Answer:
Electric field magnitude
E = K/qd
Where
K = kinetic energy of electron
d = electron distance
q = charge
Explanation:
Given the relationship between workdone and energy
Work-energy theorem:
Net workdone = Energy change
W = ∆E
In this case
W = ∆K.E
And,
∆K.E = K(final) - K(initial)
To stop the kinetic energy | K(final) = 0
K(initial) = K (given)
∆K.E = 0 - K = -K
Let the electric force on the electron has magnitude F.
And
W = -Fd = ∆K.E = -K
-Fd = -K
F = K/d .....1
The magnitude of the electric field E that can stop these electron in a distance d:
E = F/q ......2
Where q is the charge on electron.
substituting equation 1 to 2
E = (K/d)/q = K/qd
E = K/qd
A boat is traveling at 3.2 m/s in the same direction as an ocean wave of wavelength 30 m and speed 6.8 m/s. If the boat is on the crest of a wave, how much time will elapse until the boat is next on a crest?
To solve this problem we apply the kinematic equations of linear motion. For which the speed is described as the distance traveled in a time interval. This would be,
[tex]v = \frac{x}{t} \rightarrow t = \frac{x}{v}[/tex]
Our values are given as,
[tex]\text{The speed of the boat} = v_b = 3.8m/s[/tex]
[tex]\text{The speed of the ocean} = V = 6.8m/s[/tex]
[tex]\text{The wave length of the wave is the same distance traveled by boat} = d = \lambda = 30m[/tex]
[tex]\text{The relative speed of the boat} = v_r = -3.8 +6.8 = 3m/s[/tex]
Replacing we have,
[tex]t = \frac{d}{v_r}[/tex]
[tex]t = \frac{30m}{3m/s}[/tex]
[tex]t = 10s[/tex]
Therefore will take until the boat is next on a crest around to 10s
What is the electric potential energy of a system that consists of two protons 2.1×10−15 m apart-as might occur inside a typical nucleus? Express your answer using two significant figures.
The electric potential energy of a system comprising two protons 2.1 x 10^-15 m apart can be calculated using Coulomb's law, which gives us a result of approximately 0.68 MeV.
Explanation:The electric potential energy of a system depends upon the charge of the components and the distance between them. In this case, we can calculate the electric potential energy using Coulomb's law which states that the electric potential energy 'V' between two charges is given by the equation V = k*q1*q2/r where 'k' is Coulomb's constant (8.99 × 10^9 N m^2/C^2), 'q1' and 'q2' are the two charges, and 'r' is the distance between the charges. Given that the charges are two protons, they both have the same charge (1.6 × 10^-19 Coulombs). The distance 'r' is given as 2.1 × 10^-15 m. Substituting these values in, we get: V = (8.99 × 10^9 N m^2/C^2)* (1.6 × 10^-19 C) * (1.6 × 10^-19 C) / (2.1 × 10^-15 m) which results in an electric potential energy of approximately 0.68 MeV (mega electron volts).
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A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V.
Use ϵ0 = 8.85×10⁻¹² C²/N⋅m².
The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
Answer:
U_eq = 1.99 * 10^(-10) J
Explanation:
Given:
Plate Area = 10 cm^2
d = 0.01 m
k_dielectric = 3
k_air = 1
V = 15 V
e_o = 8.85 * 10 ^-12 C^2 / N .m
Equations used:
U = 0.5 C*V^2 .... Eq 1
C = e_o * k*A /d .... Eq 2
U_i = 0.5 e_o * k_i*A_i*V^2 /d ... Eq 3
For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:
U_electric = 0.5 e_o * k_1*A*V^2 /2*d
U_air = 0.5 e_o * k_2*A*V^2 /2*d
The total Energy is:
U_eq = U_electric + U_air
U_eq = 0.5 e_o * k_1*A*V^2 /2*d + 0.5 e_o * k_2*A*V^2 /2*d
U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d
Plug the given values:
U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01
U_eq = 1.99 * 10^(-10) J
The energy U2 of the partially dielectric-filled capacitor is found by calculating the total capacitance Ct when the capacitor is half-filled with the dielectric and then using the formula U2 = ½ CtV2 with the given voltage.
Explanation:To calculate the energy U2 of the partially dielectric-filled capacitor, we start by finding the capacitance when the capacitor is half-filled with the dielectric. With a dielectric constant k of 3.00 and a vacuum permittivity ε0 of 8.85×10−12 C2/N·m2, the initial capacitance Ci entirely filled with the dielectric is:
Ci = kε0A/d
When the capacitor is half-filled, the capacitance can be represented as two capacitors in parallel: one with dielectric (Cd) and one without (C0). These can be described by:
Cd = ε0kA/(2d) and C0 = ε0A/(2d)
The total capacitance Ct when the capacitor is half-filled is then:
Ct = Cd + C0
Finally, the energy stored in the capacitor U2 is given by:
U2 = ½ CtV2
This formula allows us to plug in the calculated capacitance Ct and the constant voltage V to find the desired energy U2.
In the hydrogen atom, what is the total energy of an electron that is in an orbit that has a radius of 8.784 × 10^(-10) m?
The total energy of an electron in a hydrogen atom with a given orbit radius can be determined using the formula E = -13.6 eV / n². By calculating the principal quantum number n from the given radius, we can find the total energy of the electron. In this case, the total energy is -3.4 eV.
Explanation:The total energy of an electron in the hydrogen atom can be determined using the formula:
E = -13.6 eV / n²
where E is the energy, n is the principal quantum number, and eV represents electron volts.
Given that the radius of the electron's orbit is 8.784 × 10⁽⁻¹⁰⁾ m, we can find the value of n using the formula for the radius of the nth orbit:
r = 0.529 × n² / Z Å
where r is the radius, n is the principal quantum number, and Z is the atomic number. Assuming the atomic number for hydrogen is 1, solving for n gives us:
n = √(r * Z / 0.529)
Substituting the given values, we find that n = 2.
Finally, plugging in n into the energy formula, we find:
E = -13.6 eV / (2²) = -3.4 eV
This is a computer program, why is there an error between the computer values and the nominal values of R?
Answer:
the difference is due to resistance tolerance
Explanation:
In mathematical calculations, either done by hand or in a computer program, the heat taken from the resistors is the nominal value, which is the writing in its color code, so all calculations give a result, but the Resistors have a tolerance, indicated by the last band that is generally 5%, 10%, 20% and in the expensive precision resistance can reach 1%.
This tolerance or fluctuation in the resistance value is what gives rise to the difference between the computation values and the values measured with the instruments, multimeters.
Another source of error also occurs due to temperature changes in the circuit that affect the nominal resistance value, there is a very high resistance group that indicates the variation with the temperature, they are only used in critical circuits, due to their high cost
In summary, the difference is due to resistance tolerance.
(II) At t =0, an 885-9 mass at rest on the end of a horizontal Spring (K: 184 N/m) is struck by a hammer which gives it aninitial speed of 226 m/s. Determine (a) the period and frequency ofthe motion, (b) the amplitude, (c) the maximumacceleration, (d) the total energy, and (e) the kinetic energy when x =O.4O A where A is the amplitude.
Answer:
A. T = 0.4358s and f = 2.29hz
B. A = 15.67m
C. amax = 3258.71m/s
D. amax = 22601J
E. Ek = 3616.16J
Explanation:
A. The period of the motion, T = 2pi*(sqrt(m/k))
Where m is the mass of the body in motion = 885g = 0.885kg
k = the spring constant = 184N/m2
T = 2pi*(sqrt(0.885/184))
= 0.4358s
Frequency of the motion, f = 1/T
T = 0.4358s
f = 2.2949hz
B. Maximum speed, Vmax = A*(sqrt(k/m))
Where A = amplitude of the motion
Making amplitude subject of formula,
A = Vmax(sqrt(m/k))
= 226*(sqrt(0.885/184))
= 15.6739m
C. Maximum acceleration, amax = A*(k/m)
= 15.6739*(184/0.885)
= 3258.71m/s
D. Total energy, Etotal = 1/2*(m * Vmax)2
= 1/2 * 0.885 * (226)2
= 22601J
E. Kinetic energy, Ek = Etotal - mechanical energy
Ek = 1/2*(k*A2) - 1/2*(k*x2)
Where x = 0.40A
Ek = 1/2*((k*A2) - (k*0.40A)2)
= 1/2*k*A2*(1 - 0.16)
= 1/2*k*A2*0.16
But 1/2*k*A2 = 22601J
Therefore, Ek = 22601*0.16
= 3616.16J
A force of 35 N acts on a ball for 0.2 s. If the ball is initially at rest:
a. What is the impulse on the ball?
b. What is the final momentum of the ball?
To solve this problem we will apply the concepts related to momentum and momentum on a body. Both are equivalent values but can be found through different expressions. The impulse is the product of the Force for time while the momentum is the product between the mass and the velocity. The result of these operations yields equivalent units.
PART A ) The Impulse can be calculcated as follows
[tex]L= F\Delta t[/tex]
Where,
F = Force
[tex]\Delta t =[/tex]Change in time
Replacing,
[tex]L = (35N)(0.2s)[/tex]
[tex]L= 7N\cdot s[/tex]
PART B) At the same time the momentum follows the conservation of momentum where:
Initial momentum= Final momentum
And the change in momentum is equal to the Impulse, then
[tex]\Delta p = L[/tex]
And
[tex]\Delta p = p_f - p_i[/tex]
There is not initial momentum then
[tex]\Delta p = p_f[/tex]
[tex]L = p_f[/tex]
[tex]p_f = 7N\cdot s = 7kg\cdot m/s[/tex]
Final answer:
The impulse on the ball, which is the product of the force and the time interval, is 7 N·s. Since the ball is initially at rest, the final momentum of the ball is also 7 N·s.
Explanation:
To solve the student's question, we need to apply the concepts of impulse and momentum in physics.
Impulse and Momentum
Impulse is defined as the product of the force acting on an object and the time interval over which it acts. From Newton's second law, the impulse on an object is equal to the change in its momentum.
a) The impulse on the ball is calculated using the formula Impulse = Force × Time. Given that the force is 35 N and the time is 0.2 s, the impulse can be calculated as:
Impulse = 35 N × 0.2 s = 7 N·s
b) The final momentum of the ball is equal to the impulse, since the ball was initially at rest (momentum = 0). Therefore, the final momentum of the ball is:
Final momentum = Impulse = 7 N·s
In a recent airline disaster, an airliner ying at 30,000ft,550mi/h, lost power and fell to Earth. The mass of the aircraft was 255,000 lb. If the magnitude of the work done by drag force on the plane during the fall was 2.96 × 106 Btu, estimate the velocity of the aircraft at the time of impact, in mi/h. Let g = 32.08 ft/s2.
To develop this problem we will apply the energy conservation theorem and the principle of work. Basically we will have that the kinetic, potential and work energy must be equivalent to the kinetic energy just before the impact. We will start converting the units given to the British system and then proceed with the Calculations.
Remember that according to the energy balance in this case it would be balanced like this
[tex]T_1 +\sum U_{1-2} = T_2[/tex]
[tex]\frac{1}{2} mv_1^2+mgh -W_{drag} = \frac{1}{2} mv_2^2[/tex]
Here
m = mass
[tex]v_{1,2}[/tex]= Velocity at each moment
[tex]W_{drag}[/tex]= Work by drag
h = Height
g = Acceleration due to gravity
Mass
[tex]m =255000 lb (\frac{1slug}{32.174lb})[/tex]
[tex]m = 7925.654slug[/tex]
Initial Velocity
[tex]v_1 = 550 \frac{mi}{h} (\frac{5280ft}{1mi})(\frac{1hr}{3600s})[/tex]
[tex]v_1 = 806.667ft/s[/tex]
Work by drag
[tex]W_{drag} = (2.96*10^6BTU)(\frac{778.169lb\cdot ft}{1BTU})[/tex]
[tex]W_{drag} = 2303380240lb\cdot ft[/tex]
By work energy principle
[tex]\frac{1}{2} mv_1^2+mgh -W_{drag} = \frac{1}{2} mv_2^2[/tex]
Replacing,
[tex]\frac{1}{2} (7925.654slug)(806.667ft/s)^2 +(7925.654slug)(32.08ft/s^2)(30000ft) -2303380240lb\cdot ft = \frac{1}{2} (7925.654slug)v_2^2[/tex]
Solving for [tex]v_2[/tex], we have that
[tex]v_2 = 1412.2 ft/s[/tex]
Converting this value,
[tex]v_2 = 1412.2 ft/s (\frac{1mi}{5280ft})(\frac{3600s}{1h})[/tex]
[tex]v_2 = 962.85mi/h[/tex]
Therefore the velocity of the aircraft at the time of impact is 962.85mi/h
To estimate the velocity of the aircraft at the time of impact, use the work-energy principle. The estimated velocity is 256.15 mi/h.
Explanation:To estimate the velocity of the aircraft at the time of impact, we can use the work-energy principle. The work done by the drag force is equal to the change in kinetic energy of the aircraft. The work done by the drag force is given as 2.96x10^6 Btu. We need to convert this energy into foot-pounds, and then use the kinetic energy equation to find the velocity. The equation is:
K = (1/2) mv^2
Where K is the kinetic energy, m is the mass of the aircraft, and v is the velocity. Rearranging the equation, we have:
v = √((2K) / m)
Plugging in the given values, we get:
v = √((2(2.96x10^6 Btu) * (3.968x10^8 ft-lbf/Btu)) / (255,000 lb * 32.08 ft/s^2)) = 256.15 mi/h
Therefore, the estimated velocity of the aircraft at the time of impact is 256.15 mi/h.
Two runners start simultaneously from the same point on a circular 200-m track and run in the same direction. One runs at a constant speed of 6.20 m/s and the other runs at a constant speed of 5.50 m/s. g
Answer
given,
speed of the fastest runner = 6.2 m/s
speed of the slower runner = 5.5 m/s
Assume, we need to calculate time when both the runner meet for the first time.
distance cover by the fast runner = distance cover by the slow runner
6.2 t = 5.5 t + 200
0.7 t = 200
t = 285.71 s
time after which both the runner meet.
Distance they covered after starting
D = s x t
D = 6.2 x 285.71
D = 1771.43 m.
now, calculating the second time both will meet.
they will take double time to meet again
t' = 2 x 285.71
t '= 571.42 s
What is the magnitude of the force on an electron at a distance of 1.70 angstrom from the plutonium nucleus?
Answer:
Force, [tex]F=7.04\times 10^{-5}\ N[/tex]
Explanation:
Given that,
Distance between 1.70 A from the plutonium nucleus, [tex]d=1.7\times 10^{-10}\ m[/tex]
The number of electron in plutonium is 94.
To find,
The magnitude of the force on an electron.
Solution,
Total charge in the plutonium nucleus is, [tex]q=94\times 1.6\times 10^{-19}=1.504\times 10^{-17}\ C[/tex]. The electric force between charges is given by :
[tex]F=\dfrac{kq^2}{d^2}[/tex]
[tex]F=\dfrac{9\times 10^9\times (1.504\times 10^{-17})^2}{(1.7\times 10^{-10})^2}[/tex]
[tex]F=7.04\times 10^{-5}\ N[/tex]
So, the magnitude of the force on an electron is [tex]7.04\times 10^{-5}\ N[/tex]. Hence, this is the required solution.
An ideal fluid flows through a pipe of variable cross section without any friction. The fluid completely fills the pipe. At any given point in the pipe, the fluid has a constant, What?
1. potential energy
2. velocity
3. pressure
4. total energy
5. kinetic energy
Answer:
4. total energy
Explanation:
According to Bernoulli's principle at any two points along a streamline flow The total energy that is sum of pressure energy , Kinetic energy and potential energy of the liquid all taken in per unit volume remains constant. Therefore,
for ideal fluid flows through a pipe of variable cross section without any friction. The fluid completely fills the pipe. At any given point in the pipe, the fluid has a constant Total Energy.
You are driving down the highway at 65 m p h, which is 29 m/s. Your tires have a radius of 0.30 m. a. How many times per second does each tire rotate? b. What is the speed of a point at the top of a tire, relative to the ground?
To solve this problem we will apply the concepts related to linear velocity and angular velocity to perform the respective conversion with the given values. To find the velocity in the upper part of the tire we will use the mathematical relation that expresses that it is twice the linear velocity. Let's start
PART A)
[tex]\omega = \frac{v}{r}[/tex]
[tex]\omega = \frac{29}{0.3}[/tex]
[tex]\omega = 96.66 rad/s[/tex]
Now we now that [tex]2\pi rad = 1 rev[/tex], then
[tex]\omega = 96.66rad/s (\frac{1 rev}{2\pi rad})[/tex]
[tex]\omega = 15.38rev/s[/tex]
PART B)
[tex]v = 2v_0[/tex]
[tex]v = 2(29)[/tex]
[tex]v = 58m/s[/tex]
A torque of 4 Nm is required to rotate the intermediate cylinder at 30 radians/min. Calculate the viscosity of the oil. All cylinders are 450 mm long. Neglect end effects.
To calculate the viscosity of the oil, you need to utilize the concept of shear stress in fluid dynamics. You take the known values of torque, cylinder length, and rotational speed and substitute them into the rearranged shear stress formula. Ensure you convert length from millimeters to meters to maintain SI unit consistency.
Explanation:The question can be solved using the principles of fluid dynamics, specifically the concept of viscous drag in a medium. In this case, the medium is oil and the object moving through it is a cylinder. The viscosity of the oil can be calculated using the formula for shear stress (τ), given as τ = η (du/dy), where η represents viscosity, du represents the difference in velocity, and dy represents the distance between layers of fluid. As we're given torque (τ) in Newton-metres (Nm), cylinder length (L) in millimeters, and rotational speed in rad/s (radians per second), we can derive η (viscosity) using the rearranged formula: η = τ /(du/dy).
In this case, du is equivalent to the speed of the cylinder let's represent this as 'u', dy is equivalent to the length of the cylinder 'L', hence du/dy becomes u/L. Substitute these values into the rearranged formula and solve to get the viscosity. It's critical to convert the length from millimeters to meters before performing the calculation to maintain consistency in the SI units.
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What causes the mechanical advantage of the inclined plane to be less than that of the liver
Answer:
force of friction
Explanation:
the force of friction in an inclined place is less than that in lever, also a lever works on the principle that it reduces the amount of force required to to work while an inclined plane works by decreasing the amount of force required to do the work.
A vacuum chamber contains a uniform electric field directed downward. If a proton is shot horizontally into this region, it's acceleration is? Give direction and relative magnitude (Ex 45 degrees west of north, and is exponentially decreasing)
Answer:
It is directed downward. Its relative magnitude is a= qp*E/mp = 0.96*10⁸*E m/s².
Explanation:
Assuming gravitational force to be negligible, the only force acting on the proton once within the vacuum chamber, is the one due to the electric field.
As the proton is a positive charge, the electric force on it due to the field has the same direction than the field, as the direction of the field by convention, is the one that would take a positive test charge.
According to electric field definition, and to Newton's 2nd Law, we can find the acceleration of the proton as follows:
F = m*a = q*E ⇒ a= q*E / m = 1.6*10⁻¹⁹ C * E (N/C) / 1.67*10⁻²⁷ kg
⇒ a = 0.96* 10⁸ * E m/s²
A proton shot horizontally into a vacuum chamber with a uniform electric field directed downward will experience an upward acceleration.
Explanation:A proton shot horizontally into a vacuum chamber with a uniform electric field directed downward will experience an acceleration due to the force exerted by the electric field. Since the electric field is directed downward, the proton will experience an upward acceleration.
The magnitude of the proton's acceleration can be calculated using the equation:
a = qE/m
where a is the acceleration, q is the charge of the proton, E is the magnitude of the electric field, and m is the mass of the proton.
The direction of the acceleration will be opposite to the direction of the electric field, which in this case is upward.
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A horse canters away from its trainer in a straight line, moving 126 m away in 13.0 s . It then turns abruptly and gallops halfway back in 5.0 s . Calculate the (a) its average speed and (b) itsaverage velocity for the entire trip, using "away from the trainer"as the positive direction.
Final answer:
The average speed of the horse for the entire trip is 10.5 m/s, and the average velocity is 3.5 m/s, calculated based on the total distance covered and total displacement over the entire duration of the trip.
Explanation:
To calculate the average speed and average velocity of the horse for the entire trip, we consider both phases of the journey: when the horse canters away and when it gallops back.
Part (a) - Average Speed
Average speed is total distance covered divided by total time taken. The horse travels 126 m away and then comes back halfway, which is 63 m, resulting in a total distance of 126 m + 63 m = 189 m. The total time taken for the trip is 13.0 s + 5.0 s = 18.0 s. Therefore, the average speed is:
average speed = total distance / total time = 189 m / 18.0 s = 10.5 m/s
Part (b) - Average Velocity
Average velocity is the total displacement divided by the total time taken. Since the horse ends up halfway back, the displacement is 126 m - 63 m = 63 m, in the initially defined positive direction (away from the trainer). Using a total time of 18.0 s:
average velocity = total displacement / total time = 63 m / 18.0 s = 3.5 m/s
A long string is stretched and its left end is oscillated upward and downward. Two points on the string are labeled A and B.At the instant shown, orient v(A) and v(B) to correctly represent the direction of the wave velocity at points A and B.At each of the points A and B, rotate the given vector to indicate the direction of the wave velocity.Figure:A string is attached and moving up and down
In an oscillating string, the wave velocity at a point on the string is oriented opposite to the direction of the string's movement at that point. Thus, if point A is moving up, v(A) is pointed down, and if point B is moving down, v(B) is pointed up.
Explanation:
The orientation of the velocity of the wave at points A and B is determined by the direction of the wave's movement. For an upward and downward oscillating string, the wave moves perpendicular to the string. When a segment of the string is moving upwards, the wave velocity at that point is downward, and vice versa.
So, if point A is currently moving upwards, then v(A) should be pointed downwards to represent the direction of the wave velocity. Similarly, if point B is moving downwards, then v(B) should be pointed upwards. The velocity vectors v(A) and v(B) are always opposite to the direction of string's movement at those points.
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You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads 28.4 NN . If you then lower the statue into a tub of water, so that it is completely submerged, the scale reads 17.0 NN .
Answer:
2491.23 kg/m³
Explanation:
From Archimedes principle,
R.d = weight of object in air/ upthrust in water = density of the object/density of water
⇒ W/U = D/D' ....................... Equation 1
Where W = weight of the ceramic statue, U = upthrust of the ceramic statue in water, D = density of the ceramic statue, D' = density of water.
Making D the subject of the equation,
D = D'(W/U).................... Equation 2
Given: W = 28.4 N, U = lost in weight = weight in air- weight in water
U = 28.4 - 17.0 = 11.4 N,
Constant: D' = 1000 kg/m³.
Substitute into equation 2,
D = 100(28.4/11.4)
D = 2491.23 kg/m³
Hence the density of the ceramic statue = 2491.23 kg/m³
a cat is on a merry-go-round that completes 1 full rotation in 6 seconds. The cat sits at a radius of 8.4 metres from the centre.
Find the minimum coefficient of friction to prevent the cat from sliding off.
To solve this problem we will use the concepts related to the uniform circular movement from where we will obtain the speed of the object. From there we will go to the equilibrium equations so that the friction force must be equal to the centripetal force. We will clear the value of the coefficient of friction sought.
The velocity from the uniform circular motion can be described as
[tex]v = \frac{2 \pi r}{T}[/tex]
Here,
r = Radius
T = Period
Replacing,
[tex]v = \frac{2\pi (8.4)}{6}[/tex]
[tex]v =8.7964 m/s[/tex]
From equilibrium to stay in the circle the friction force must be equivalent to the centripetal force, therefore
[tex]F_f = F_c[/tex]
[tex]\mu N = \frac{mv^2}{r}[/tex]
Here,
[tex]\mu =[/tex] Coefficient of friction
N = Normal Force
m = mass
v = Velocity
r = Radius
The value of the Normal force is equal to the Weight, then
[tex]\mu(mg) = \frac{mv^2}{r}[/tex]
Rearranging to find the coefficient of friction
[tex]\mu = \frac{v^2}{gr}[/tex]
Replacing,
[tex]\mu = \frac{(8.7964)^2}{(9.8)(8.4)}[/tex]
[tex]\mu =0.9399[/tex]
Therefore the minimum coefficient of friction to prevent the cat from sliding off is 0.9399
The displacement z of a particle of rest mass m0, resulting from a constant force m0g along the z-axis is including relativistic effect. Find the displacement z as a power series in time t. Compare with the classical result,
Final answer:
The displacement z of a particle with rest mass m0 subject to a constant force m0g along the z-axis, accounting for relativistic effects, involves integrating the work-energy theorem's relativistic expression for kinetic energy and comparing it with the classical displacement z = 1/2 gt².
Explanation:
The problem given involves finding the displacement z of a particle with rest mass m0 under the influence of a constant force along the z-axis, considering relativistic effects. The comparison with the classical result implies a consideration of the relativistic factor y in kinetic energy expressions.
We recall that the relativistic kinetic energy is more complex than the classical expression K = 1/2 mu², because as the velocity u of the particle approaches the speed of light c, the kinetic energy diverges due to the increase in y.
The task involves using the work-energy theorem in a relativistic context, where the work done by the force m0g produces a change in the relativistic kinetic energy. As such, relativistic mechanics dictates that energy and momentum are dependent on the relativistic factor y, which is a function of velocity.
Therefore, to expand the displacement z as a power series in time t, one would start by expressing the relativistic momentum and kinetic energy, then integrate to find the time-dependent expressions for velocity and displacement.
Comparing with the classical result, where z is simply obtained by z = 1/2 gt² (ignoring air resistance), shows how relativistic effects modify the simple parabolic trajectory of a particle in a gravitational field. The relativistic displacement would involve a series with higher-order terms in time, accounting for the varying mass with velocity.
A ball of mass 0.7 kg flies through the air at low speed, so that air resistance is negligible. (a) What is the net force acting on the ball while it is in motionWhich components of the ball's momentum will be changed by this force? What happens to the x component of the ball's momentum during its flight? What happens to the y component of the ball's momentum during its flight? It decreases. What happens to the z component of the ball's momentum during its flight?
With negligible air resistance and low speed, the only significant net force on a 0.7 kg ball is gravity, affecting the ball's y component of momentum. The x component remains constant, and z component changes are not discussed without additional forces.
Explanation:When a ball of mass 0.7 kg flies through the air at low speed with air resistance negligible, the net force acting on the ball while it is in motion is primarily due to gravity, which will be impacting the y component of the ball's momentum. The x component of the ball's momentum remains unchanged because no horizontal force is applied, while the y component changes due to gravity, and the z component would only change if there were forces acting in a direction out of the horizontal plane, which are not mentioned in the scenario. As for the Earth-ball system, momentum is conserved in the vertical direction because the system experiences no net external vertical force.
Which of the following quantities are vectors? Select all that apply. Which of the following quantities are vectors? Select all that apply. 27 J/s 1980 kg m/s due south 35 m/s 8.2 m/s2 north-west 3.2 mi straight up 2.9 m/s2 293 K 200 s
Vector quantities have magnitude and direction. The quantities that are vectors from the given list are:
1980 kg m/s due south,
8.2 m/s^2 north-west,
3.2 mi straight up.
Explanation:Quantities that have a magnitude and direction are called vectors. Based on this definition, the following quantities from the list are vectors:
1980 kg m/s due south: This quantity has both magnitude (1980 kg m/s) and direction (south).8.2 m/s2 north-west: This quantity represents acceleration with both magnitude (8.2 m/s2) and direction (north-west).3.2 mi straight up: This quantity has a magnitude (3.2 mi) and a specific direction (upward).Therefore, these three quantities are vectors.
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The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a uniform 20∘C) is 1482 m/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/s ? Assume that the ship is at rest in the water.
Your target variable is Δf, the magnitude of the difference in frequency between the waves emitted from the sonar device and the waves received by the device after reflecting off the whale. Write an expression for Δf in terms of the relevant frequencies using the subscript notation introduced above.
Express your answer in terms of some or all of the variables fLe, fLr, fSe, and fSr.
Δf =
Answer:
[tex]\Delta f=f_{Lr}-f_{Se}[/tex]
147.45 Hz
Explanation:
v = Speed of sound in water = 1482 m/s
[tex]v_w[/tex] = Speed of whale = 4.95 m/s
Frequency of the wave in stationary condition
[tex]f_{Lr}=f\dfrac{v+v_w}{v-v_w}[/tex]
Ship's frequency which is reflected back
[tex]f_{Se}=f\dfrac{v}{v-v_w}[/tex]
The difference in frequency is given by
[tex]\Delta f=f\dfrac{v+v_w}{v-v_w}-f\dfrac{v}{v-v_w}\\\Rightarrow \Delta f=f_{Lr}-f_{Se}[/tex]
[tex]\mathbf{\Delta f=f_{Lr}-f_{Se}}[/tex]
[tex]f_{Lr}=22\times \dfrac{1482+4.95}{1482-4.95}\\\Rightarrow f_{Lr}=22.14745\ kHz[/tex]
[tex]f_{Se}=22\ kHz[/tex]
[tex]\Delta f=f_{Lr}-f_{Se}\\\Rightarrow \Delta f=22.14745-22\\\Rightarrow \Delta f=0.14745\ kHz\\\Rightarrow \Delta f=147.45\ Hz[/tex]
The difference in wavelength is 147.45 Hz
Water from a 100 ft x 200 ft lawn converges into a gully. During a 1-hour storm, discharge into the gully is 0.37 cfs (ft3/s). The interception by grass is 0.02 acre-ft. Overall, 70% of the rain is infiltrated. What is the amount of rainfall?
Answer:
7344 ft^3
Explanation:
rain flown = 0.37 ft^3/s * 3600 s/hr = 1332 ft^3
rain intercepted by grass = 0.02 acre-ft = 0.02 * 43559.9 = 871.2 ft^3
AS 1 acre-ft = 43559.9 ft^3
LET THE TOTAL RAINFALL BE 'X'
Thus the equations become
x = 0.7x + 871.2 +1332 (as 70 % of rain in infiltrated)
x = 7344 ft^3
Two charged particles are a distance of 1.62 m from each other. One of the particles has a charge of 7.10 nc, and the other has a charge of 4.42 nc. (a) What is the magnitude (in N) of the electric force that one particle exerts on the other? (b) is the force attractive or repulsive? O attractive O repulsive
Answer:
A. F=107.6nN
B. Repulsive
Explanation:
According to coulombs law, the force between two charges is express as
F=(Kq1q2) /r^2
If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.
Note the constant K has a value 9*10^9
Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m
If we substitute values we have
F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)
F=(282.4×10^-9)/2.6244
F=107.6×10^-9N
F=107.6nN
B. Since the charges are both positive, the force is repulsive
The magnitude of the electric force between the two particles is approximately 0.012 N, according to Coulomb's Law. The force could be either attractive or repulsive, depending on whether the charges are the same or opposite, respectively.
Explanation:The subject of this question is Physics, specifically about Coulomb’s Law, which deals with the electric force between two charges.
(a) According to Coulomb’s Law, the electric force (F) between two charges is directly proportional to the product of their charges (q1 and q2) and inversely proportional to the square of the distance (d) between them. This is mathematically represented as F = k*q1*q2/d^2, where k is Coulomb's constant (8.988 x 10^9 Nm^2/C^2). Substituting in the given values, we find F = (8.988 * 10^9 Nm^2/C^2) * (7.10 * 10^-9 C) * (4.42 * 10^-9 C) / (1.62 m)^2, resulting in an electric force of approximately 0.012 N.
(b) Whether the force is attractive or repulsive depends on the nature of the charges. Same charges repel each other, while opposite charges attract each other. As the charges are not specified, we cannot definitively answer this part of the question.
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When two point charges are 2.0 cm apart, each one experiences a 1.0-N electric force due to the
other charge. If they are moved to a new separation of 8.0 cm, the electric force on each of them
is closest to
A) 1.0 N.
B) 4.0 N.
C) 16 N.
D) 0.25 N.
E) 0.063 N.
Answer:
option (E)
Explanation:
F = 1 N
r = 2 cm
r' = 8 cm
F' = ?
According to the Coulomb's law
[tex]F \alpha \frac{1}{r^{2}}[/tex]
So, [tex]\frac{F'}{F}=\left (\frac{r}{r'} \right )^{2}[/tex]
[tex]\frac{F'}{1}=\left (\frac{2}{8} \right )^{2}[/tex]
F' = 0.063 N
When they are moved to a new separation of 8.0 cm, the electric force on each of them will be closest to 0.063 N. Option E is correct.
Given:
The point charges are initially at a distance of [tex]d=2[/tex] cm.
The initial force experienced by them is [tex]f=1[/tex] N.
The electrostatic force experienced by two charges q and Q is defined as,
[tex]f=K\dfrac{qQ}{r^2}[/tex]
where K is the constant and r is the distance between the charges.
The initial force experienced by them will be,
[tex]f=K\dfrac{qQ}{d^2}=1\rm\; N[/tex]
Now, the final distance between the charges is changed to 8 cm which is equal to 4d.
So, the new force on the charges will be,
[tex]F=K\dfrac{qQ}{(4d)^2}\\F=K\dfrac{qQ}{d^2}\times \dfrac{1}{16}\\F=f\times \dfrac{1}{16}\\F=0.0625\approx0.063\rm\;N[/tex]
Therefore, when they are moved to a new separation of 8.0 cm, the electric force on each of them will be closest to 0.063 N. Option E is correct.
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In fair weather, the ground may become charged such that there is an electric field just above the surface of the Earth, pointing down toward the ground. What is the electric charge of the ground in this situation?
The electric field is defined as the electric force per unit of charge. The direction of the field is taken as the direction of the force it would exert on a positive test load. An electric field is always directed from its positive charge to the negative. Under this condition if the terrestrial electric field is pointing towards the ground, the terminal that is pointing is the negative and that of its surface is the positive. This implies that the sign of the charge of the electric field of the soil will be negative.
In a particular crash test, an automobile of mass 1271 kg collides with a wall and bounces back off the wall. The x components of the initial and final speeds of the automobile are 19 m/s and 3.3 m/s, respectively. If the collision lasts for 0.16 s, find the magnitude of the impulse due to the collision. Calculate the magnitude of the average force exerted on the automobile during the collision.
Answer:
28343.3 kgm/s
177145.625 N
Explanation:
u = Initial velocity = 19 m/s
v = Final velocity = -3.3 m/s (opposite direction)
m = Mass of car = 1271 kg
t = Time taken = 0.16 s
Impulse is given by
[tex]J=m(v-u)\\\Rightarrow J=1271(-3.3-19)\\\Rightarrow J=-28343.3\ kgm/s[/tex]
The magnitude of the impulse due to the collision is 28343.3 kgm/s
Force given by
[tex]F=\dfrac{J}{t}\\\Rightarrow F=\dfrac{-28343.3}{0.16}\\\Rightarrow F=-177145.625\ N[/tex]
The magnitude of the average force exerted on the automobile is 177145.625 N
Final answer:
The magnitude of the impulse during the collision is 28345.3 kg·m/s, and the magnitude of the average force exerted on the automobile is 177158 N.
Explanation:
In the crash test scenario described, the impulse experienced by the automobile can be determined by using the change in momentum due to the collision. Momentum (p) is the product of mass (m) and velocity (v), and impulse is the change in momentum which can be calculated using the initial (vi) and final velocities (vf) of the car:
Impulse = Change in momentum
= m × (vf - vi)
For the automobile:
Mass (m) = 1271 kgInitial velocity (vi) = 19 m/s (positive in the direction of motion)Final velocity (vf) = -3.3 m/s (negative since the car bounces back)Impulse = 1271 kg × (-3.3 m/s - 19 m/s)
= 1271 kg × (-22.3 m/s)
= -28345.3 kg·m/s (to one decimal placing)
The magnitude of the impulse is the absolute value:
28345.3 kg·m/s.
To calculate the average force exerted on the automobile, we use the fact that impulse also equals the average force (Favg) multiplied by the time of impact (t):
Impulse = Favg × t
Thus, the average force can be found by dividing the impulse by the collision time.
Favg = Impulse / t
= 28345.3 kg·m/s / 0.16 s
= 177158.1 N (to one decimal placing)
The magnitude of the average force is the absolute value: 177158 N.
Two satellites A and B of the same mass are orbiting Earth in concentric orbits. The distance of satellite B from Earth’s center is twice that of satellite A. What is the ratio of the tangential speed of B to that of A?
Answer:
ratio of tangential velocity of satellite b and a will be 0.707
Explanation:
We have given distance of satellite B from satellite A is twice
So [tex]r_b=2r_a[/tex]
Tangential speed of the satellite is given by
[tex]v=\sqrt{\frac{GM}{r}}[/tex], G is gravitational constant. M is mass of satellite and r is distance from the earth
We have to find the ratio of tangential velocities of b and a
From the relation we can see that tangential velocity is inversely proportional to square root of distance from earth
So [tex]\frac{v_b}{v_a}=\sqrt{\frac{r_a}{r_b}}[/tex]
[tex]\frac{v_b}{v_a}=\sqrt{\frac{r_a}{2r_a}}[/tex]
[tex]\frac{v_b}{v_a}=\sqrt{\frac{1}{2}}[/tex]
[tex]\frac{v_b}{v_a}=0.707[/tex]
So ratio of tangential velocity of satellite b and a will be 0.707
The ratio of tangential velocity of satellite B to satellite A is 0.707.
Tangential Speed of Satellite
The Tangential velocity is the linear speed of any object moving along a circular path. The tangential speed of the satellite is given below.
[tex]v = \sqrt{\dfrac{Gm}{r}}[/tex]
Where v is the velocity, m is the mass and r is the circular distance. G is the gravitational constant.
Given that mass of both the satellite is the same. Let us consider the mass of both satellites as m. The distance of satellite B from Earth’s center is twice that of satellite A.
Let us consider that the distance of satellite A from the center of the earth is r. The distance of satellite B from the center of the earth is 2r.
The tangential speed of satellite A is,
[tex]v_a = \sqrt{\dfrac {Gm}{r}}[/tex]
The tangential speed of satellite B is,
[tex]v_b = \sqrt{\dfrac {Gm}{2r}}[/tex]
In the ratio form, the tangential speed of both satellites is given below.
[tex]\dfrac {v_b}{v_a} = \dfrac {\sqrt{\dfrac {GM}{2r}} }{\sqrt{\dfrac {Gm}{r}} }[/tex]
[tex]\dfrac {v_b}{v_a} = \sqrt{\dfrac{1}{2}}[/tex]
[tex]\dfrac {v_b}{v_a} = 0.707[/tex]
Hence we can conclude that the ratio of tangential velocity of satellite B to satellite A is 0.707.
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If y(t)y(t) describes the position with time, what is the proper formula for velocity with time? (Recall velocity is related to the derivative of position with time.)
Answer:
Although this question is not complete, I would give a general solution to this kind of problems.
If y(t) describes the position of a body with time such that
y(t) = at^(n) + bt^(m) + C
Then
V(t) = dy(t)/dt = ant^(n-1) + bmt^(m-1)
Explanation:
As an example supplies the position of a particle is given by
y(t) = 4t³- 3t² + 9
V(t) = 4x3t²- 3x2t¹
V(t) = d(t)/dt = 12t² - 6t.
Another example,
If y(t) = 15t³ - 2t² + 30t -80
V(t) = d(t)/dt = 15x3t² - 4t +30 = 45t² + 4t + 30.
Basically, in the equations above the powers of t reduces by one when computing the velocity function from y(t) by differentiation (calculating the derivative of y(t)). The constant term C (9 and 80 in the functions of y(t) in examples 1and 2 above) reduces to zero because the derivative of a constant (and ordinary number without the t attached to it) is always zero.
One last example,
y(t) = 2t^6 -3t²
V(t) = d(t)/dt = 12t^5 - 6t