Answer:
The charge on the wool after rubbing is - 10 C
Explanation:
Every uncharged body is electrically neutral, if the plastic rod acquires 10 Coulombs of charge after been rubbed with wool, then the wool will be left with an equal but opposite charge. This shows that the initial charge on the wool is 10 protons and 10 electrons and when the plastic acquires 10 C (10 protons), the wool will be left with excess 10 electrons.
Therefore, the charge on the wool after rubbing is - 10 C (negative 10 Coulombs).
An elevator filled with passengers has a mass of 1,700 kilograms and accelerates upward from rest at a rate of 1 meters/seconds 2 for 1.8 seconds. Calculate the tension in the cable (in Newtons) supporting the elevator during this time.
Answer:
The tension in the cable is 18371.9 newtons.
Explanation:
Physically speaking, the tension can be calculated with the help of the Second Newton's Law. The upward acceleration means that magnitude of tension must be greater than weight of elevator, whose equation of equilibrium is described below:
[tex]\Sigma F = T - m\cdot g = m \cdot a[/tex]
Where:
[tex]T[/tex] - Tension in the cable, measured in newtons.
[tex]m[/tex] - Mass of the elevator, measured in kilograms.
[tex]g[/tex] - Gravity constant, measured in meters per square second.
[tex]a[/tex] - Net acceleration of the elevator, measured in meters in square second.
Now, tension is cleared and resultant expression is also simplified:
[tex]T = m \cdot (a + g)[/tex]
If [tex]m = 1700\,kg[/tex], [tex]a = 1\,\frac{m}{s^{2}}[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], the tension in the cable is:
[tex]T = (1700\,kg)\cdot \left(1\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]T = 18371.9\,N[/tex]
The tension in the cable is 18371.9 newtons.
a wave with a high amplitude______?
. . . is carrying more energy than a wave in the same medium with a lower amplitude.
The efficiency of a carnot cycle is 1/6. If on reducing the temperature of the sink 75 degree Celsius, the efficiency becomes 1/3, determine the initial and final temperature between which the cycle is working.
Answer:
375 and 450
Explanation:
The computation of the initial and the final temperature is shown below:
In condition 1:
The efficiency of a Carnot cycle is [tex]\frac{1}{6}[/tex]
So, the equation is
[tex]\frac{1}{6} = 1 - \frac{T_2}{T_1}[/tex]
For condition 2:
Now if the temperature is reduced by 75 degrees So, the efficiency is [tex]\frac{1}{3}[/tex]
Therefore the next equation is
[tex]\frac{1}{3} = 1 - \frac{T_2 - 75}{T_1}[/tex]
Now solve both the equations
solve equations (1) and (2)
[tex]2(1 - T_2/T_1) = 1 - (T_2 - 75)/T_1\\\\2 - 1 = 2T_2/T_1 - (T_2 - 75)/T_1\\\\ = (T_2 + 75)/T_1T_1 = T_2 + 75\\\Now\ we\ will\ Put\ the\ values\ into\ equation (1)\\\\1/6 = 1 - T_2/(T_2 + 75)\\\\1/6 = (75)/(T_2 + 75)[/tex]
T_2 + 450 = 75
T_2 = 375
Now put the T_2 value in any of the above equation
i.e
T_1 = T_2 + 75
T_1 = 375 + 75
= 450
Red light from three separate sources passes through a diffraction grating with 5.60×105 slits/m. The wavelengths of the three lines are 6.56 ×10−7m (hydrogen), 6.50 ×10−7m (neon), and 6.97 ×10−7m (argon).
Part A
Calculate the angle for the first-order diffraction line of first source (hydrogen).
Express your answer using three significant figures.
θH = ∘
Part B
Calculate the angle for the first-order diffraction line of second source (neon).
Express your answer using three significant figures.
θNe = ∘
Part C
Calculate the angle for the first-order diffraction line of third source (argon).
Express your answer using three significant figures.
θAr = ∘
Answer:
I can help
Explanation:
A uniform solid disk has a mass of 1.00 kg and a radius of 1.00 m. The disk is mounted on frictionless bearings and is used as a turntable. The turntable is initially rotating at 7.00 rad/s. A uniform rod with a length of 3.00 m and a mass of 0.500 kg is released from rest, just above the turntable, such that the axis of the rod is the same as the axis of the disk. The rod slips on the turntable until it acquires the same final angular velocity.
a. Find the final angular velocity of the system.
b. Find the amount of mechanical energy lost due to friction.
Answer:
a) The final angular velocity of the system is 4 radians per second, b) The amount of mechanical energy lost due to friction is 5.25 joules.
Explanation:
a) The problem is a clear representation of the Principle of the Angular Momentum Conservation, where moment of inertia of the system is increased by the adding of the uniform rod and there are no external forces exerted on the system. This system is represented by the following model:
[tex]I_{d} \cdot \omega_{o} = (I_{d} + I_{r})\cdot \omega_{f}[/tex]
Where:
[tex]\omega_{o}[/tex], [tex]\omega_{f}[/tex] - Initial and final angular velocities, measured in radians per second.
[tex]I_{d}[/tex], [tex]I_{r}[/tex] - Moments of inertia of the uniform solid disk and uniform rod, measured in [tex]kg\cdot m^{2}[/tex].
Now, the final angular speed is cleared:
[tex]\omega_{f} = \frac{I_{d}}{I_{d}+I_{r}}\cdot \omega_{o}[/tex]
The moments of inertia of the uniform solid disk and uniform rod are modelled by these formulas:
Solid Disk
[tex]I_{d} = \frac{1}{2}\cdot m_{d}\cdot r_{d}^{2}[/tex]
Where:
[tex]m_{d}[/tex] - Mass of the disk, measured in kilograms.
[tex]r_{d}[/tex] - Radius of the disk, measured in meters.
Given that [tex]m_{d} = 1\,kg[/tex] and [tex]r_{d} = 1\,m[/tex], the moment of inertia of the disk is:
[tex]I_{d} = \frac{1}{2}\cdot (1\,kg)\cdot (1\,m)^{2}[/tex]
[tex]I_{d} = 0.5\,kg\cdot m^{2}[/tex]
Rod (rotating about its center)
[tex]I_{r} = \frac{1}{12}\cdot m_{r}\cdot l_{r}^{2}[/tex]
Where:
[tex]m_{r}[/tex] - Mass of the rod, measured in kilograms.
[tex]l_{r}[/tex] - Length of the rod, measured in meters.
Given that [tex]m_{r} = 0.5\,kg[/tex] and [tex]l_{r} = 3\,m[/tex], the moment of inertia of the rod is:
[tex]I_{r} = \frac{1}{12}\cdot (0.5\,kg)\cdot (3\,m)^{2}[/tex]
[tex]I_{r} = 0.375\,kg\cdot m^{2}[/tex]
Now, knowing that [tex]\omega_{o} = 7\,\frac{rad}{s}[/tex], the final angular velocity is:
[tex]\omega_{f} = \left(\frac{0.5\,kg\cdot m^{2}}{0.5\,kg\cdot m^{2}+0.375\,kg\cdot m^{2}}\right)\cdot \left(7\,\frac{rad}{s} \right)[/tex]
[tex]\omega_{f} = 4\,\frac{rad}{s}[/tex]
The final angular velocity of the system is 4 radians per second.
b) According to the Principle of Energy Conservation, the inclusion of the uniform rod on the turntable is represented by the following expression:
[tex]K_{1} = K_{2} + \Delta E_{loss}[/tex]
Where:
[tex]K_{1}[/tex] - Rotational kinetic energy of the uniform disk, measured in joules.
[tex]K_{2}[/tex] - Rotational kinetic energy of the system (uniform disk + uniform rod), measured in joules.
[tex]\Delta E_{loss}[/tex] - Mechanical energy lost due to friction, measured in joules.
The mechanical energy lost due to friction is cleared:
[tex]\Delta E_{loss} = K_{1} - K_{2}[/tex]
Now, the expression is expanded and mechanical energy losses is calculated:
[tex]\Delta E_{loss} = \frac{1}{2}\cdot I_{d}\cdot \omega_{o}^{2} - \frac{1}{2}\cdot (I_{d}+I_{r})\cdot \omega_{f}^{2}[/tex]
[tex]\Delta E_{loss} = \frac{1}{2}\cdot (0.5\,kg\cdot m^{2})\cdot \left(7\,\frac{rad}{s} \right)^{2} - \frac{1}{2}\cdot (0.5\,kg\cdot m^{2} + 0.375\,kg\cdot m^{2})\cdot \left(4\,\frac{rad}{s} \right)^{2}[/tex]
[tex]\Delta E_{loss} = 5.25\,J[/tex]
The amount of mechanical energy lost due to friction is 5.25 joules.
Find the pressure difference (in kPa) on an airplane wing if air flows over the upper surface with a speed of 125 m/s, and along the bottom surface with a speed of 109 m/s. [Express answer in TWO decimal places]
Answer:
P= 2414.9 Pa
Explanation:
given
density of air , p = 1.29 kg/m³
speed of air over the upper surface , v₁ = 125 m/s
speed of air over the lower surface , v₂ = 109 m/s
the pressure difference on an airplane wing , P = 0.5 × p × ( v₁² - v₂²)
P = 0.5 × 1.29 × ( 125² - 109²)
P= 0.645(3744)
P = 2414.9 Pa
the pressure difference on an airplane wing is 2414.9 Pa
g A launched rocket is not initially considered a projectile. Explain why not and describe the point at which a rocket becomes a projectile during its flight.
Explanation:
A projectile is usually launched by an initial force, is influenced by gravity forces and air resistance, and takes a parabolic, or more accurately, an elliptical path. At the beginning of the launch of a rocket, the rocket is mostly under the effect of its engine thrust, and takes a vertical flight path upwards; using engine thrust to maneuver itself to remain in this vertical flight till it is almost at its required orbit. At this stage, the rocket can't be said to follow a projectile path. When the rocket gets to the pitchover stage, at which it is almost ready to enter its atmosphere, the engine thrust is maneuvered to turn the rocket by an angle of attack. At this stage, the flight path is no longer vertical. At this point after pitchover, in which the flight path is no longer vertical, gravity tends to pull the rocket down in a parabolic path. This is the point where the rocket acts as a projectile, but is countered by engine thrust.
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will Group of answer choices
Answer:
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.
Explanation:
The power dissipation by an electrical circuit is given by the following formula:
Power Dissipation = (Voltage)(Current)
P = VI
but, from Ohm's Law, we know that:
Voltage = (Current)(Resistance)
V = IR
Substituting this in formula of power:
P = (IR)(I)
P = I²R ---------------- equation 1
Now, if we double the current , then the power dissipated by that circuit will be:
P' = I'²R
where,
I' = 2 I
Therefore,
P' = (2 I)²R
P' = 4 I²R
using equation 1
P' = 4 P
Therefore, the power dissipated by the circuit will becomes four times of its initial value.
What is Ohm's Law, and how does it work in real life.
Explanation:
Ohms law states that the electrical current present in a metallic conductor is directly proportional to the potential difference between the metallic conductor and inversely proportional to the resistance therefore if the voltage is increased resistance also increases provided that temperature and other physical properties remains constant V=IR
A current carrying wire is oriented along the y axis It passes through a region 0.45 m long in which there is a magnetic field of 6.1 T in the z direction The wire experiences a force of 15.1 N in the x direction.1. What is the magnitude of the conventional current inthe wire?I = A2. What is the direction of the conventional current in thewire?-y+y
Answer:
The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.
Explanation:
- To find the direction of the conventional current in the wire you use the following formula:
[tex]\vec{F}=i\vec{l}\ X\ \vec{B}[/tex] (1)
i: current in the wire = ?
F: magnitude of the magnetic force on the wire = 15.1N
B: magnitude of the magnetic field = 6.1T
l: length of the wire that is affected by the magnetic field = 0.45m
The direction of the magnetic force is in the x direction (+^i) and the direction of the magnetic field is in the +z direction (+^k).
The direction of the current must be in the +y direction (+^j). In fact, you have:
^j X ^k = ^i
The current and the magnetic field are perpendicular between them, then, you solve for i in the equation (1):
[tex]F=ilBsin90\°\\\\i=\frac{F}{lB}=\frac{15.1N}{(0.45m)(6.1T)}=5.5A[/tex]
The magnitude of the current in the wire is 5.5A, and the direction of the current is in the positive y direction.
Two cars start moving from the same point. One travels south at 28 mi/h and the other travels west at 21 mi/h. At what rate is the distance between the cars increasin
Complete question:
Two cars start moving from the same point. One travels south at 28 mi/h and the other travels west at 21 mi/h. At what rate is the distance between the cars increasing four hours later.
Answer:
The rate at which the distance between the cars is increasing four hours later is 35 mi/h.
Explanation:
Given;
speed of one car, dx/dt = 28 mi/h South
speed of the second car, dy/dt = 21 mi/h West
The distance between the cars is the line joining west to south, which forms a right angled triangle with the two positions.
Apply Pythagoras theorem to evaluate this distance;
let the distance between the cars = z
x² + y² = z² -------- equation (1)
Differentiate with respect to time (t)
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt}[/tex] ----- equation (2)
Since the speed of the cars is constant, after 4 hours their different distance will be;
x: 28(4) = 112 mi
y: 21(4) = 84 mi
[tex]z = \sqrt{x^2 + y^2} \\\\z = \sqrt{112^2 + 84^2} \\\\z = 140 \ mi[/tex]
Substitute in the value of x, y, z, dx/dt, dy /dt into equation (2) and solve for dz/dt
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \\\\2(112)(28) + 2(84)(21) = 2(140)\frac{dz}{dt} \\\\9800 = 280\frac{dz}{dt} \\\\\frac{dz}{dt} = \frac{9800}{280} \\\\\frac{dz}{dt} = 35 \ mi/h[/tex]
Therefore, the rate at which the distance between the cars is increasing four hours later is 35 mi/h
The gravitational potential has a zero value gravitational field in the same place is? Maximum or null indeterminate
Answer:maximum
Explanation:
A closely wound, circular coil with radius 2.50 cmcm has 740 turns. Part A What must the current in the coil be if the magnetic field at the center of the coil is 0.0760 TT?
Answer:
The current in the coil is 4.086 A
Explanation:
Given;
radius of the circular coil, R = 2.5 cm = 0.025 m
number of turns of the circular coil, N = 740 turns
magnetic field at the center of the coil, B = 0.076 T
The magnetic field at the center of the coil is given by;
[tex]B = \frac{N\mu_o I}{2R}[/tex]
where;
μ₀ is permeability of free space = 4 x 10⁻⁷ m/A
I is the current in the coil
R is radius of the coil
N is the number of turns of the coil
The current in the circular coil is given by
[tex]B = \frac{N\mu_o I}{2R} \\\\I = \frac{2BR}{N\mu_o} \\\\I =\frac{2*0.076*0.025}{740*4\pi*10^{-7}} \\\\I = 4.086 \ A[/tex]
Therefore, the current in the coil is 4.086 A
A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation between the plates is increased, what will happen to the charge on the capacitor and the electric potential across it
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. [tex]Q[/tex], the amount of charge stored in this capacitor, will stay the same.
The formula [tex]\displaystyle Q = C\, V[/tex] relates the electric potential across a capacitor to:
[tex]Q[/tex], the charge stored in the capacitor, and[tex]C[/tex], the capacitance of this capacitor.While [tex]Q[/tex] stays the same, moving the two plates apart could affect the potential [tex]V[/tex] by changing the capacitance [tex]C[/tex] of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
[tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex],
where
[tex]\epsilon[/tex] is the permittivity of the material between the two plates.[tex]A[/tex] is the area of each of the two plates.[tex]d[/tex] is the distance between the two plates.Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of [tex]\epsilon[/tex]. Neither will that change the area of the two plates.
However, as [tex]d[/tex] (the distance between the two plates) increases, the value of [tex]\displaystyle C = \frac{\epsilon\, A}{d}[/tex] will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula [tex]\displaystyle Q = C\, V[/tex] can be rewritten as:
[tex]V = \displaystyle \frac{Q}{C}[/tex].
The value of [tex]Q[/tex] (charge stored in this capacitor) stays the same. As the value of [tex]C[/tex] becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
An electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. What is the direction of the magnetic force on the electron?
Answer: the direction of the magnetic force on the electron will be moving out of the screen, perpendicular to the magnetic field.
Explanation:
The magnetic force F on a moving electron at right angle to a magnetic field is given by the formula:
F = BqVSinØ
If an electron moves in the plane of this screen toward the top of the screen. A magnetic field is also in the plane of the screen and directed toward the right. Then, the direction of the magnetic force on the electron will be perpendicular to the magnetic field
According to the Fleming's left - hand rule, the direction of the magnetic force on the electron will be moving out of the plane of the screen.
The position of a particle is r(t)= (4.0t'i+ 2.4j- 5.6tk) m. (Express your answers in vector form.) (a) Determine its velocity (in m/s) and acceleration (in m/s2) as functions of time. (Use the following as necessary: t. Assume t is seconds, r is in meters, and v is in m/s, Do not include units in your answer.) v(t)= ________m/s a(t)= ________m/s2 (b) What are its velocity (in m/s) and acceleration (in m/s2) at time t 0? v(0) =_______ m/s a(0)=_______ m/s2
The position of a particle is r(t)= (4.0t²i+ 2.4j- 5.6tk) m. (Express your answers in vector form.)
(a) Determine its velocity (in m/s) and acceleration (in m/s²) as functions of time. (Use the following as necessary: t. Assume t is seconds, r is in meters, and v is in m/s, Do not include units in your answer.)
v(t)= ________m/s
a(t)= ________m/s²
(b) What are its velocity (in m/s) and acceleration (in m/s²) at time t 0?
v(0) =_______ m/s
a(0)=_______ m/s²
Answer:
(a)
v(t)= [tex]8ti - 5.6k[/tex] m/s
a(t)= 8i m/s²
(b)
v(0) = -5.6k m/s
a(0)= 8i m/s²
Explanation:From the question, the position of the particle is given by;
r(t)= (4.0t²i+ 2.4j- 5.6tk) -----------------(i)
(a)
(i)To get the velocity, v(t), of the particle, we'll take the first derivative of the position of the particle (given by equation (i)) with respect to time, t, as follows;
v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]\frac{d(4.0t^2i + 2.4j - 5.6tk)}{dt}[/tex]
v(t) = [tex]\frac{dr(t)}{dt}[/tex] = [tex]8ti +0j - 5.6k[/tex]
v(t) = [tex]8ti - 5.6k[/tex] --------------------(ii)
(ii) To get the acceleration, a(t), of the particle, we'll take the first derivative of the velocity of the particle (given by equation (ii)) with respect to time, t, as follows;
a(t) = [tex]\frac{dv(t)}{dt}[/tex] = [tex]\frac{d(8ti - 5.6k)}{dt}[/tex]
a(t) = 8i --------------------(iii)
(b)
(i) To get the velocity of the particle at time t = 0, substitute the value of t = 0 into equation (ii) as follows;
v(t) = [tex]8ti - 5.6k[/tex]
v(0) = 8(0)i - 5.6k
v(0) = 0 - 5.6k
v(0) = -5.6k
(ii) To get the acceleration of the particle at time t = 0, substitute the value of t = 0 into equation (iii) as follows;
a(t) = 8i
a(0) = 8i
if the input work on a machine is equal to it's output work. the machine has _____ efficiency.
(fill in the blank)
Answer:
100% efficiency
Explanation:
the machine has _100%_ efficiency.
(fill in the blank)
Answer:
100%
Explanation: A p e x
What is surface tension??
Answer:
Surface tension is the tendency of liquid surfaces to shrink into the minimum surface area possible. Surface tension allows insects (e.g. water striders), usually denser than water, to float and slide on a water surface.
Explanation:
Answer:
It is the tension of the surface film of a liquid caused by the attraction of the particles in the surface layer by the bulk of the liquid, which tends to minimize surface area.
You push a hockey puck that is initially at rest on slick ice by applying a constant force until the puck reaches a final velocity of 1 m/s. On the second attempt, you want the hockey puck to reach the same final velocity by applying a force that is twice as large.
1. Therefore, you must exert the force for a time interval that is
A. shorter than the time interval of your first attempt.
B. longer than the time interval of your first attempt.
C. the same as the time interval of your first attempt.
2. After the hockey puck has reached the final velocity, you suddenly stop pushing it. The hockey puck:
A. stops abruptly
B. reduces speed gradually
C. continues at constant velocity
D. increases speed gradually
E. reduces speed abruptly
Answer:
1. A
2. B or C
Explanation:
1.
F=ma, meaning that if you use two times more force on a constant mass, the acceleration must double. Acceleration is change in velocity, which means that if you are aiming for the same final velocity the change must happen in half of the time. Therefore, the correct answer is choice A.
2.
By Newton's first law, an object in motion will stay in motion unless an external force acts on it. Since there is nothing pushing the puck in the other direction, the puck will either keep on going for at a constant velocity or will reduce its speed gradually, depending on whether or not this ice is considered to be frictionless. Hope this helps!
(1) You must exert the force for a time interval that is shorter than the time interval of your first attempt.
(2) The hockey puck reduces speed abruptly.
According to Newton's second law of motion; the force applied to an object is directly proportional to the mass and acceleration of the object.
F = ma
[tex]F = \frac{mv}{t}[/tex]
The force applied to an object is directly proportional to the velocity of the object and inversely proportional to the time of motion of the object.To double the force, you must halve the time interval.Thus, to apply a force that is twice as large as the first while maintaining the same velocity, you must exert the force for a time interval that is shorter than the time interval of your first attempt.
(2) The force applied to an object is direct directly proportional to the velocity of the object. Once you stop applying force to the hockey puck, it moves for a short with initial momentum gained before it will stop.
Thus, the magnitude of the velocity (speed) will drop sharply once the force on the object is removed.
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A 300-foot cable weighing 5 pounds per foot is hanging from a winch 300 feet above ground level. Find the work (in ft-lb) done in winding up the cable when there is a 300-pound load attached to the end of the cable.
Answer:
315,000 ft·lb
Explanation:
At 300 ft and 5 lb/ft, the weight of the cable is (300 f)(5 lb/ft) = 1500 lb. The work done to raise it is equivalent to the work done to raise the cable's center of mass. Since the cable is of uniform density, its center of mass is half the cable length below the winch.
total work done = work to raise cable + work to raise load
= (1500 lb)(150 ft) +(300 lb)(300 ft) = 315,000 ft·lb
A lunar module weighs 12 metric tons on the surface of the Earth. How much work is done in propelling the module from the surface of the moon to a height of 80 miles
Answer:
149.05J
Explanation:
Hello,
Data;
Weight = 12N
x = 80miles
Radius of the moon = 1079.4mi
W = mg
But gravity at moon = ⅙ the gravity on earth
12 = ⅙ × m
M = 2kg
Mass of the module = 2kg
K = F.x
F(x) = k / x²
2 = k / (1079.4)²
k = 2.33×10⁶
Work = ∫f.dx
work = ∫₁₀₇₉.₄¹¹⁵⁹ . (2.33×10⁶/x²).dx
Work = (-2.33×10⁶) ×(¹/₁₁₅₉ - ¹/₁₀₇₉)
work = 149.05J
Two children, Ahmed and Jacques, ride on a merry-go-round. Ahmed is at a greater distance from the axis of rotation than Jacques. Which of the following are true statements?
A. Ahmed has a greater tangential speed than Jacques.
B. Jacques has a greater angular speed than Ahmed.
C. Jacques has a smaller angular speed than Ahmed.
D. Jacques and Ahmed have the same angular speed.
Answer:
a
Explanation:
Wanda exerts a constant tension force of 12 N on an essentially massless string to keep a tennis ball (m = 60 g) attached to the end of the string traveling in uniform circular motion above her head at a constant speed of 9.0 m/s. What is the length of the string between her hand and the tennis ball? You may ignore gravity in this problem (assume the motion of the tennis ball and string happen in a purely horizontal plane). A. 41 m B. 0.24 m C. 3.2 cm D. 0.41 m
Answer:
r = 0.405m = 40.5cm
Explanation:
In order to calculate the length of the string between Wanda and the ball, you take into account that the tension force is equal to the centripetal force over the ball. So, you can use the following formula:
[tex]F_c=ma_c=m\frac{v^2}{r}[/tex] (1)
Fc: centripetal acceleration (tension force on the string) = 12N
m: mass of the ball = 60g = 0.06kg
r: length of the string = ?
v: linear speed of the ball = 9.0m/s
You solve for r in the equation (1) and replace the values of the other parameters:
[tex]r=\frac{mv^2}{F_c}=\frac{(0.06kg)(9.0m/s)^2}{12N}=0.405m[/tex]
The length of the string between Wanda and the ball is 0.405m = 40.5cm
If you place a 50 Volt power source across a 10 Ohm resistor, what will the current
be?
Answer:
use voltage=current*resistance
v=i*r
50=i*10
50/10=i
i=5
hope it's clear
Given the following data:
Voltage = 50 Volt
Resistance = 10 Ohm
To determine the current flowing through the circuit, we would apply Ohm's law:
Mathematically, Ohm's law is given by the formula;
[tex]V = IR[/tex]
Where;
V is voltage measured in voltage.
I is current measured in amperes.
R is resistance measured in ohms.
Making I the subject of formula, we have:
[tex]I = \frac{V}{R}[/tex]
Substituting the given parameters into the formula, we have;
[tex]I = \frac{50}{10}[/tex]
Current, I = 5 Ampere.
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If a key is pressed on a piano, the frequency of the resulting sound will determine the ________, and the amplitude will determine the ________ of the perceived musical note.
Answer:
If a key is pressed on a piano, the frequency of the resulting sound will determine the ___PITCH_____, and the amplitude will determine the _____LOUDNESS___ of the perceived musical note.
Explanation:
The frequency of a vibrating string is primarily based on three factors:
The sounding length (longer is lower, shorter is higher)
The tension on the string (more tension is higher, less is lower)
The mass of the string, normally based on a uniform density per unit length (higher mass is lower, lower mass is higher)
To make a shorter string (such as in an upright piano) sound the same fundamental frequency as a longer string (such as in a 9' grand piano), either the thickness of the string must be increased (which increases the density and the mass) or the tension must be decreased, and usually it's a bit of both.
Thicker strings are often stiffer and that creates more inharmonic partials, and lower tension is associated with other problems, so the best way to make a string sound lower is the make it longer, but it is not practical to make a piano from strings that are all the same density and tension, because the lowest strings would have to be ridiculously long. Nine feet is already a great demand on space for a single musical instrument, and of course those pianos are extremely expensive and difficult to move.
And alsoBesides the pitch of a musical note, perhaps the most noticeable feature in how loud the note is. The loudness of a sound wave is determined from its amplitude. While loudness is only associated with sound waves, all types of waves have an amplitude. Waves on a calm ocean may be less than 1 foot high. Good surfing waves might be 10 feet or more in amplitude. During a storm the amplitude might increase to 40 or 50 feet.
Many things can influence the amplitude.
What is producing the sound?
How far are you from the source of the sound? The farther away the smaller the amplitude.
Intervening material. Sound does not travel through walls as well as air.
Depends on what is detecting the wave sound. Ear vs. microphone.
Answer:
The frequency will determine the pitch
the amplitude will determine the loudness
Explanation:
The frequency of a sound refers to the number of vibrations made by the sound wave produced in a unit of time. This usually affects how high or how low a note is perceived in music. High-frequency sounds have higher pitches, while low-frequency sounds have lower pitches.
The amplitude of a sound wave refers to the height between the wave crests and the equilibrium line in a sound wave. It shows how loud a sound will be. High amplitude sounds are loud while low amplitude sounds are quiet.
What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 18.0 m/s in the same direction
What is the velocity of a 900-kg car initially moving at 30.0 m/s, just after it hits a 150-kg deer initially running at 18.0 m/s in the same direction? Assume the deer remains on the car.
Answer:28.29m/s
Explanation:
In this situation, linear momentum is conserved. And since the deer remains on the car after collision, the linear momentum is given as;
([tex]m_{C}[/tex] x [tex]u_{C}[/tex]) + ([tex]m_{D}[/tex] x [tex]u_{D}[/tex]) = ([tex]m_{C}[/tex] + [tex]m_{D}[/tex]) v -----------------(i)
Where;
[tex]m_{C}[/tex] = mass of car
[tex]u_{C}[/tex] = initial velocity of car before collision
[tex]m_{D}[/tex] = mass of deer
[tex]u_{D}[/tex] = initial velocity of the deer before collision
v = common velocity with which the car and the deer move after collision
From the question;
[tex]m_{C}[/tex] = 900kg
[tex]u_{C}[/tex] = +30.0m/s (direction of the motion of the car taken positive)
[tex]m_{D}[/tex] = 150kg
[tex]u_{D}[/tex] = +18.0m/s (relative to the direction of the car, the velocity of the deer is also positive )
Substitute these values into equation (i) as follows;
(900 x 30.0) + (150 x 18.0) = (900 + 150)v
27000 + 2700 = 1050v
29700 = 1050v
v = [tex]\frac{29700}{1050}[/tex]
v = 28.29m/s
Therefore, the velocity of the car after hitting the deer is 28.29m/s. This is also the velocity of the deer after being hit by the car.
how does current change under different polarity?
Answer:
Due to flipping of polarity
Explanation:
During the changing of polarity, the current on the one side is maximum as the polarity change then the current is gradually reducing toget from another end.
A circuit contains an EMF source, a resistor R, a capacitor C, and an open switch in series. The capacitor initially carries zero charge. How long after the switch is closed will it take the capacitor to reach 2/3 its maximum charge?
Answer:
t = 1.098*RC
Explanation:
In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:
[tex]Q=Q_{max}[1-e^{-\frac{t}{RC}}][/tex] (1)
Qmax: maximum charge capacity of the capacitor
t: time
R: resistor of the circuit
C: capacitance of the circuit
When the capacitor has 2/3 of its maximum charge, you have that
Q=(2/3)Qmax
You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:
[tex]Q=\frac{2}{3}Q_{max}=Q_{max}[1-e^{-\frac{t}{RC}}]\\\\\frac{2}{3}=1-e^{-\frac{t}{RC}}\\\\e^{-\frac{t}{RC}}=\frac{1}{3}\\\\-\frac{t}{RC}=ln(\frac{1}{3})\\\\t=-RCln(\frac{1}{3})=1.098RC[/tex]
The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC
A ranger needs to capture a monkey hanging on a tree branch. The ranger aims his dart gun directly at the monkey and fires the tranquilizer dart. However, the monkey lets go of the branch at exactly the same time as the ranger fires the dart. Will the monkey get hit or will it avoid the dart?
Answer:
Yes the monkey will get hit and it will not avoid the dart.
Explanation:
Yes, the monkey will be hit anyway because the dart will follow a hyperbolic path and and will thus fall below the branches, so if the monkey jumps it will be hit.
No, the monkey will not avoid the dart because dart velocity doesn't matter. The speed of the bullet doesn’t even matter in this case because a faster bullet will hit the monkey at a higher height and while a slower bullet will simply hit the monkey closer to the ground.
Two charged particles of equal magnitude (+Q and +Q) are fixed at opposite corners of a square that lies in a plane. A test charge +q is placed at the third corner of the square. What is the direction of force on the test charge due to other two charges?
Answer:
The test charge will take the south-west direction indicated in option 6.
Explanation:
The image is shown below.
Since all the charges are positively charged, they will all repel each other. If we consider the force on +q due to +Q and +Q, then we can proceed as follows
The +Q particle at the top left corner of the cube will exert a vertical downward force on +q in the -ve y-axis.
The +Q particle at the bottom right corner of the cube will exert a force on +q towards the horizontal left on the -ve x-axis.
Both of these forces will act at angle of 90°, and therefore, the resultant force will act at an angle of 45° to horizontal and vertical forces.
The result is that the +q charge will move in a south-west direction of the cube.