Answer:
[tex]v(1.5)=0.7648\ m/s[/tex]
Explanation:
Dynamics
When a particle of mass m is subject to a net force F, it moves at an acceleration given by
[tex]\displaystyle a=\frac{F}{m}[/tex]
The particle has a mass of m=2.37 Kg and the force is horizontal with a variable magnitude given by
[tex]F=2cos1.1t[/tex]
The variable acceleration is calculated by:
[tex]\displaystyle a=\frac{F}{m}=\frac{2cos1.1t}{2.37}[/tex]
[tex]a=0.8439cos1.1t[/tex]
The instant velocity is the integral of the acceleration:
[tex]\displaystyle v(t)=\int_{t_o}^{t_1}a.dt[/tex]
[tex]\displaystyle v(t)=\int_{0}^{1.5}0.8439cos1.1t.dt[/tex]
Integrating
[tex]\displaystyle v(1.5)=0.7672sin1.1t \left |_0^{1.5}[/tex]
[tex]\displaystyle v(1.5)=0.7672(sin1.1\cdot 1.5-sin1.1\cdot 0)[/tex]
[tex]\boxed{v(1.5)=0.7648\ m/s}[/tex]
Final answer:
To find the x-component of velocity at t= 1.5 seconds for a particle under a sinusoidally varying force, we derive the equation of motion, integrate the acceleration, and calculate the velocity, resulting in [tex]v_x[/tex] = 0.567 m/s.
Explanation:
The question pertains to finding the x-component of velocity (vx) of a particle at t= 1.5 seconds, given the mass and the sinusoidally varying force. Since the force applied on the particle varies as Fx = F0cos(ωt), where F0 = 2 N and ω = 1.1 rad/s, we can find the acceleration and then integrate it with respect to time to find velocity. The acceleration ax is given by Fx/m = (2cos(1.1t))/2.37. To find the change in velocity, we integrate ax with respect to time, giving us vx = ∫ ax dt = ∫ (2cos(1.1t))/2.37 dt. Evaluating this integral from 0 to 1.5 s, we use the definite integral which simplifies to vx(t) = (2/2.37)(sin(1.1(1.5))-sin(1.1(0)))/1.1. The calculation yields vx = 0.567 m/s, indicating the particle's velocity in the x-direction at 1.5 seconds.
A Rankine oval is formed by combining a source-sink pair, each having a strength of 36 ft2/s and separated by a distance of 15 ft along the x axis, with a uniform velocity of 12 ft/s (in the positive x direction). Determine the length of the oval.
Answer:
0.28 ft
Explanation:
We are given that
Strength=m=[tex]36ft^2/s[/tex]
Distance between source and sink=15 ft
Distance between the sink of the source and origin=[tex]a=\frac{15}{2}[/tex] ft
Uniform velocity, U=12 ft/s
We have to find the length of the oval.
Formula to find the half length of the body
[tex]\frac{l}{a}=(\frac{m}{\pi Ua}+1)^{\frac{1}{2}}[/tex]
Where a=Distance between sink of source and origin
U=Uniform velocity
m=Strength
l=Half length
Using the formula
[tex]\frac{l}{\frac{15}{2}}=(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}[/tex]
[tex]l=\frac{2}{15}(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}[/tex]
[tex]l=0.14[/tex]
Length of oval=[tex]2l=2(0.14)=0.28 ft[/tex]
A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed. The ball is in contact with the wall 0.050 s. What is the magnitude of the average force exerted on the wall by the ball?
Answer:
[tex]F=800N[/tex]
Explanation:
The average force is defined as the mass of the body multiplied by its average velocity over the contact time. According to the third Newton's law, the magnitude of the average force exerted on the wall by the ball is equal to the magnitude of the average force exerted on the ball by the wall. Thus:
[tex]F=m\frac{v_f-v_i}{t}\\F=0.80kg\frac{25\frac{m}{s}-(-25\frac{m}{s})}{0.05s}\\F=800N[/tex]
Answer:
800 N
Explanation:
Force: This can be defined as the product of mass and acceleration. The S.I unit of force is Newton(N).
From the question,
F = m(v-u)/t ....................... Equation 1
Where F = force exerted on the wall by the ball, m = mass of the ball, v = final velocity, u = initial velocity, t = time.
Note: Assuming the direction of the initial motion to be negative
Given: m = 0.80 kg, v = 25 m/s ( bounce back), u = -25 m/s, t = 0.05 s
Substitute into equation 1
F = 0.8[25-(25)]/0.05
F = 0.8(50)/0.05
F = 0.8(1000)
F = 800 N
When a certain ideal gas thermometer is placed in water at the triple point, the mercury level in the right arm is 846 mm above the reference mark. Part A How far is the mercury level above the reference mark when this thermometer is placed in boiling water at a location where the atmospheric pressure is 1.00 atm
Answer:
P =1156mmHg
Explanation:
Given
Ptriple = 846mmHg
Ttriple = 273.16K
T = temperature of boiling water = 100°C = (273.16 +100)K = 373.16K
P/Ptriple = T/Ttriple
P = T/Ttriple × Ptriple
P = 373.16/×273.6 × 846
P = 1156mmHg.
The temperature used in this solution is the absolute temperature because the gas thermometer use the absolute temperature scale. On this scale (Kelvin temperature scale) the zero point, Absolute zero is 273.16K.
Answer:
The mercury level is [tex]1155.68 mm[/tex] above the reference levelExplanation:
Ideal gas obeys the law , [tex]PV = RT[/tex]
When volume remains constant
[tex]\frac{P2}{P1} = \frac{T2}{T1}[/tex]
Temperature of triple point of water is [tex]T1 = 273.16K[/tex]
Temperature of normal boiling point of water is [tex]T2 = 373.15 K[/tex]
therefore,
[tex]P2 = 846 * \frac{373.15}{273.16}\\\\P2 = 1155.68 mm[/tex]
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7–14 Engine oil at 80°C flows over a 12-m-long flat plate whose temperature is 30°C with a velocity of 3 m/s. Determine the total drag force and the rate of heat transfer over the entire plate per unit width.
Answer:
Total drag force is 88.74 N
Rate of heat transfer over the entire plate per unit width is 26451.6 W
Explanation:
The rate of heat transfer over the entire plate per unit width can be obtained by applying the Newton's law of cooling. The pictures attached are step by step solution to the question;
A large fraction of the ultraviolet (UV) radiation coming from the sun is absorbed by the atmosphere. The main UV absorber in our atmosphere is ozone, O3. In particular, ozone absorbs radiation with frequencies around 9.38×1014 Hz . What is the wavelength λ lambda of the radiation absorbed by ozone?
Answer:
[tex]3.2\times 10^{-7}\ m[/tex] or 0.32 μm.
Explanation:
Given:
The radiations are UV radiation.
The frequency of the radiations absorbed (f) = [tex]9.38\times 10^{14}\ Hz[/tex]
The wavelength of the radiations absorbed (λ) = ?
We know that, the speed of ultraviolet radiations is same as speed of light.
So, speed of UV radiation (v) = [tex]3\times 10^8\ m/s[/tex]
Now, we also know that, the speed of the electromagnetic radiation is related to its frequency and wavelength and is given as:
[tex]v=f\lambda[/tex]
Now, expressing the above equation in terms of wavelength 'λ', we have:
[tex]\lambda=\frac{v}{f}[/tex]
Now, plug in the given values and solve for 'λ'. This gives,
[tex]\lambda=\frac{3\times 10^8\ m/s}{9.38\times 10^{14}\ Hz}\\\\\lambda=3.2\times 10^{-7}\ m\\\\\lambda=3.2\times 10^{-7}\times 10^{6}\ \mu m\ [1\ m=10^6\ \mu m]\\\\\lambda=3.2\times 10^{-1}=0.32\ \mu m[/tex]
Therefore, the wavelength of the radiations absorbed by the ozone is nearly [tex]3.2\times 10^{-7}\ m[/tex] or 0.32 μm.
The current supplied by a battery in a portable device is typically about 0.151 A. Find the number of electrons passing through the device in five hours.
Answer:
n = 1.7*10²² electrons.
Explanation:
As the current, by definition, is the rate of change of charge, assuming that the current was flowing at a steady rate of .151 A during the 5 hours, we can find the total charge that passed perpendicular to the cross-section of the circuit, as follows:[tex]I =\frac{\Delta q}{\Delta t} \\ \\ \Delta q = I* \Delta t \\ \\ \Delta t = 5hs*\frac{3600s}{1h} = 18000 s[/tex]
⇒ Δq = I * Δt = 0.151 A * 18000 s = 2718 C
As this charge is carried by electrons, we can express this value as the product of the elementary charge e (charge of a single electron) times the number of electrons flowing during that time, as follows:Δq = n*e
Solving for e:[tex]n = \frac{\Delta q}{e} =\frac{2718C}{1.6-19C} = 1.7e22 electrons.[/tex]
Explanation:
Below is an attachment containing the solution.
To understand the terms in Faraday's law and to be able to identify the magnitude and direction of induced emf. Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux. Mathematically, it can be written as
Answer: [tex]V_{\epsilon}\propto \frac{d\phi_{B}}{dt}[/tex]
Explanation:
A direct proportionality means a linear relationship between two variables and rate of change means an application of derivatives. Hence, the mathematical model is:
[tex]V_{\epsilon}\propto \frac{d\phi_{B}}{dt}[/tex]
Faraday's Law of Induction, primarily focused on Physics, reveals that the induced electromotive force in a coil is proportional to the rate of change of magnetic flux through the coil and is also dependent on the number of coil turns.
Explanation:Understanding Faraday's Law of Induction
Faraday's Law of Induction is a critical concept in electromagnetism, underlying the working principles of various electrical devices. The induced electromotive force (emf) is directly proportional to the time rate of change of magnetic flux. In simpler terms, when the magnetic environment of a circuit changes, an emf is induced. Faraday's Law is given by the equation EMF = -N(ΔΦ/Δt), where 'N' represents the number of turns in a coil, and (ΔΦ/Δt) is the rate of change of magnetic flux over time. This equation highlights three main factors influencing the induced EMF:
The induced EMF is proportional to the number of loops in the coil.It is proportional to the change in magnetic flux, ΔΦ.It is inversely proportional to the time interval, Δt, over which the change occurs.The negative sign in Faraday's Law is a reflection of Lenz's Law, which indicates that the induced emf always works to oppose the change in magnetic flux that produces it.
Consider an electron on the surface of a uniformly charged sphere of radius 1.2 cm and total charge 1.4 10-15 C. What is the "escape speed" for this electron
Answer:
Explanation:
electric potential of electron
= k Qq / r
= - 9 x 10⁹ x 1.4 x 10⁻¹⁵ x 1.6 x 10⁻¹⁹ / ( 1.2 x 10⁻² )
= - 16.8 x 10⁻²³ J
If v be the escape velocity
1/2 m v² = potential energy of electron
= 1/2 x 9.1 x 10⁻³¹ x v² = 16.8 x 10⁻²³
v² = 3.69 x 10⁸
v = 1.92 x 10⁴ m /s
Electroplating uses electrolysis to coat one metal with another. In a copper-plating bath, copper ions with a charge of 2ee move through the electrolyte from the copper anode to the cathode; the metal object to be plated.
If the current through the system is 1.2 A, how many copper ions reach the cathode each second?
Answer:
1.2 A of current will send (3.744 × 10¹⁸) ions to the cathode per second.
Explanation:
According to Faraday's second law of electrolysis, the amount of ions/mass of substance deposited at an electrode depends on its equivalent weight.
For a divalent ion, it will require 2F of electricity per mole.
1 F = 96500 C
Amount of electricity that passes through the electrolyte per second = (magnitude of current) × (time) = It = (1.2 × 1) = 1.2 C
2F (2×96500C) of electricity will deposit 1 mole of Copper ions
That is,
193000 C of electricity will deposit 1 mole of Copper.
1.2 C will deposit (1.2×1/193000); 0.0000062176 mole of Copper.
1 mole of Copper contains (6.022 × 10²³) ions according to the Avogadro's constant.
0.0000062176 mole of Copper will contain (0.0000062176 × 6.022 × 10²³) ions = (3.744 × 10¹⁸) ions.
Therefore, 1.2 A of current will send (3.744 × 10¹⁸) ions to the cathode per second.
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A proton (being 1836 times heavier than an electron) gains how much energy when moving through a potential increase of one volt?
Answer:
1.6*10⁻¹⁹ J = 1 eV
Explanation:
As the potential is defined as energy per unit charge, the increase in the electrical potential energy of a proton, can be written as follows:[tex]\Delta U_{e} = e*\Delta V[/tex]
where e = 1.6*10⁻¹⁹ C, and ΔV = 1 V = 1 J/C⇒ [tex]\Delta U_{e} = e*\Delta V = 1.6e-19 C* 1.0 V = 1.6e-19 J = 1 eV[/tex]
As the electrical potential energy only depends on the charge and the voltage, the mass has no influence.Explanation:
Below is an attachment containing the solution.
A battery has a terminal voltage of 12.0 V when no current flows. Its internal resistance is 2.0 Ω. If a 7.2 Ω resistor is connected across the battery terminals, what is the terminal voltage and what is the current through the 7.2 Ω resistor?
Answer:
Voltage= 9.4 V
Current= 1.3 A
Explanation:
Equivalent resistance will be the sum of two resistors hence
[tex]R_{equ}[/tex]=2+7.2=9.2Ω
From Ohm's law, V=IR where I is current, R is equivalent resistance and V is voltage
Terminal voltage will be given by V=e-Ir where r is internal resistance, I is current and e is electromotive force.
From Ohm's law, current is given by dividing emf of the battery by the equivalent resistance hence
[tex]I=\frac {e}{R_{equ}}\\I=\frac {12}{9.2}=1.304347826A\approx 1.3A[/tex]
This is the current through external resistor
Terminal voltage will be given by V=e-Ir hence V=12-(1.3*2)=9.4 V
Final answer:
The terminal voltage across the 7.2
Ω resistor when connected to a 12.0 V battery with 2.0
Ω internal resistance is approximately 9.392 V, and the current flowing through the resistor is approximately 1.304 A.
Explanation:
The student is asking about the calculation of terminal voltage and current through a resistor when connected to a battery that has internal resistance. This is a problem related to the topic of electrical circuits in Physics.
To find the current through the 7.2
Ω resistor, use Ohm's law which states that V = IR, where V is voltage, I is current, and R is resistance. Since the battery has an internal resistance, we need to consider the total resistance in the circuit which is the sum of the internal resistance and the resistance of the 7.2
Ω resistor.
Total resistance (Rtotal) = Internal resistance (r) + External resistance (R) = 2.0
Ω + 7.2
Ω = 9.2
Ω
Using Ohm's Law, the current (I) through the circuit can be calculated as follows:
I = V / Rtotal = 12.0 V / 9.2
Ω = 1.304
A (approximately).
To find the terminal voltage (Vterminal) across the 7.2
Ω resistor when the current is flowing, we account for the voltage drop across the internal resistance:
Vterminal = emf - I × r = 12.0 V - (1.304
A × 2.0
Ω) = 12.0 V - 2.608 V = 9.392 V (approximately).
The terminal voltage across the 7.2
Ω resistor when it is connected to the battery is 9.392 V, and the current flowing through the resistor is 1.304
A.
If blocks A and B of mass 10 kg and 6 kg respectively, are placed on the inclined plane and released, determine the force developed in the link. The coefficients of kinetic friction between the blocks and the inclined plane are mA = 0.1 and mB = 0.3. Neglect the mass of the link.
Answer:
The force developed in the link is 6.36 N.
Explanation:
Given that,
Mass of block A = 10 kg
Mass of block B = 6 kg
Coefficients of kinetic friction [tex]\mu_{A}= 0.1[/tex]
Coefficients of kinetic friction [tex]\mu_{B}= 0.3[/tex]
Suppose the angle is 30°
We need to calculate the acceleration
Using formula of acceleration
[tex]a=\dfrac{m_{A}g\sin\theta+m_{B}g\sin\theta-\mu_{A}m_{A}g\cos\theta-\mu_{A}m_{A}g\cos\theta}{m_{A}+m_{B}}[/tex]
Put the value into the formula
[tex]a=\dfrac{10\times9.8\sin30+6\times9.8\sin30-0.1\times10\times9.8\times\cos30-0.3\times6\times9.8\times\cos30}{16}[/tex]
[tex]a=3.415\ m/s^2[/tex]
We need to calculate the force developed in the link
For block A,
Using balance equation
[tex]ma=m_{A}g\sin\theta-\mu m_{A}g\cos\theta-T[/tex]
[tex]T=ma+\mu m_{A}g\cos\theta-m_{A}g\sin\theta[/tex]
Put the value into the formula
[tex]T=10\times3.415+0.1\times10\times9.8\times\cos30-10\times9.8\times\sin30[/tex]
[tex]T=-6.36\ N[/tex]
Negative sign shows the opposite direction of the force.
Hence, The force developed in the link is 6.36 N.
In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop.
(c)How much work does the gravitational force do on the ball while it is compressing the mattress?
(d)How much work does the mattress do on the ball?
(e)Now model the mattress as a single spring with an unknown spring constant k, and consider the whole system formed by the ball, the earth and the mattress. By how much does the potential energy of the mattress increase as it compresses?
(f)What is the value of the spring constant k?
Answer:
(c) 10.29 J
(d) 113.19 J
(e) 113.19 J
(f) 10061 N/m
Explanation:
15 cm = 0.15 m
Let g = 9.8 m/s2
(c) The work done by gravitational force is the product of gravity force and the distance compressed
[tex]E_p = mgx = 7*9.8*0.15 = 10.29 J[/tex]
(d) By using law of energy conservation with potential energy reference being 0 at the maximum compression point. As the ball falls and come to a stop at the compression point, its potential energy is transferred to elastic energy, which is the work that the mattress does on the ball:
[tex]E_p = E_e[/tex]
[tex]E_e = mgh[/tex]
where h = 1.5 + 0.15 = 1.65 m is the vertical distance that it falls.
[tex]E_e = 7*9.8*1.65 = 113.19 J[/tex]
(e) Before the compression, the potential energy of the mattress is 0. After the compression, the potential energy is 113.19J. So it has increased by 113.19J due to the potential energy transferred from the falling ball.
(f) [tex]E_e = 113.19 = kx^2/2[/tex]
[tex]k0.15^2/2 = 113.19[/tex]
[tex]k = 10061 N/m[/tex]
Answer:
(C) Wg = 113.2J
(D) Wm = 10.3J
(E) E = 1/2kx² + mgh where h is the height above the mattress and x is the compressed distance in the mattress.
(F) k = 457N/m.
Explanation:
See attachment below.
You are given three iron rods. Two of the rods are magnets but the third one is not. How could you use the three rods to definitively identify the unmagnetized rod?
a. Test the rods a pair at a time in all possible orientations to determine which two rods have the strongest force between them; the leftover rod is unmagnetized.
b. Find one pair of rods that attract when placed end‑to‑end, regardless of orientation; the leftover rod is unmagnetized.
c. Test different pairs of rods in all possible orientations to find a rod that is not attracted to either of the other two rods; that rod is unmagnetized.
d. Test the rods one pair at a time in each possible end‑to‑end orientation until finding a pair that repels; the leftover rod is unmagnetized.
Answer:
d. Test the rods one pair at a time in each possible end‑to‑end orientation until finding a pair that repels; the leftover rod is unmagnetized.
Explanation:
Repulsion is the surest test of magnetisation . If two objects repel each other , they must be magnets . If two objects attract each other , they may not be magnets. In later case , one may be non - magnet. So the last test will help us identify the unmagnetised rod.
A carbon resistor is to be used as a thermometer. On a winter day when the temperature is 4.00 ∘C, the resistance of the carbon resistor is 217.0 Ω . What is the temperature on a spring day when the resistance is 215.1 Ω ? Take the temperature coefficient of resistivity for carbon to be α = −5.00×10−4 C−1 .
Answer:
21.5 °C.
Explanation:
Given:
α = −5.00 × 10−4 °C−1
To = 4°C
Ro = 217 Ω
Rt = 215.1 Ω
Rt/Ro = 1 + α(T - To)
215.1/217 = 1 + (-5 × 10^−4) × (T - 4)
-0.00876 = -5 × 10^−4 × (T - 4)
17.5 = (T - 4)
T = 21.5 °C.
The temperature on a spring day can be calculated by using the relationship between temperature and resistance. Given the initial resistance, the change in resistance and the temperature coefficient of resistivity for carbon, the spring day temperature can be deduced to be 21.51 °C.
Explanation:The change in temperature can be determined by using the equation ΔR = R0αΔT, where ΔR is the change in resistance, R0 is the initial resistance, α is the temperature coefficient of resistivity and ΔT is the change in temperature. The initial resistance of the carbon resistor is given as 217.0 Ω, the resistance on a spring day was 215.1 Ω, so the change in resistance ΔR is 215.1 - 217.0 = -1.9 Ω. The temperature coefficient for carbon is given as -5.00×10-4 C-1, and α is -5.00×10-4 C-1. We can rearrange the above equation to solve for ΔT: ΔT = ΔR / (R0 * α) = -1.9 Ω / (217.0 Ω * -5.00×10-4 C-1) = 17.51 C.
Thus, the temperature on the spring day would be the winter day temperature plus the calculated change in temperature: T_spring = T_winter + ΔT = 4.00 °C + 17.51 °C = 21.51 °C.
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A 2-kg disk is constrained horizontally but is free to move vertically. The disk is struck from below by a vertical jet of water. The speed and diameter of the water jet are 10 m/s and 25 mm at the nozzle exit. Obtain a general expression for the speed of the water jet as a function of height, h. Find the height to which the disk will rise and remain stationary
Answer:
h = 4.281 m
Explanation
Given Data,
Weight of disk = 2kg
Speed of water jet (V) = 10 m/s
diameter of water jet = 25 mm
Cal. the velocity of the water jet as function height by applying Bernoulli's eqtn of water surface to the jet
[tex]\frac{P}{\rho } +\frac{V^{2}}{2} + gz[/tex] = Constant
[tex]\frac{V_{0}^{2}}{2} + g(0) = \frac{V^{2}}{2} + g[/tex]
[tex]V=\sqrt{V_{0}^{2}-2gh}[/tex]
Relation between [tex]V_{0}[/tex] & V
[tex]m=\rho V_{0}A_{0}[/tex]
[tex]\rho VA=\rho V_{0}A_{0}[/tex]
[tex]VA= V_{0}A_{0}[/tex]
Momentum
[tex]F_{w}+F_{d}= \frac{\partial }{\partial t}\int_{cv} w\rho dA + \int_{cs} w\rho V dA[/tex]
[tex]-mg = w_{1}[-\rho VA] +w_{2}[\rho VA][/tex]
[tex]w_{1}[/tex] = V
[tex]w_{2}[/tex] = 0
[tex]mg = \rho V^{2}A[/tex]
[tex]mg = \rho VV_{0} A[/tex]
[tex]mg = \rho VV_{0} A_{0}[/tex]
[tex]mg = \rho V_{0} A_{0} \sqrt{V_{0} ^{2}-2gh }[/tex]
Solving for h
[tex]h = \frac{1}{2g}[V_{0}^{2}-\frac{mg}{\rho V_{0}A_{0}}][/tex]
g is gravitational acc.
[tex]= \frac{1}{2\times 9.81}[10^{2}-(\frac{2\times 9.81}{999\times10\times\frac{\Pi }{4}\times(0.025)^{2}})^{2} ][/tex]
[tex]= \frac{1}{19.62}[100-(\frac{19.68}{4.9038})^{2}][/tex]
[tex]= \frac{100-16.0078}{19.62}[/tex]
h = 4.281 m
h of disk on which it remains stationary.
A physics professor is pushed up a ramp inclined upward at an angle 33.0° above the horizontal as he sits in his desk chair that slides on frictionless rollers. The combined mass of the professor and chair is 90.0 kg. He is pushed a distance 2.00 m along the incline by a group of students who together exert a constant horizontal force of 600 N. The professor's speed at the bottom of the ramp is 2.30 m/s.
A) Find velocity at the top of the ramp.
Answer:
2.51 m/s
Explanation:
Parameters given:
Angle, A = 33°
Mass, m = 90kg
Inclined distance, D = 2m
Force, F = 600N
Initial speed, u = 2.3m/s
From the relationship between work and kinetic energy, we know that:
Work done = change in kinetic energy
W = 0.5m(v² - u²)
We also know that work done is tẹ product of force and distance, hence, net work done will be the sum of the total work done by the force from the students and gravity.
Hence,
W = F*D*cosA - w*D*sinA
w = m*9.8 = weight
=> W = 600*2*cos33 - 90*9.8*2*sin33
W = 45.7J
=> 45.7 = 0.5*m*(v² - u²)
45.7 = 0.5*90*(v² - 2.3²)
45.7 = 45(v² - 5.29)
=> v² - 5.29 = 1.016
v² = 6.306
v = 2.51 m/s
The final velocity is 2.51 m/s
Answer: v = 3.26 m/s
Explanation:
Force applied (F) = 900 N,
mass of professor and chair (m) = 90kg,
g = acceleration due to gravity = 9.8 m/s²,
Fr = frictional force = 0 ( since from the question, the rollers the chair and professor are moving is frictionless),
θ = angle of inclination = 33°
a = acceleration of object =?
From newton's second law of motion, we have that
F - (mg sinθ +Fr) = ma
Where mg sinθ is the horizontal component of the weight of the mass of professor and chair due to the inclination of the ramp.
But Fr = 0
Hence, we have that
F - mg sinθ = ma
600 - (90×9.8×sin30) = 90 (a)
600 - 480.371= 90a
119.629 = 90a
a = 119.629/ 90
a = 1.33 m/s².
But the body started the morning (at the bottom of the ramp) with a velocity of 2.30 m/s²
Hence u = initial velocity = 2.30 m/s², a = acceleration = 1.33m/s², v = final velocity =?, s = distance covered = 2m
By using equation of motion for a constant acceleration, we have that
v² = u² + 2as
v ² = (2.3)² +2(1.33)×(2)
v² = 5.29 + 5.32
v² = 10.61
v = √10.61
v = 3.26 m/s
A 20-kg child starts at the center of a playground merry-go-round that has a radius of 1.5 mm and rotational inertia of 500kg⋅m2 and walks out to the edge. The merry-go-round has a rotational speed of 0.20 s−1 when she is at the center.
What is its rotational speed when she gets to the edge?
Answer:
w = 0.189 rad/ s
Explanation:
This exercise we work with the conservation of the moment, the system is made up of the merry-go-round and the child, for which we write the moment of two instants
Initial
L₀ = I₀ w₀
Final
[tex]L_{f}[/tex] = I w
L₀ = L_{f}
I₀ w₀ = I_{f} w
.w = I₀/I_{f} w₀
The initial moment of inertia is
I₀ = 500 kg. m2
The final moment of inertia
[tex]I_{f}[/tex] = 500 + m r²
I_{f} = 500 + 20 1.5
I_{f} = 530 kg m²
Initial angular velocity
w₀ = 0.20 rad / s
Let's calculate
w = 500/530 0.20
w = 0.189 rad / s
My Notes (a) A 41 Ω resistor is connected in series with a 6 µF capacitor and a battery. What is the maximum charge to which this capacitor can be charged when the battery voltage is 6 V? (When entering units, use micro for the metric system prefix µ.)
Explanation:
The give data is as follows.
C = 6 [tex]\mu F[/tex] = [tex]6 \times 10^{-6} F[/tex]
V = 6 V
Now, we know that the relation between charge, voltage and capacitor for series combination is as follows.
Q = CV
= [tex]6 \times 10^{-6} F \times 6 V[/tex]
= [tex]36 \times 10^{-6} C[/tex]
or, = 36 [tex]\mu C[/tex]
Thus, we can conclude that maximum charge of the given capacitor is 36 [tex]\mu C[/tex].
Answer:
Explanation:
capacitance, C = 6 μF
Voltage, V = 6 V
Let the maximum charge is Q.
Q = C x V
Q = 6 x 6 = 36 μC
Two long parallel wires are a center-to-center distance of 4.90 cm apart and carry equal anti-parallel currents of 4.10 A. Find the magnetic field intensity at the point P which is equidistant from the wires. (R = 6.00 cm).
Answer:
The magnetic field intensity at the point P, equidistant from the wires is
9.566 x [tex]10^{-5}[/tex] T
Explanation:
The magnetic field intensity can be obtained by resolving the vertical and horizontal components of the magnetic field. The step by step calculation is contained in the pictures attached;
Human eyes detect only a very small band of the electromagnetic spectrum. However, some animals and insects see in higher and lower frequency bands than humans do. For this discussion, first do some research to find three examples of animals or insects that see in higher or lower frequency bands.Then, in your initial discussion post, address the following:What were the examples that you found? Did they see in lower or higher frequency bands? List the specific wavelengths used by the animals/insects you chose.Were you surprised by the information that you found? Why or why not?
Answer:
Squids = 450 - 490 nm (Moderate Frequency) (Blue)
Bees = 300 - 650 nm (Lower Frequency Bands)
Frogs = 280 - 580 nm (Very Low Frequency)
Explanation:
All of the above mentioned ranges are compared to that of humans.
I'm just surprised a little bit in the imagination that how these organisms see the world through their unique eyes. On the other hands, they are evolved like this just like we do so that may not be surprising enough. SIKE
Calculate the force required to pull a copper ball of radius 1.69 cm upward through a viscous fluid at a constant speed of 9.3 cm/s. Take the damping constant of the fluid to be 0.884 kg/s\.\*
Answer:
[tex] m = \rho V[/tex]
Since we have an ball we can consider this like a sphere and the volume is given by [tex] V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.69cm)^3 = 20.218 cm^3 = 0.00002022 m^3[/tex]
The density for the copper is approximately [tex] \rho = 8940 kg/m^3[/tex]
So then the mass is :
[tex] m =8940 kg/m^3 * 0.00002022m^3 = 0.1808 Kg[/tex]
And now we have everything in order to replace into the formula for F, like this:[tex] F = 0.1808 Kg *9.8 m/s^2 + 0.884 kg/s * 0.093 m/s= 1.772 +0.975 N = 2.747 N[/tex]And that would be the final answer for this case.
Explanation:
For this case if we assume that we have a damping motion the force action on the vertical direction would be:
[tex] F = mg + bv[/tex]
Where F represent the upward force on the copper ball
m represent the mass
g = 9.8 m/s^2 represent the gravity
b = 0.884 kg/s represent the proportionality constant
v = 9.3 cm/s = 0.093 m/s represent the velocity
We can solve for the mass from the following expression:
[tex] m = \rho V[/tex]
Since we have an ball we can consider this like a sphere and the volume is given by [tex] V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.69cm)^3 = 20.218 cm^3 = 0.00002022 m^3[/tex]
The density for the copper is approximately [tex] \rho = 8940 kg/m^3[/tex]
So then the mass is :
[tex] m =8940 kg/m^3 * 0.00002022m^3 = 0.1808 Kg[/tex]
And now we have everything in order to replace into the formula for F, like this:[tex] F = 0.1808 Kg *9.8 m/s^2 + 0.884 kg/s * 0.093 m/s= 1.772 +0.975 N = 2.747 N[/tex]And that would be the final answer for this case.
A dielectric material is inserted between the charged plates of a parallel-plate capacitor. Do the following quantities increase, decrease, or remain the same as equilibrium is reestablished?
1. Charge on plates (plates remain connected to battery)
2. Electric potential energy (plates isolated from battery before inserting dielectric)
3.Capacitance (plates isolated from battery before inserting dielectric)
4. Voltage between plates (plates remain connected to battery)
5. Charge on plates (plates isolated from battery before inserting dielectric)
6. Capacitance (plates remain connected to battery)
7. Electric potential energy (plates remain connected to battery)
8. Voltage between plates (plates isolated from battery before inserting dielectric)
Answer:
1. Charge on plates (plates remain connected to battery) increases.
2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases.
3.Capacitance (plates isolated from battery before inserting dielectric) increases.
4. Voltage between plates (plates remain connected to battery) remains the same.
5. Charge on plates (plates isolated from battery before inserting dielectric) remains the same.
6. Capacitance (plates remain connected to battery) increases.
7. Electric potential energy (plates remain connected to battery) increases.
8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases.
Explanation:
When a dielectric material is inserted between the plates of a capacitor, the capacitance is increase by a factor K, the dielectric constant.
[tex]C = KC_0[/tex]
By the capacitance formula, the other factors change accordingly.
1. Charge on plates (plates remain connected to battery) increases, because charge and capacitance are directly proportional.
2. Electric potential energy (plates isolated from battery before inserting dielectric) decreases, because potential is inversely proportional to capacitance, and potential energy is given by the following formula
[tex]U = \frac{1}{2}CV^2[/tex]
3.Capacitance (plates isolated from battery before inserting dielectric) increases.
4. Voltage between plates (plates remain connected to battery) stays the same, because the voltage is applied by the battery.
5. Charge on plates (plates isolated from battery before inserting dielectric) remains constant. If the plates isolated from the battery, then the total charge is conserved.
6. Capacitance (plates remain connected to battery) increases when a dielectric is inserted.
7. Electric potential energy (plates remain connected to battery) increases, because when plates remain connected to battery the voltage remains the same. But the capacitance increases. Therefore, electric potential energy increases.
8. Voltage between plates (plates isolated from battery before inserting dielectric) decreases, because voltage is inversely proportional to capacitance.
Final answer:
When a dielectric is inserted into a capacitor, the capacitance increases regardless of whether the plates are connected to a battery or isolated. If connected to a battery, charge on the plates increases while voltage remains the same; if isolated, charge remains the same while voltage decreases.
Explanation:
When inserting a dielectric material into a parallel-plate capacitor, the effect on different parameters depends on whether the capacitor is connected to a battery or isolated:
Charge on plates (plates remain connected to battery): The charge on the plates will increase.
Electric potential energy (plates isolated from the battery before inserting the dielectric): The electric potential energy will decrease due to the polarization of the dielectric.
Capacitance (plates isolated from the battery before inserting the dielectric): The capacitance will increase.
Voltage between plates (plates remain connected to battery): The voltage will remain the same, as it is maintained by the battery.
Charge on plates (plates isolated from battery before inserting dielectric): The charge will remain the same as isolation prevents any charge exchange.
Capacitance (plates remain connected to battery): The capacitance will increase.
Electric potential energy (plates remain connected to the battery): The electric potential energy will increase, as more charge is accumulated due to the dielectric.
Voltage between plates (plates isolated from battery before inserting dielectric): The voltage will decrease because the same charge is now across a larger capacitance.
It is important to note that the introduction of a dielectric material affects the capacitance of a parallel-plate capacitor by reducing the effective electric field and permitting greater charge storage capability for a given voltage.
When an object slides down an inclined plane, the angle between the displacement and the force of gravity is the same as the angle that the surface makes with the horizontal. Suppose you have a 2-kg cart rolling a distance of 1.5 m down a 30° incline. According to Eq. (7.8), what is the work done on it by the force of gravity? Select One of the Following:
(a) 3J
(b) 25.5 J
(c) 2.6J
(d) 14.7 J
Answer:
(d) 14.7 J.
Explanation:
Using the equation,
W = mghsin∅................. equation 1
Where W = work done on the cart by the force of gravity, g = acceleration due to gravity, h = height, ∅ = angle between the displacement and the force of gravity.
Given: m = 2 kg, g = 9.8 m/s², h = 1.5 m, ∅ = 30°
Substitute into equation 1
W = 2(9.8)(1.5)sin30
W = 2×9.8×1.5×0.5
W = 14.7 J.
The right option is (d) 14.7 J
The work done by gravity on a 2-kg cart rolling down a 1.5 m, 30° incline is 14.7 J, corresponding to option (d).
Explanation:When an object slides down an inclined plane, the work done by gravity can be calculated using the formula W = mgh, where W is the work done, m is the mass of the object, g is the acceleration due to gravity (9.8 m/s2), and h is the vertical height. The vertical height can be determined from the incline's length and the angle of inclination using the sine function (h = l · sin(θ)). For a 2-kg cart rolling a distance of 1.5 m down a 30° incline, the work done by gravity would be W = m · g · l · sin(θ) = 2 kg · 9.8 m/s2 · 1.5 m · sin(30°) = 2 · 9.8 · 1.5 · 0.5 = 14.7 J, which corresponds to option (d).
Teams red and blue are having a tug-of-war. According to Newton's third law, the force with which the red team pulls on the blue team exactly equals the force with which the blue team pulls on the red team.
Answer: The statement is TRUE.
Explanation: The Newton's third law states that a body that experiments an external force responds with an opposite force with same magnitude. Therefore, the statement is true.
3 In a tempering process, glass plate, which is initially at a uniform temperature Ti , is cooled by suddenly reducing the temperature of both surfaces to Ts. The plate is 20 mm thick, and the glass has a thermal diffusivity of 6 107 m2 /s.
Answer:
t=63s
tempersture gradient -2.36 x 10⁴
Explanation:
A client with a head injury is being monitored for increased intracranial pressure (ICP). The client's blood pressure is 90/60 mm Hg and the ICP is 18 mm Hg; therefore their cerebral perfusion pressure (CPP) is
Answer:
[tex]CPP=52mm \ Hg[/tex]
Explanation:
Cerebral Perfusion Pressure is obtained by subtracting IntraCranial Pressure(ICP) from the Mean Arterial Pressure(MAP). Adequate cerebral perfusion requires a minimum goal of [tex]70mm \ Hg[/tex]. MAP is obtained using the formula:-
[tex]MAP=\frac{(diastolic \ blood \ pressure\times2)+\ systolic \ blood \ pressure)}{3}\\\\MAP=\frac{2\times60+90}{3}\\\\MAP=70mm \ Hg\\CPP=MAP-ICP\\CPP=70-18\\CPP=52mm \ Hg[/tex]
The 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 975 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 75 volts
Answer: 23.92 ns
Explanation:
The capacitance, C = (e(0) * A) / d
C = [8.85*10^-12 * 2*10^-2 * 10*10^-2] / 1*10^-3
C = (1.77*10^-14) / 1*10^-3
C = 1.77*10^-11
C = 17.7 pF
Vc = V * [1 - e^-(t/CR)]
75 = 100 * [1 - e^-(t/CR)]
75/100 = 1 - e^-(t/CR)
e^-(t/CR) = 1 - 0.75
e^-(t/CR) = 0.25
If we take the log of both sides, we then have
Log e^-(t/CR) = Log 0.25
-t/CR = In 0.25
t = -CR In 0.25
t = - 17.7*10^-12 * 975 * -(1.386)
t = 2.392*10^-8 s
t = 23.92 ns
Calculate the time required (in years) for water to penetrate a layer of clay that is 40 cm deep when exposed to a hydraulic gradient of 1 cm/cm. The permeability of clay is 1x10-8 cm/sec. If there is 30 cm static head of water on the clay layer, how long will it take for moisture to penetrate the 40 cm clay layer (in years)?
Answer:
The time required to penetrate the 40 cm clay layer is 126.82 years
Explanation:
Given:
Hydraulic gradient = 1 [tex]\frac{cm}{cm}[/tex]
Permeability of clay [tex]K = 1 \times 10^{-8}[/tex] [tex]\frac{cm}{sec}[/tex]
Time required to penetrate 40 cm clay layer,
[tex]= \frac{40}{K}[/tex]
[tex]= \frac{40}{1 \times 10^{-8} }[/tex]
[tex]= 40 \times 10^{8}[/tex] sec
But we have to find time in years,
[tex]= \frac{40 \times 10^{8} }{24 \times 60 \times 60 \times 365}[/tex] years
[tex]= 126.82[/tex] years
Therefore, the time required to penetrate the 40 cm clay layer is 126.82 years
Answer:
its a
Explanation:
Four objects are situated along the y axis as follows: a 1.93-kg object is at +2.93 m, a 3.06-kg object is at +2.58 m, a 2.41-kg object is at the origin, and a 3.96-kg object is at -0.498 m. Where is the center of mass of these objects?
Answer: x, y (0, 1)
Explanation:
the X coordinate of the center mass is
X(c) = Σm(i)*x(i) / Σx(i)
X(c) = (0 + 0 + 0 + 0) / (1.93 + 3.06 + 2.41 + 3.96)
X(c) = 0 / 11.36
X(c) = 0
The y coordinate of the center mass is
Y(c) = Σm(i)y(i) / Σm(i)
Y(c) = [(1.93)(2.93) + (3.06)(2.58) + (2.41)(0) + (3.96)(-0.498)] / (1.93 + 3.06 + 2.41 + 3.96)
Y(c) = (5.6549 + 7.8948 + 0 - 1.97208) / 11.36
Y(c) = 11.57762 / 11.36
Y(c) = 1.02
Therefore, the center of masses is at x, y (0, 1)