaller than the earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8 m
Incomplete question.The complete one is here
Spacecraft have been sent to Mars in recent years. Mars is smaller than Earth and has correspondingly weaker surface gravity. On Mars, the free-fall acceleration is only 3.8m/s2. What is the orbital period of a spacecraft in a low orbit near the surface of Mars?
Answer:
[tex]T=5900s=99min[/tex]
Explanation:
Given
[tex]r_{satelite}=r_{mars}=3.37*10^{6}m\\ g_{mars}=3.8m/s^{2}\\[/tex]
To find
orbital period of a spacecraft T
Solution
An the initial calculating is computing the angular velocity of satellite :
[tex]w=\frac{2\pi }{T}\\ w=\frac{2\pi }{110min}(1min/60s)\\ w=9.52*10^{-4}rad/s[/tex]
Computing T
[tex]T=\frac{2\pi }{w}\\ as\\w=\sqrt{\frac{a}{r} }\\ So\\T=\frac{2\pi }{\sqrt{\frac{a}{r} }} \\T=\frac{2\pi }{\sqrt{\frac{3.8m/s^{2} }{3.37*10^{6} m} }}\\T=5900s=99min[/tex]
A child is standing on the edge of a merry-goround that is rotating with frequency f. The child then walks towards the center of the merry-go-round. For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards th
The angular velocity of the merry-go-round increases as a child walks towards the center, due to the conservation of angular momentum and a decrease in the system's moment of inertia.
Explanation:Angular Momentum and Angular Velocity on a Merry-Go-RoundWhen a child walks from the outer edge of a rotating merry-go-round towards the center, the system's angular momentum remains constant due to the conservation of angular momentum. However, the moment of inertia of the system decreases as the child moves closer to the center. Since angular momentum is the product of the moment of inertia and angular velocity (L = I∙ω), and it is conserved, the angular velocity of the merry-go-round must increase to compensate for the decrease in moment of inertia.
This concept can be compared to a figure skater pulling in their arms to spin faster. As the skater's arms come closer to the body, their moment of inertia decreases, causing an increase in their spinning speed, or angular velocity, with no external torques acting on the system.
The same principle applies to the merry-go-round: as the child moves inward, the system's angular velocity increases to conserve angular momentum. This can be observed in playgrounds and is a practical example of conservation laws in rotational motion.
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As the child walks towards the center of the rotating merry-go-round, the system's angular momentum remains constant due to the conservation of angular momentum. The moment of inertia decreases, causing the angular velocity to increase.
A child is standing on the edge of a merry-go-round that is rotating with frequency f.
As the child walks towards the center of the merry-go-round, the angular momentum of the system (child plus merry-go-round) remains constant.
This is due to the conservation of angular momentum, which states that if no external torque acts on the system, the total angular momentum remains unchanged.
Angular momentum (L) is given by the formula:
L = I * ωwhere I is the moment of inertia and ω is the angular velocity.
As the child moves towards the center, the moment of inertia (I) decreases because it depends on the distance from the axis of rotation. To keep angular momentum constant, the angular velocity (ω) must increase.Suppose the initial angular momentum is Lo = Io * ωo. As the child walks inward, the new moment of inertia In becomes smaller, and thus the new angular velocity ωn must be greater to satisfy Lo = In * ωn.complete question:
A child is standing on the edge of a merry-goround that is rotating with frequency f. The child then walks towards the center of the merry-go-round. For the system consisting of the child plus the merry-go-round, what remains constant as the child walks towards the center of the rotating merry-go-round?
The electric field of a sinusoidal electromagnetic wave obeys the equation E = (360V/m) sin[ (6.00×1015rad/s)t + (1.96×107rad/m)x ]. What is the amplitude of the magnetic field of this wave? A) 0.06 μT B) 0.23 μT C) 1.10 μT D) 1.20 μT
Answer:
Option D is correct.
Explanation:
Bmax = Emax / c
The general form for electromagnetic wave equation is
E = jEmax ×cos(kx-wt)
We were given
(360V/m) sin[ (6.00×1015rad/s)t + (1.96×107rad/m)x ].
So from the equation above
Emax = 360V/m
Bmax = 360/(3×10⁸) = 1.2 ×10‐⁶ T.
Answer
Option D
Amplitude of Magnetic field = B = 1.2×10⁻⁶ T
Explanation:
The relationship between electric field and magnetic field of an electromagnetic wave is given by
B = E/c
Where B is the amplitude of magnetic field and E is the amplitude of electric field and c is the speed of light
The amplitude of electric field is given as 360 V/m
B = (360 V/m)/(3×10⁸ m/s)
B = 1.2×10⁻⁶ V.s/m²
Since 1 Tesla is equal to 1 V.s/m²
B = 1.2×10⁻⁶ T
Therefore, option D is correct
What is the measure of the ability of a force to rotate or accelerate an object around an axis?
A. Centripetal Force
B. Level Arm
C. Axis of Rotation
D. Torque
D. Torque
Explanation:
Torque is a pattern of the force that can make a victim to revolve on an axis. The torque's direction vector based on the force's direction on the axis. The SI unit for torque is the Newton-meter. Torque can be both static or dynamic.
A static is one that does not generate an angular acceleration.The Torque's magnitude vector ζ for a torque generated by a given force F is
ζ = F .r sin(θ)
where,
r is the width of the moment arm
θ is the angle within the force vector and the moment arm.
In rotational kinematics, torque is measured as
ζ = Iα
Where,
α is the angular acceleration
I is the rotational inertia
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Suppose you want to operate an ideal refrigerator with a cold temperature of − 13.5 °C , and you would like it to have a coefficient of performance of at least 7.75. What is the maximum hot reservoir temperature for such a refrigerator?
Answer:
Th = 50°C
Explanation:
See the attachment below.
K = 7.75
Tc = 13.5°C = 273+13.5 = 286.5K
Answer:
292.984 K or 19.98 °C
Explanation:
Using,
η = Tc/(Th-Tc)................ Equation 1
Where η = coefficient of performance, Tc = cold temperature of the refrigerator, Th = hot temperature of the refrigerator.
Make Th the subject of the equation
η (Th-Tc) = Tc
η Th-η Tc = Tc
Th = (Tc+ηTc)/η ....................... Equation 2
Given: Tc = -13.5°C+273 = 259.5 °C, η = 7.75
Substitute into equation 2
Th = (259.5+7.75×259.5)/7.75
Th = (259.5+2011.125)/7.75
Th = 2270.625/7.75
Th = 292.984 °C or 19.98 °C
Hence the maximum hot reservoir temperature = 292.984 °C or 19.98 °C
11 Initially, a single resistor R 1 is wired to a battery. Then resistorR 2 is added in parallel. Are (a) the potential difference across R 1and (b) the current i 1 through R 1 now more than, less than, or thesame as previously?
Answer:
Explanation:
Since , resistances are joined in parallel , potential difference across either of R₁ or R₂ will be same , and it will be equal to emf of the cell provided no internal resistance exists in the cell.
Since potential difference across R₁ is equal to emf and resistance is R₁ , current through it too will be same as that in the previous case.
The potential difference across resistor R1 remains the same when R2 is added in parallel, and the current through R1 also remains unchanged because the voltage across it does not change.
Explanation:When R2 is added in parallel to R1, the potential difference (voltage) across R1 remains the same because in a parallel circuit, all components have the same potential difference across them. However, the current i1 through R1 remains the same as previously as well since the potential difference across it hasn't changed and Ohm's Law (I = V/R) dictates that the current through a resistor will depend only on the potential difference across it and its resistance, which remains unchanged.
Adding R2 in parallel with R1 has the effect of decreasing the overall resistance of the circuit as found using the equation 1/Rp = 1/R1 + 1/R2. This means that the total current drawn from the battery will increase, but it will be divided among R1 and R2. Since the voltage across R1 remains constant, the current through R1 is unaffected by the addition of R2.
A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects beats such that the
frequency of the echo is 900 Hz higher than the frequency the bat is emitting. What is the speed of the bat? The
speed of sound in air is 340 m/s.
Answer:
5.02 m/s
Explanation:
given,
frequency of sound emitted by bat, [tex]f_1[/tex] = 30 kHz = 30000 Hz
beat frequency detected by bat[tex]f_b[/tex] = 900 Hz
Speed of sound, v = 340 m/s
[tex]f_b = f_1 -f_2[/tex]
[tex]900 = 30000-f_2[/tex]
[tex]f_2 = 30900[/tex]
Using Doppler's effect formula
[tex]f_2=f_1(\dfrac{v+v_b}{v-v_b})[/tex]
[tex]v_b = velocity\ of\ bat[/tex]
[tex]30900=30000(\dfrac{340+v_b}{340-v_b})[/tex]
[tex]1.03 =(\dfrac{340+v_b}{340-v_b})[/tex]
[tex]v_b = 5.02 m/s[/tex]
Hence, the speed of bat is equal to 5.02 m/s
Find a unit vector in the direction in which f increases most rapidly at P and give the rate of chance of f in that direction; find a unit vector in the direction in which f decreases most rapidly at P and give the rate of change of f in that direction.
Answer:
Check attachment for complete question
Question
Find a unit vector in the direction in which
f increases most rapidly at P and give the rate of change of f
in that direction; Find a unit vector in the direction in which f
decreases most rapidly at P and give the rate of change of f in
that direction.
f (x, y, z) = x²z e^y + xz²; P(1, ln 2, 2).
Explanation:
The function, z = f(x, y,z), increases most rapidly at (a, b,c) in the
direction of the gradient and decreases
most rapidly in the opposite direction
Given that
F=x²ze^y+xz² at P(1, In2, 2)
1. F increases most rapidly in the positive direction of ∇f
∇f= df/dx i + df/dy j +df/dz k
∇f=(2xze^y+z²)i + (x²ze^y) j + (x²e^y + 2xz)k
At the point P(1, In2, 2)
Then,
∇f= (2×1×2×e^In2+2²)i +(1²×2×e^In2)j +(1²e^In2+2×1×2)
∇f=12i + 4j + 6k
Then, unit vector
V= ∇f/|∇f|
Then, |∇f|= √ 12²+4²+6²
|∇f|= 14
Then,
Unit vector
V=(12i+4j+6k)/14
V=6/7 i + 2/7 j + 3/7 k
This is the increasing unit vector
The rate of change of f at point P is.
|∇f|= √ 12²+4²+6²
|∇f|= 14
2. F increases most rapidly in the positive direction of -∇f
∇f=- (df/dx i + df/dy j +df/dz k)
∇f=-(2xze^y+z²)i - (x²ze^y) j - (x²e^y + 2xz)k
At the point P(1, In2, 2)
Then,
∇f= -(2×1×2×e^In2+2²)i -(1²×2×e^In2)j -(1²e^In2+2×1×2)
∇f=-12i -4j - 6k
Then, unit vector
V= -∇f/|∇f|
Then, |∇f|= √ 12²+4²+6²
|∇f|= 14
Then,
Unit vector
V=-(12i+4j+6k)/14
V= - 6/7 i - 2/7 j - 3/7 k
This is the increasing unit vector
The rate of change of f at point P is.
|∇f|= √ 12²+4²+6²
|∇f|= 14
There's a part of the question missing and it is:
f(x, y) = 4{x(^3)}{y^(2)} ; P(-1,1)
Answer:
A) Unit vector = 4(3i - 2j)/ (√13)
B) The rate of change;
|Δf(1, - 1)|= 4/(√13)
Explanation:
First of all, f increases rapidly in the positive direction of Δf(x, y)
Now;
[differentiation of the x item alone] to get;
fx(x, y) = 12{x(^2)}{y^(2)}
So at (1,-1), fx(x, y) = 12
Similarly, [differentiation of the y item alone] to get; fy(x, y) =
8{x(^3)}{y}
At (1,-1), fy(x, y) = - 8
Therefore, Δf(1, - 1) = 12i - 8j
Simplifying this, vector along gradient = 4(3i - 2j)
Unit vector = 4(3i - 2j)/ (√(3^2) + (-2^2) = 4(3i - 2j)/ (√13)
Therefore, the rate of change;
|Δf(1, - 1)|= 4/(√13)
A 67 kg person climbs up a uniform 12 kg ladder. The ladder is 5 m long; its lower end rests on a rough horizontal floor (static
friction coefficient 0.39) while its upper end rests against a frictionless vertical wall. The angle between the ladder and the horizontal is 43◦.
Let d denote the climbing person’s distance from the bottom of the ladder (see the above diagram). When the person climbs too far (d > dmax), the ladder slips and falls down (kaboom!). Calculate the maximal distance dmax the person will reach before the ladder slips. The
acceleration of gravity is 9.8 m/s*s. Answer in two decimal places max.
Answer:
1.7 m
Explanation:
Draw a free body diagram of the ladder. There are 5 forces:
Normal force N pushing up at the base of the ladder.
Friction force Nμ pushing right at the base of the ladder.
Weight force mg pushing down a distance d up the ladder.
Weight force Mg pushing down a distance L/2 up the ladder.
Reaction force R pushing left at the top of the ladder.
Sum of forces in the x direction:
∑F = ma
Nμ − R = 0
Sum of forces in the y direction:
∑F = ma
N − mg − Mg = 0
Sum of moments about the base of the ladder:
∑τ = Iα
mg (d cos θ) + Mg (L/2 cos θ) − R (L sin θ) = 0
Use the first equation to substitute for R:
mg (d cos θ) + Mg (L/2 cos θ) − Nμ (L sin θ) = 0
Use the second equation to substitute for N:
mg (d cos θ) + Mg (L/2 cos θ) − (mg + Mg) μ (L sin θ) = 0
Simplify and solve for d:
m (d cos θ) + M (L/2 cos θ) − (m + M) μ (L sin θ) = 0
m (d cos θ) = (m + M) μ (L sin θ) − M (L/2 cos θ)
d = [ (m + M) μ (L sin θ) − M (L/2 cos θ) ] / (m cos θ)
Plug in values and solve:
d = [ (67 kg + 12 kg) (0.39) (5 m sin 43°) − (12 kg) (2.5 m cos 43°) ] / (67 kg cos 43°)
d = 1.70 m
Rounded to two significant figures, the maximum distance is 1.7 m.
The maximal distance, [tex]d_{max[/tex], the person will reach before the ladder slips will be 3.41 m.
Here, we need to consider the forces acting and torque equilibrium. Let's start by noting down the given data:
Mass of the person, [tex]m_p[/tex] = 67 kgMass of the ladder, [tex]m_l[/tex] = 12 kgLength of the ladder, L = 5 mCoefficient of static friction, μ = 0.39Angle with the horizontal, θ = 43°The forces acting on the ladder are:
The weight of the ladder ([tex]W_l[/tex]) acting at its center of mass, which is at L/2.The weight of the person ([tex]W_p[/tex]) acting at distance d from the bottom.The normal force from the ground (N) and the frictional force from the ground (f).The normal force from the wall ([tex]F_w[/tex]), which is horizontal since the wall is frictionless.Using Newton’s second law for horizontal and vertical equilibrium and setting torques about the base of the ladder:
1. Horizontal forces:
[tex]\[F_w = f\][/tex]2. Vertical forces:
[tex]N=[/tex] [tex]W_l + W_p[/tex]For torque equilibrium about the base:
[tex]\[N \times L \sin(\theta) = W_l \left(\frac{L}{2}\right) \cos(\theta) + W_p d \cos(\theta)\][/tex]Substitute: [tex]\[N = (m_l + m_p)g\][/tex]
And frictional force: f = μN
We solve for d:
[tex]\[0.39(m_l + m_p)g \cdot L \sin(\theta) = m_l g \left(\frac{L}{2}\right) \cos(\theta) + m_p g \cdot d \cos(\theta)\][/tex]Simplify:
[tex]\[d = \frac{0.39(m_l + m_p)L \sin(\theta) - \left(m_l \frac{L}{2}\right) \cos(\theta)}{m_p \cos(\theta)}\][/tex]Substituting values:
d = [tex]\[\frac{0.39(12 + 67) \times 5 \sin(43^\circ) - \left(12 \times \frac{5}{2}\right) \cos(43^\circ)}{67 \cos(43^\circ)}\][/tex]d ≈ 3.41 mIf an astronaut can throw a certain wrench 15.0 m vertically upward on earth, how high could he throw it on our moon if he gives it the same starting speed in both places?
Final answer:
The maximum height a wrench can be thrown on the Moon with the same starting speed as on Earth is 6 times higher.
Explanation:
To determine how many times higher an astronaut could jump on the Moon compared to Earth, we need to compare the gravitational accelerations of both locations. The gravitational acceleration on the Moon is about 1/6 of the acceleration on Earth, denoted as g.
When an object is thrown vertically upward, its maximum height is determined by the initial velocity and the gravitational acceleration. Since the astronaut gives the wrench the same starting speed in both places, the maximum height it can reach on the Moon will be 6 times higher than on Earth. This is because the weaker gravitational acceleration on the Moon allows the object to reach a greater height before being pulled back down.
Therefore, if the astronaut can throw the wrench 15.0 m vertically upward on Earth, they could throw it about 90.0 m (15.0 m * 6) on the Moon with the same starting speed.
A heart defibrillator is used to enable the heart to start beating if it has stopped. This is done by passing a large current of 12.0 A through the body at 25.0 V for a very short time, usually about 3.00 ms.
(a) What power does the defibrillator deliver to the body?
(b) How much energy is transferred?
To solve this problem we will apply the concepts related to Power, such as the product of voltage and current. Likewise, Power is defined as the amount of energy per unit of time, therefore, using this relationship we will solve the second part.
PART A)
We know that Power is,
[tex]P = VI[/tex]
Substitute [tex]25.0V[/tex] for V and 12.0 A for I in the expression,
[tex]P=(25.0V)(12.0A)[/tex]
[tex]P = 300W[/tex]
Therefore the power deliver to the body is 300W
PART B) Converting the time to SI,
[tex]t = 3ms = 0.003s[/tex]
Therefore if we have that the expression for energy is,
[tex]P = \frac{E}{t} \rightarrow E = Pt[/tex]
Here,
E = Energy,
P = Power,
t = Time,
Replacing,
[tex]E = (300W)(0.003s)[/tex]
[tex]E = 0.9J[/tex]
Therefore the energy transferred is equal to 0.9J
This question involves the concepts of electrical power and energy.
(a) The power delivered by the defibrillator is "300 W".
(b) The energy trasferred is "0.9 J".
(a) POWERThe electrical power delivered by the defibrillator can be given by the following formula:
[tex]P=IV[/tex]
where,
P = electrical power = ?I = electric current = 12 AV = voltage = 25 VTherefore,
[tex]P=(12\ A)(25\ V)[/tex]
P = 300 W
(b) Energy
The transferred energy can be given by the following formula:
[tex]P=\frac{E}{t}\\\\E=Pt[/tex]
where,
E = Energy = ?P = Power = 300 Wt = time = 3 ms = 3 x 10⁻³ sTherefore,
[tex]E=(300\ W)(3\ x\ 10^{-3}\ s)\\[/tex]
E = 0.9 J
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A 60-N box rests on a rough horizontal surface with a coefficient of static friction of 0.5. A horizontal force of 23 N acts on the box but the box is observed to be at rest. What is the value of the static friction force
Answer:23 N
Explanation:
Given
Weight of box [tex]W=60\ N[/tex]
Coefficient of static friction is [tex]\mu _s=0.5[/tex]
Applied force [tex]F=23\ N[/tex]
When Force is applied box is observed to be at rest i.e. static friction is overcoming the applied force.
Thus Static friction [tex]F_s[/tex]=applied force
[tex]F_s=23\ N[/tex]
Although it maximum value can go up to [tex](F_s)_{max}=\mu _sN[/tex]
[tex]F_s=0.5\times 60[/tex]
[tex]F_s=30\ N[/tex]
At what minimum speed must a roller coaster be traveling so that passen- gers upside down at the top of the circle (Fig. 5–48) do not fall out? Assume a radius of curvature of 8.6 m.
Answer: [tex]v \approx 18.37 \frac{m}{s}[/tex]
Explanation:
Let assume that system is conservative. From application of the Principle of Energy Conservation, it is noticed that initial linear kinetic energy must be equal to the gravitational energy at the top of the circle. That is to say:
[tex]K_{1} = U_{2}\\[/tex]
[tex]\frac{1}{2} \cdot m \cdot v^{2} = m \cdot g \cdot (2\cdot R) \\v = \sqrt{4 \cdot g \cdot R}[/tex]
Where [tex]g = 9.81 \frac{m}{s^{2}}[/tex].
[tex]v = \sqrt{4 \cdot (9.81 \frac{m}{s^2})\cdot(8.6 m)} \\v \approx 18.37 \frac{m}{s}[/tex]
"Sort the following objects as part of the system or not."An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast will he be moving backward just after releasing the ball?
Answer:
0.080625 m/s
Explanation:
Horizontally, the total momentum must be conserved, meaning the momentum before and after the throw must be the same.
The total momentum before the throw is 0, and so is after the throw:
mv + MV = 0
where m, M are the masses of the ball and the quarterback, respectively. v and V are the velocities of the ball and the quarterback, respectively.
0.43*15 + 80V = 0
80V = -6.45
V = -6.45 / 80 = -0.080625 m/s
So he will be moving in the opposite direction with the ball at the rate of 0.080625 m/s
Final answer:
To determine the quarterback's movement after throwing the football, one would need to use conservation of momentum principles. Initial velocity, time in the air, and the maximum height of the football once thrown can be calculated using the equations of projectile motion.
Explanation:
The scenario presented in the question can be analyzed using the principles of conservation of momentum. Since the situation described involves horizontal motion and the exchange of momentum between the quarterback and the football, we will apply the concepts of momentum conservation and the horizontal aspect of projectile motion.
Conservation of Momentum
For part of the scenario where we determine the quarterback's movement after throwing the ball, we use the conservation of momentum which states that the total momentum of a system is conserved if there are no external forces acting on the system. The formula for momentum is p = mv, where p is momentum, m is mass, and v is velocity. Just before the quarterback throws the ball, both he and the ball are part of the same system, and their combined momentum is zero if he jumps straight up in the air. After throwing the ball, the quarterback will have a backward velocity to conserve the total momentum of the system; we calculate this by equating the momentum of the thrown ball and the momentum of the moving quarterback.
Projectile Motion
In a situation where the football is thrown in a certain manner, the motion of the football can be separated into horizontal and vertical components. The initial velocity of the football can be calculated based on the given range, height, and angle of throw using kinematic equations. The time it takes for the ball to reach the receiver and its maximum height can also be determined through these equations.
A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.25, and the differential height between the two arms of the manometer is 28 in. determine the density
Answer:
77.88 lbm/ft³
Explanation:
Given,
Specific gravity, SG = 1.25
Density of water, ρ = 62.30 lbm/ft³
density of the fluid =
= S.G x ρ_{water}
= 62.30 x 1.25
= 77.88 lbm/ft³
Density of the fluid is equal to 77.88 lbm/ft³
Two objects, A with charge +Q and B with charge +4Q, are separated by a distance r. The magnitude of the force exerted on the second object by the first is F. If the first object is moved to a distance 2r from the second object, what is the magnitude of the electric force on the second object?
Final answer:
The magnitude of the electric force on the second object remains the same when the first object is moved to a distance 2r from it.
Explanation:
To find the magnitude of the electric force on the second object (B) when the first object (A) is moved to a distance 2r from B, we can use Coulomb's law. Coulomb's law states that the magnitude of the electric force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Given that the charge of A is Q and the charge of B is 4Q, the force exerted on B when A is at distance r is F. Now, when A is moved to a distance 2r from B, the new distance between A and B is 2r. Using Coulomb's law and the proportionalities mentioned earlier, the magnitude of the new electric force on B can be calculated as:
F' = (k * Q * 4Q) / (2r)^2
Where k is the constant in Coulomb's law. Simplifying the equation gives:
F' = (4 * F) / 4 = F
Therefore, the magnitude of the electric force on the second object B remains the same when the first object A is moved to a distance 2r from B.
A diver goes under water and measures the pressure. At some point his instruments read a pressure of 50,000 Pa. How deep did the diver go in meters? The density of water is 1000 kg/m^3. (Write the number only with 1 significant figure)
Answer:
[tex]y\approx 5\ m[/tex]
Explanation:
The pressure of a Fluid
A fluid of density [tex]\rho[/tex] exerts pressure at a distance y (deep) given by
[tex]P=\rho\cdot y\cdot g[/tex]
Where g is the acceleration of gravity or [tex]g=9.8\ m/s^2[/tex]
This formula computes the pressure assuming the initial pressure is 0 at fluid (water in this case) level.
Knowing the measured pressure, we can know how deep the diver went by solving the equation for y
[tex]\displaystyle y=\frac{P}{\rho\cdot g}[/tex]
Let's plug in the given values
[tex]P=50,000\ Pa= 50,000\ N/m^2[/tex]
[tex]\rho=1,000\ kg/m^3[/tex]
[tex]g=9.8\ m/s^2[/tex]
Thus
[tex]\displaystyle y=\frac{50,000\ N/m^2}{1,000\ kg/m^3\cdot 9.8\ m/s^2}[/tex]
[tex]y\approx 5\ m[/tex]
A steel tube (G 11.5 106 psi) has an outer diameter d2 2.0 in. and an inner diameter d1 1.5 in. When twisted by a torque T, the tube develops a maximum normal strain of 170 106 . What is the magnitude of the applied torque T?
Explanation:
Given data:
G = 11.5×10⁶psi
d₂ = 2.0 inch
d₁ = 1.5 inch
ε[tex]_{max}[/tex] = 170 × 10⁻⁶
Y[tex]_{max}[/tex] = 2ε
T/J = τ[tex]_{max}[/tex] /R
[tex]\frac{Td_{2} }{2J}[/tex] = τ[tex]_{max}[/tex] (1)
τ[tex]_{max}[/tex] = G Y
from 1 and 2
[tex]\frac{T d_{2} }{2J} = G Y_{max}[/tex]
T = [tex]\frac{2 G Y_{max}J }{d_{2} }[/tex]
[tex]\frac{2* 11.5*10^{6}*0.006895*10^{6}*340*10*^{-6} *\frac{\pi }{32}[2^{4}-1.5^{4}]*(0.0254)^{4} }{2* 0.0254}[/tex]
= 474.14 Nm
What is the motion of the negative electrons and positive atomic nuclei caused by the external field?
Answer:
The negative electrons moves to the left While the positively charged current moves in the opposite direction to the right when an external field is applied.
The positive atomic nuclei remains almost motionless.
In an external field, electrons and atomic nuclei move differently. In an electric field, electrons tend towards the positive field direction, whereas nuclei tend towards the negative. In a magnetic field, their paths are perpendicular to the field lines.
Explanation:The motion of negative electrons and positive atomic nuclei under the influence of an external field depends on the nature of the external field. This field could be of many types such as electric, magnetic, or electromagnetic. In an electric field, electrons move towards the positive field direction while the nuclei move towards the negative field direction. In a magnetic field, the motion of electrons and nuclei is governed by the right-hand rule, which means they move in a path perpendicular to the magnetic field lines.
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A constant force of 12 N in the +x direction acts on a 4 kg block as it moves from the origin to the point (6 − 8) m. How much work is done on the block by this force during this displacement?
Answer:
72 Nm²
Explanation:
work = F . d
work = (12i ,0j) . (6i ,-8j )
work = 72 J
The 12N force is a vector acting in the direction of the positive x axis, so its vector notation is 12i + 0j.
It's not exactly clear what the point's location is but I interpret it to be x= m*6, y= -m*8 (I hope the "m" wasn't a typo). The direction vector from the origin is then m*6i -m*8j
When a force vector F acts in a nonparallel direction d, the work is given by:
W = |F|*|d|*cos(theta) where theta is the angle between the vectors.
alternatively you can use the dot product of the two vectors to get:
W = (12N)*(m*6) + (0N)*(-m*8) = 72 Nm²
Final answer:
The work done on the block by the constant force of 12 N in the +x direction is 24 J.
Explanation:
The work done on the block can be calculated by multiplying the force applied to the block by the displacement of the block in the direction of the force. In this case, the force applied is 12 N in the +x direction and the displacement is (6 - 8) m = -2 m. Since the force and displacement are in opposite directions, the work done will be negative. The formula for work is given by:
Work = Force * Displacement * cos(theta)
where theta is the angle between the force and the displacement.
In this case, the angle between the force and the displacement is 180 degrees, so cos(theta) = -1. Substituting the values into the formula:
Work = 12 N * -2 m * (-1) = 24 J
A cave rescue team lifts an injured spelunker directly upward and out of a sinkhole by means of a motor-driven cable. The lift is performed in three stages, each requiring a vertical distance of 10.0 m: (a) the initially stationary spelunker is accelerated to a speed of 4.70 m/s; (b) he is then lifted at the constant speed of 4.70 m/s; (c) finally he is decelerated to zero speed. How much work is done on the 56.0 kg rescue by the force lifting him during each stage
Answer:
(a) the initially stationary spelunker is accelerated to a speed of 4.70 m/s - 6106 J
(b) he is then lifted at the constant speed of 4.70 m/s - 5488 J
(c) finally he is decelerated to zero speed. How much work is done on the 56.0 kg rescue by the force lifting him during each stage - 4869 J
Explanation:
knowing
d = 10 m
m = 56 kg
The work done by the applied force to pull the spelunker is given by
Wa + Wg = Kf - Ki
Wg = -mgd
First
Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]
Wa = (56*9.8*10) + (0.5*56*[tex]4.7^{2}[/tex])
Wa = 6106 J
Second
Kf = Ki
Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]
Wa = 56*9.8*10
Wa = 5488 J
Third
Kf = 0
Wa = mgd + 0.5 [tex]mv^{2}_{f}[/tex] - 0.5 [tex]mv^{2}_{i}[/tex]
Wa = (56*9.8*10) - (0.5*56*[tex]4.7^{2}[/tex])
Wa = 4869 J
A solid nonconducting sphere of radius R carries a uniform charge density throughout its volume. At a radial distance r1 = R/4 from the center, the electric field has a magnitude E0. What is the magnitude of the electric field at a radial distance r2 = 2R?
Answer:
Explanation:
Given that,
Sphere of radius R
At radial distance ¼R it has an electric field of Eo
Magnitude of electric field E at r=2R
Solution
electric field is given as
E=KQ/r²
Now at R,
E=KQ/R²
When, at r=R/4 inside the sphere is given as
Eo=kQ/R³ • R/4
Eo=kQ/4R²
When, r =2R
Then,
E=kQ/(2R)²
E=kQ/4R²
Comparing this to Eo
Then, E=Eo
Then, the electric field at r=2R is equal to the electric field at r=R/4
green light in the visible portion of the electromagnetic radiation sepectrum has a wave length around 550nm.Express this wavelength in meters using exponential notation
The wavelength of green light in meters using exponential notation is 5.5 × 10-7 m.
Explanation:The green light in the visible portion of the electromagnetic radiation spectrum has a wavelength of around 550 nm (nanometers).
To express this wavelength in meters using exponential notation, we can convert nanometers to meters by dividing by 109. So, the wavelength of green light is 5.5 × 10-7 m (meters).
The speed of a 3.5-N hockey puck sliding across a level icy surface decreases at a rate of 0.45 m/s2. The coefficient of kinetic friction between the puck and the ice is _____. Round your answer to the nearest hundredth.
Answer:
The coefficient of kinetic friction between the puck and the ice is 0.05.
Explanation:
Given that,
Weight of the hockey puck, W = mg = 3.5 N
Acceleration of the hockey puck, [tex]a=0.45\ m/s^2[/tex]
The frictional force opposing the motion of the puck is given by
[tex]F=\mu N[/tex]
This force is balanced by the force due to motion of the hockey puck. So,
[tex]\mu N=ma\\\\\mu mg=ma\\\\\mu=\dfrac{a}{g}\\\\\mu=\dfrac{0.45}{9.8}\\\\\mu=0.045[/tex]
or
[tex]\mu=0.05[/tex]
So, the coefficient of kinetic friction between the puck and the ice is 0.05. Hence, this is the required solution.
According to the Ideal Gas Law, , where P is pressure, V is volume, T is temperature (in Kelvins), and k is a constant of proportionality. A tank contains 2500 cubic inches of nitrogen at a pressure of 36 pounds per square inch and a temperature of 700 K. Write P as a function of V and T after evaluating k.
Answer:
P = 128.6 T / V
Explanation:
The ideal gas equation is
P V = n R T
Where the pressure is
P = 36 pounds / in²
V = 2500 in³
T = 700 K
PV = k T
k = PV / T
k = 36 2500/700
k = 128.6
P = 128.6 T / V
A student sees her physical science professor approaching on the sidewalk that runs by her dorm. She gets a water balloon and waits. When the professor is 2.0s from being directly under the window about 11m from the sidewalk, she drops the balloon. Finish the story.
Answer:
The balloon falls to the ground before the professor gets there. The student is DEFINITELY in for some TROUBLE!
Explanation:
The balloon picks up speed due to gravity and we can calculate the time taken for it to fall to the ground as follows:
Gravity (g) = 9.81 m/s^2
Height or distance (s) = 11 meters
Initial Speed (u) = 0 m/s
[tex]s = u*t + 0.5 * (a*t^2)[/tex]
[tex]11 = 0*t + 0.5 (9.81*t^2)[/tex]
[tex]t= 1.4975 s[/tex]
So we can see that the balloon takes 1.4975 seconds to fall to the ground, and since the professor takes 2 seconds to get to that place, the balloon hits the ground right before the professor gets there.