Answer:
D. Iron nail
Explanation:
________ is the si unit of angular momentum
Answer:
kg m2/s
Explanation:
I think it is :)
The SI unit of angular momentum is kg·m²/s, reflecting an object's rotational motion dynamics. Fundamental to rotational dynamics, angular momentum is calculated based on an object's moment of inertia and angular velocity, adhering to the dimensions ML²T-¹.
Explanation:The SI unit of angular momentum is the kilogram meter squared per second (kg·m²/s). Angular momentum, denoted by the symbols “l” for an individual particle and “L” for a system of particles or a rigid body, represents the rotational equivalent of linear momentum. It is fundamentally connected to the concepts of moment of inertia and angular velocity, where the angular momentum (“l” or “L”) can be calculated by the product of these two quantities. This unit emphasizes the momentum of an object in rotational motion about an axis and is derived from taking the product of the object's moment of inertia and its angular velocity.
The calculation of angular momentum involves factors such as the mass (m) of the particle, its velocity (v) perpendicular to the line joining it to the axis of rotation, and its distance (r) from the axis. This relationship is encapsulated in the formula L=mvr, highlighting the linear momentum's component (mv) and its lever arm distance (r) from the rotation axis. The SI units and dimensions (ML²T-¹) of angular momentum validate its role in describing the rotational dynamics of objects.
A sinusoidal wave is traveling on a string with speed 40 cm/s. The displacement of the particles of the string at x = 10 cm varies with time according to
y=(5.0cm)sin[1.0−(4.0s−1)t]
. The linear density of the string is 4.0 g/cm. What are (a) the frequency and (b) the wavelength of the wave? If the wave equation is of the form y(x, t) = y_m sin(kx ± ωt), what are (c) y_m, (d) k, (e) ω, and (f ) the correct choice of sign in front of ω? (g) What is the tension in the string?
Answer:
a) [tex]f=0.64 Hz[/tex]
b) [tex]\lambda=62.5 cm[/tex]
c) [tex]y_{m}=5 cm[/tex]
d) [tex]k=0.1 cm^{-1}[/tex]
e) [tex]\omega=4 s^{-1}[/tex]
g) [tex]T=0.064 N[/tex]
Explanation:
We know that the wave equation is of the form:
[tex]y(x,t) = y_{m}sin(kx \pm \omega t)[/tex]
Comparing with the equation of the sinusoidal wave [tex](y=(5.0cm)sin[1.0−(4.0s^{-1})t])[/tex] we will have:
[tex]\omega = 4 s^{-1}[/tex]
a) ω is the angular frequency and it can writes in terms of frequency as:
[tex]\omega = 2\pi f[/tex] , where f is the frequency.
[tex]f=\frac{\omega}{2\pi}[/tex]
[tex]f=0.64 Hz[/tex]
b) Let's recall that the speed of the wave is the product between the wave length and the frequency, so we have:
[tex]v=\lambda f[/tex]
[tex]\lambda=\frac{v}{f}[/tex]
v is 40 cm/s
[tex]\lambda=\frac{40}{0.64}[/tex]
[tex]\lambda=62.5 cm[/tex]
If we compare each equation we can find y(m), k and ω:
c) [tex]y_{m}=5 cm[/tex]
d) [tex]kx=1[/tex] and we know that x = 10 cm, so:
[tex]k=\frac{1}{x}=\frac{1}{10}[/tex]
[tex]k=0.1 cm^{-1}[/tex]
e) [tex]\omega=4 s^{-1}[/tex]
f) The minus sign in front of the angular frequency in the equation is the correct choice, just by comparing.
g) We have to use the equation of the speed in terms of tension.
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
T is the tensionμ is the linear densityv is the speed of the wave[tex]T=\mu*v^{2}[/tex]
[tex]T=4*40^{2}=6400 g*cm*s^{-2}[/tex]
[tex]T=0.064 N[/tex]
I hope it helps you!
(a) The frequency of the wave is 6.37 Hz and (b) the wavelength is 2π cm. (c) The values for y_m, k, and ω are 5.0 cm, (d) 1 cm^-1, and (e) 4.0 s^-1 respectively. (f) The correct choice of sign in front of ω is positive. (g) The tension in the string is 12533.5 g*cm²/s².
Explanation:(a) To find the frequency of the wave, we can use the formula f = v/λ, where f is the frequency, v is the velocity, and λ is the wavelength. In this case, v = 40 cm/s and we need to find λ. Since y(x,t) = y_m sin(kx ± ωt), we can compare it to the given equation y=(5.0cm)sin[1.0−(4.0s−1)t] to find k and ω. From the comparison, we know k = 1 cm^-1 and ω = 4.0 s^-1. Therefore, the wavelength is given by λ = 2π/k, and plugging in the values, we get λ = 2π cm. The frequency can then be calculated as f = v/λ = 40 cm/s / 2π cm = 6.37 Hz.
(b) the wavelength is 2π cm.
(c) y_m is the amplitude of the wave, which is 5.0 cm.
(d) k is the wave number, which is 1 cm^-1.
(e) ω is the angular frequency, which is 4.0 s^-1.
(f) The correct choice of sign in front of ω depends on the direction of wave propagation. If the wave is traveling in the positive x-direction, the sign should be positive, so ωt is correct.
(g) To find the tension in the string, we can use the formula T = λ²μf², where T is the tension, λ is the wavelength, μ is the linear density, and f is the frequency. Plugging in the values, we get T = (2π cm)² * (4.0 g/cm) * (6.37 Hz)² = 12533.5 g*cm²/s².
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The center of gravity is defined as: a. The part of the skeleton composed of the bones of the vertebral column, ribs, and skull b. A plane the passes through the midpoint of the body c. State of an object as a result of forces pushing on it d. Imaginary point through which the resultant force of gravity acts on an object
Answer:
the corect answer it b and c
Explanation:
b. the density of lines shows the strength of the force.
c. the arrows on the lines of force show which way a posative object will move.
youre welcome
A football player with the force of 500 N leaps 2 m into the air. How much work is done?
Explanation:
Given
Force (F) = 500 N
Distance (d) = 2m
Now
Work done = F * d
= 500 * 2
= 1000 joule
Therefore 1000 joule of work is done.
Hope this helps.
The work done by a football player leaping 2 m into the air with a force of 500 N is calculated using the equation W = F * d, resulting in 1000 joules of work.
When calculating work done on an object, it's essential to understand that work (W) is defined as the product of the force (F) applied to an object and the distance (d) that the object moves in the direction of the force. The equation for work is:
W = F * d,
where work (W) is measured in joules (J), force (F) in newtons (N), and distance (d) in meters (m).
For a football player leaping vertically into the air with a force of 500 N over a distance of 2 m:
W = (500 N) * (2 m),
W = 1000 J.
The work done by the football player in leaping 2 meters into the air is therefore 1000 joules.
A horizontal vinyl record of mass 0.115 kg and radius 0.0896 m rotates freely about a vertical axis through its center with an angular speed of 5.58 rad/s and a rotational inertia of 4.84 x 10-4 kg·m2. Putty of mass 0.0484 kg drops vertically onto the record from above and sticks to the edge of the record.What is the angular speed of the record immediately afterwards?
Answer:
Explanation:
Angular momentum is the product of inertial and angular frequency
L = I × ω
Where
L is angular momentum
I is inertia
ω is angular frequency
So, given that
Vinyl record has a massz
M = 0.115kg
Radius R = 0.0896m
Angular velocity of vinyl record
ω(initial) = 5.58 rad/s
Rotational inertial of vinyl record.
I(initial) = 4.84 × 10^-4 kgm²
Putty drop on the record
Mass of putty M' = 0.0484kg
Angular speed after putty drop ω'
Using conversation of angular momentum
Initial angular momentum is equal to final angular momentum
I(initial) × ω(initial) = I(final) × ω(final)
So, we need to find I(final)
Inertia log putty can be determine using MR² by assuming a thin loop
I(putty) = M'R² = 0.0484 × 0.0896
I(putty) = 3.89 × 10^-4 kgm²
I(final) = I(initail) + I(putty)
I(final) = 4.84 × 10^-4 + 3.89 × 10^-4
I(final) = 8.73 × 10^-4 kgm²
Therefore,
I(initial) × ω(initial) = I(final) × ω(final)
ω(final) = I(initial) × ω(initial) / I(final)
ω(final) = 4.84 × 10^-4 × 5.58 / 8.73 × 10^-4
ω(final) = 3.1 rad/s
What happens to the resistance of a filament if it is replaced by a shorter
wire?
Answer:the resistance decrease
Explanation:
What are the three bone "shelves" inside the nasal cavity called? *
Vour ancier
Answer: the nasal conchae also known as the turbinates
Explanation:
Technician A says that when checking a voltage drop across an open switch, a measurement of 12 volts means the circuit is open. Technician B says that when checking voltage drop across an open switch, a measurement of 12 volts means the switch contacts are closed. Who is correct
Answer:
Technician B is correct
Explanation:
Electrical Switches makes or brakes the flow of signal (current or voltage) in a circuit.
The Technician B that says the switch contacts are closed is correct because it is until when the circuit is continues that voltage flows across the switch and voltmeter will read the amount of voltage present in the circuit. This simply means closed circuit.
Technician A is not correct because when the switch contacts are open, voltage does not flow and no reading will be measured.
Answer:
Technician B is correct
Explanation:
When a circuit is open there is no energy that allows the flux of charge in the cable. Voltage determines the capability of doing a work over charges to generate a charge flux. An open circuit is not able to generate a current in the circuit and concequently, ther is no voltage in the circuit.
Hence, a measurement of 12V in a circuit means that the circuit is closed, store energy and can generate a flux of charge.
Technician B is correct.
An electron moving with a velocity = 5.0 × 10 7 m/s enters a region of space where perpendicular electric and a magnetic fields are present. The electric field is = . What magnetic field will allow the electron to go through the region without being deflected?
Answer:
The magnetic field is [tex]2 \times 10^{-4}[/tex] T
Explanation:
Given:
Velocity of electron [tex]v = 5 \times 10^{7} \frac{m}{s}[/tex]
Electric field [tex]E = 10^{4} \frac{V}{m}[/tex]
The force on electron in magnetic field is given by,
[tex]F = qvB \sin \theta[/tex] ......(1)
The force on electron in electric field is given by,
[tex]F = qE[/tex] ......(2)
Compare both equation,
[tex]qE = qvB \sin \theta[/tex]
Here [tex]\sin \theta = 1[/tex]
[tex]E = vB[/tex]
[tex]B= \frac{E}{v}[/tex]
[tex]B = \frac{10^{4} }{5 \times 10^{7} }[/tex]
[tex]B = 2 \times 10^{-4}[/tex] T
Therefore, the magnetic field is [tex]2 \times 10^{-4}[/tex] T
A child, hunting for his favorite wooden horse, is running on the ground around the edge of a stationary merry-go-round. The angular speed of the child has a constant value of 0.233 rad/s. At the instant the child spots the horse, one-quarter of a turn away, the merry-go-round begins to move (in the direction the child is running) with a constant angular acceleration of 0.0136 rad/s2. What is the shortest time it takes for the child to catch up with the horse?
Final answer:
The shortest time it takes for the child to catch up with the horse can be found by analyzing the angular motion of the merry-go-round and the child. Using the given values of angular speed and angular acceleration, we can set up an equation and solve for time using the quadratic formula.
Explanation:
To find the shortest time it takes for the child to catch up with the horse, we need to analyze the angular motion of the merry-go-round and the child.
Given that the child's angular speed is 0.233 rad/s and the merry-go-round starts rotating with an angular acceleration of 0.0136 rad/s^2, we can use the equation:
Δθ = ω_0 * t + (1/2) * α * t^2
where Δθ is the angle the child needs to catch up with the horse (π/2 radians in this case), ω_0 is the initial angular speed of the merry-go-round, α is the angular acceleration, and t is the time.
Plugging in the values, we have:
π/2 = 0.233 * t + (1/2) * 0.0136 * t^2
This equation is quadratic in form, so we can solve it using the quadratic formula. The positive root will give us the time it takes for the child to catch up with the horse.
Two technicians are discussing exhaust check valves used in SAI systems. Technician A says that they are used to prevent
the output from the SAI pump from entering the intake manifold. Technician B says the check valves are used to keep the
exhaust from entering the AIR pump. Which technician is correct?
a. Technician A only
b. Technician B only
c. Both Technicians A and B
d. Neither Technician A nor B
Answer: Technical B is right.
Explanation:
The first systems injected air very close to the engine, either in the cylinder head's exhaust ports or in the exhaust manifold
However, the extra heat of recombustion, particularly with an excessively rich exhaust caused by misfiring or a maladjusted carburetor, tended to damage exhaust valves and could even be seen to cause the exhaust manifold to incandesce.
A fisherman notices that one wave passes the bow of his anchored boat every 3 seconds. He measured the wavelength to be 8.5 meters. How fast are the waves traveling?
Answer:
2.3 m/s
Explanation:
A fisherman notices that one wave passes the bow of his anchored boat every 3 seconds. The speed of the wave traveling is 2.83 m/s. The correct option is 1.
What is velocity?Velocity is a vector expression of an object's or particle's displacement with respect to time. The meter per second (m/s) is the standard unit of velocity magnitude (also known as speed). The meter per second (m/s) is a unit of velocity magnitude (also known as speed).
Speed is the rate at which an object moves along a path in time, whereas velocity is the rate and direction of movement. In other words, velocity is a vector, whereas speed is a scalar value.
Frequency = 1/time
Time = 3 seconds
λ = 8.5 meters.
Putting the value in the formula:
v = f x λ
v = 1/3 x 8.5 = 2.83 m/s
Therefore, the speed of the wave traveling is 2.83 m/s. The correct option is 1.
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The question is incomplete. Your most probably complete question is given below:
1.)2.83m/s
2.)0.283m/s
3.)28.3m/s
A mass of 230 g, hanging on a spring, vertically oscillates with a period of 1 sec (the spring itself has no mass). After adding a mass, m, to the 230 g, we find that the period of oscillation of this mass-spring system becomes 2 sec. The value of m is equal to________.
Answer:
The added mass m= 0.7kg
Explanation:
This problem bothers on the simple harmonic motion of a spiral spring
We know that the period of a simple harmonic motion of a spring is given as
T=2π√m/k
We need to solve first for the spring constant k
Given data
Mass m =230g - - - - - kg
=230/1000= 0.230kg
Period T = 1sec
Substituting we have
1= 2*3.142√0.230/k
1=6.284√0.230/k
1/6.284=√0.230/k
Square both sides
(0.159)²=0.230/k
0.025=0.230/k
k=0.230/0.025
k= 9.2N/m
Now we find that the period of oscillation is 2 after adding mass m to 230g.. Let's solve for the new mass
Using the formula for the period T=2π√m/k
2=2*3.142√m/9.2
2=6.284√m/9.2
2/6.284=√m/9.2
Square both sides
(0.318)²=m/9.2
0.10=m/9.2
m= 0.930kg
Therefore the added mass is
0.930kg-0.230kg
The added mass m= 0.7kg
Identify the type of force described (contact or noncontact):
The force that is exerted when a shopping cart is pushed:
The force that causes a metal ball to move toward a magnet:
Answer:
1.contact force
2. Non contact
Explanation:
1. Because bodies are in contact
2.Bodies are not in contact.
Answer:
1. contact
2. non contact
Explanation:
just did it
A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of refraction in the plastic?
Answer:
they are both equal
Explanation:
check the laws of reflection
A student hooks up a voltmeter and an ammeter in a circuit to find the resistance of a light bulb. The ammeter read 3 amps and the voltmeter reads 6 volts. What is the resistance of the light bulb?
Answer:
[tex]2\Omega[/tex]
Explanation:
(Assuming the cell in the circuit has 0 internal resistance)
Ohm's Law is given as:
[tex]V=IR[/tex]
Voltage is Current multiplied by Resistance.
We can rearrange this formula to give us:
[tex]R=\frac{V}{I}[/tex]
Now we can plug in our values
[tex]R=\frac{6}{3}=2\Omega[/tex]
I= 3A
V= 6V
R=?
V=IR
R=V/I
R=6/3
Therefore R = 2ohms
Hope this will help u mate :)
Suppose the universe contained only low-mass stars. Would elements heavier than carbon exist?
Answer:
No, since they don't have the necessary mass and a high temperature in their core.
Explanation:
There are two forces that play an important role in the stars: the force of gravity in the inward direction due to stars' own mass and the radiation pressure in the upward direction as a consequence of the nuclear reaction in their core.
The superficial layers of the stars compress the core as an effect of their own gravity. Therefore, atoms will be closer to each other in the core, allowing them to combine, increasing the density and temperature.
A nuclear reaction occurs when light elements combine into heavier elements (that is known as nucleosynthesis). To get fusion reactions that generate heavier elements than carbon high temperature is necessary, which can be gotten by a more massive star for what was already explained in the first paragraph.
Calculate the linear acceleration of a car, the 0.260-m radius tires of which have an angular acceleration of 14.0 rad/s2. Assume no slippage and give your answer in m/s2. 3.64 Correct: Your answer is correct. m/s2 (b) How many revolutions do the tires make in 2.50 s if they start from rest
Final answer:
A car with tires of radius 0.280 m decelerating at 7.00 m/s² has an angular acceleration of 25 rad/s². To find the number of revolutions before stopping, kinematic equations for rotational motion are used with an initial angular velocity of 95.0 rad/s.
Explanation:
A car decelerating at 7.00 m/s² with tires of radius 0.280 m requires an understanding of the relationship between linear acceleration and angular acceleration. The formula linking these two is α = a/r, where α is the angular acceleration, a is the linear acceleration, and r is the radius of the tire. Using these values, the angular acceleration is calculated to be 25 rad/s².
For part (b), to find out how many revolutions the tires make before coming to rest, we need to use kinematic equations for rotational motion. Initially, we have an angular velocity (ω) of 95.0 rad/s and we want to find the total angle (θ) covered by the tires. Using the equation θ = (ω0²)/(2*α), we calculate the angle in radians and then convert this to revolutions by dividing by 2π. After finding the revolutions, the car's tires will have stopped rotating completely.
Calculate the speed of the ball, vo in m/s, just after the launch. A bowling ball of mass m = 1.5 kg is launched from a spring compressed by a distance d = 0.21 m at an angle of θ = 32° measured from the horizontal. It is observed that the ball reaches a maximum height of h = 4.4 m, measured from the initial position of the ball. Let the gravitational potential energy be zero at the initial height of the bowling ball.
Answer:
[tex]v_0=17.3m/s[/tex]
Explanation:
In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:
[tex]E_1=E_2=E_3\\\\U_e_1=U_g_2+K_2=U_g_3+K_3\\\\\frac{1}{2}kd^2=mg(d\sin\theta)+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}m(v_0\cos\theta)^2[/tex]
Since we only care about the velocity [tex]v_0[/tex], we can keep only the second and third parts of the equation and solve:
[tex]mgd\sin\theta+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}mv_0^2\cos^2\theta\\\\\frac{1}{2}mv_0^2(1-\cos^2\theta)=mg(h-d\sin\theta)\\\\v_0=\sqrt{\frac{2g(h-d\sin\theta)}{1-\cos^2\theta}}\\\\v_0=\sqrt{\frac{2(9.8m/s^2)(4.4m-(0.21m)\sin32\°)}{1-\cos^232\°}}\\\\v_0=17.3m/s[/tex]
So, the speed of the ball just after the launch is 17.3m/s.
We all depend on electricity. Most electricity is created by electromagnetic generators at large power plants and distributed through an electric grid. The electric grid in the United States is interconnected and includes more than 400,000 miles of electric transmission lines. However, the grid has not been well maintained, and the number and severity of power outages has been rising over the past decade. Because the grid is interconnected, a severe power outage in one area can quickly spread to other areas.
How long could you survive without electricity? What parts of your life would be affected by loss of electricity? Should you prepare for an electricity outage, and if so, how would you prepare? What backup system could your family or community install to generate limited amounts of electricity during an outage? How does this system create an electric force field and generate electric current?
Answer:
1) not so long (maybe an hour or two)
2) access to information through the internet will be most affected if my computer and mobile phone run out of battery power.
3) yes, one should prepare for power outage. This can be done by having a standby alternative source of power like the use of inverters that stores electrical energy in form of chemical energy, and small internal combustion engine powered electric generators.
4) solar panels can be used to draw power from incident sun rays, this power can be stored in an inverter for future use in case of a power outage.
5) energy from the sun is converted into direct current which is then supplied to an accumulator in the opposite direction to its flow of current. When the energy is needed, it can be used directly, or converted to an alternating current. This is achieved by connecting its terminal to the supply. Electric field is generated by flow of ions and electrons within the working chemical (e.g lithium).
Explanation:
Answer:
If there was a power outage, you could survive normally using older ways of life and with proper preparations; some ways to prepare is having canned foods that don't require heating or refrigeration, since you wouldn't have electronic means of preparing it, and supplies for starting camp fires. Some ways your life would be affected is lack of simple methods of day-to-day tasks that you have become accustomed to: no warm water, to refrigerators, or heating systems such as electronic kettles, microwaves or ovens. A backup system you and/or your family/community could install would be solar panels, to collect power during the day to then be used at night or when needed. Solar panels collect solar energy by catching the sun-rays in the cells of the panels and turning the energy into a direct current (DC), then the panels convert the direct current into usable alternating current (AC) energy with the help of inverter technology. That current is then distributed throughout the building accordingly.
A 12.0 V voltage is applied to a wire with a resistance of 3.50 ohms. what is the magnetic field 0.150 m from the wire?
____x10^____T
Answer:
4.57x10^-6
Explanation:
So first you'll need to find the current, which is
I = voltage / resistance
which equals 3.43
Then just plug your numbers into the magnetic field equation to get your final answer
4.57x10^-6
Also it was right on Acellus lol
Hope this helps :)
The magnetic field 0.150 m from the wire is 2.86 x 10⁻⁵ T.
The magnetic field created by a current-carrying wire is inversely proportional to the distance from the wire. The closer you get to the wire, the stronger the magnetic field. The distance from the wire is 0.150 m. The voltage applied to the wire is 12.0 V and the resistance of the wire is 3.50 ohms.
We can use the following equation to calculate the magnetic field:
B = μ0 * I / 2πr
B = magnetic field (T)
μ0 = permeability of free space (4π x 10^-7 T m/A)
I = current (A)
r = distance from wire (m)
Plugging in the values, we get:
B = (4π x 10⁻⁷ T m/A) * (12.0 V / 3.50 ohms) / 2π * 0.150 m
B = 2.86 x 10⁻⁵ T.
As a result, the magnetic field 0.150 m away from the wire is 2.86 x 10⁻⁵ T.
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A cue ball of mass m1 = 0.34 kg is shot at another billiard ball, with mass m2 = 0.575 kg, which is at rest. The cue ball has an initial speed of v = 7.5 m/s in the positive direction. Assume that the collision is elastic and exactly head-on.
a) write an expression for the horizontal component of the billiard ball's velocity, vr after the collision, in terms of the other variables of the problem.
b) what is this velocity, in meters per second?
c) Write an expression for the horizontal component of the cue ball's velocity, vr, after the collision.
d) what is the horizontal component of the cue ball's final velocity, in meters per second?
Answer:
Part(a): The expression for the velocity of the billiard ball is [tex]\bf{v_{2f} = \dfrac{2m_{1}v}{m_{1}+m_{2}}}[/tex]
Part(b): The value of the velocity of the billiard ball is [tex]\bf{5.57~m/s}[/tex].
Part(c): The expression for the velocity of the cue ball is [tex]\bf{v_{1f} = \dfrac{(m_{1} - m_{2})v}{m_{1}+m_{2}}}[/tex]
Part(d): The value of the velocity of the cue ball is [tex]\bf{1.93~m/s}[/tex].
Explanation:
Given:
The mass of the cue ball, [tex]m_{1} = 0.34~kg[/tex].
The mass of the billiard ball, [tex]m_{2} = 0.575~kg[/tex].
The initial velocity of the cue ball, [tex]u_{1} = 7.5~m/s[/tex]
The initial velocity of the billiard ball, [tex]u_{2} = 0[/tex]
(a)
Consider the final velocity of the cue ball be [tex]v_{1}[/tex] and the final velocity of the billiard ball be [tex]v_{2}[/tex].
From the conservation of linear momentum , we can write
[tex]~~~~&& m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\\&or,& m_{1}(u_{1} - v_{1}) = m_{2}(v_{2} - u_{2})~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
From the conservation of energy, we can write
[tex]~~~&& \dfrac{1}{2}m_{1}u_{1}^{2} + \dfrac{1}{2}m_{2}u_{2}^{2} = \dfrac{1}{2}m_{1}v_{1}^{2} + \dfrac{1}{2}m_{2}v_{2}^{2}\\&or,& \dfrac{1}{2}m_{1}(u_{1}^{2} - v_{1}^{2}) = \dfrac{1}{2}m_{2}(v_{2}^{2} - u_{2}^{2})~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Dividing equation (1) by equation (2), we have
[tex]~~~&& \dfrac{m_{1}(u_{1} - v_{1}) }{m_{1}(u_{1}^{2} - v_{1}^{2})} = \dfrac{m_{2}(v_{2} - u_{2})}{m_{2}(v_{2}^{2} - u_{2}^{2})}\\&or,& u_{1} + v_{1} = u_{2} + v_{2}~~~~~~~~~~~~~~~~~~~~~~~~(3)[/tex]
Rearranging equation (3) for [tex]v_{1}[/tex], we have
[tex]v_{1} = u_{2} + v_{2} - u_{1}~~~~~~~~~~~~~~~~~~~~(4)[/tex]
Substitute equation (4) in equation (1), we can write
[tex]v_{2} = \dfrac{2m_{1}u_{1}}{m_{1}+m_{2}} + \dfrac{m_{2} - m_{1}}{m_{1} + m_{2}}u_{2}~~~~~~~~~~~~~~(5)[/tex]
Substituting [tex]0[/tex] for [tex]u_{2}[/tex], [tex]v[/tex] for [tex]u_{1}[/tex] and [tex]v_{2f}[/tex] for [tex]v_{2}[/tex] in equation (5), we have
[tex]v_{2f} = \dfrac{2m_{1}v}{m_{1}+m_{2}}~~~~~~~~~~~~~~~~~~~~~~~~(6)[/tex]
(b)
Substituting [tex]0.34~kg[/tex] for [tex]m_{1}[/tex], [tex]7.5~m/s[/tex] for [tex]v[/tex] and [tex]0.575~kg[/tex] for [tex]m_{2}[/tex] in equation (6), we have
[tex]v_{2f} &=& \dfrac{2(0.34~kg)(7.5~m/s)}{(0.34 + 0.575)~kg}\\~~~~~&=& 5.57~m/s[/tex]
(c)
Rearranging equation(3) for [tex]v_{2}[/tex], we have
[tex]v_{2} = u_{1} + v_{1} – u_{2}~~~~~~~~~~~~~~~~~~~~(7)[/tex]
Substitute equation (7) in equation (1), we can write
[tex]v_{1} = \dfrac{2m_{2}u_{2}}{m_{1}+m_{2}} + \dfrac{m_{1} - m_{2}}{m_{1} + m_{2}}u_{2}~~~~~~~~~~~~~~(8)[/tex]
Substituting [tex]0[/tex] for [tex]u_{2}[/tex], [tex]v[/tex] for [tex]u_{1}[/tex] and [tex]v_{1f}[/tex] for [tex]v_{1}[/tex] in equation (8), we have
[tex]v_{1f} = \dfrac{(m_{1} - m_{2})v}{m_{1}+m_{2}}~~~~~~~~~~~~~~~~~~~~~~~~(9)[/tex]
(d)
Substituting [tex]0.34~kg[/tex] for [tex]m_{1}[/tex], [tex]7.5~m/s[/tex] for [tex]v[/tex] and [tex]0.575~kg[/tex] for [tex]m_{2}[/tex] in equation (9), we have
[tex]v_{1f} &=& \dfrac{(0.34 - 0.575)~kg(7.5~m/s)}{(0.34 + 0.575)~kg}\\~~~~~&=& -1.93~m/s[/tex]
Negative sign indicates that the cue ball will bounce back.
The expressions for the horizontal component of the billiard ball's velocity after the collision and the cue ball's velocity after the collision are given. The velocities can be calculated using the provided equations and given values.
Explanation:a) The expression for the horizontal component of the billiard ball's velocity, vr after the collision is given by:
vr = (m1 - m2) * (v1 / (m1 + m2))
b) Substituting the given values into the equation, we find that the horizontal component of the billiard ball's velocity after the collision is approximately 4.02 m/s.
c) The expression for the horizontal component of the cue ball's velocity, vr after the collision is given by:
vr = (2 * m2 * v1) / (m1 + m2)
d) Substituting the given values into the equation, we find that the horizontal component of the cue ball's final velocity is approximately 2.48 m/s.
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is it possible that the enzymes in our bodies use quantum tunneling?
Answer:
Yes it is possible
Explanation:
These are the workhorses of the living world, speeding up chemical reactions so that processes that would otherwise take thousands of years happen inside living cells in seconds. How they achieve this speed-up – often more than a trillion-fold – has long been an enigma. But now, research by Judith Klinman at the University of California, Berkeley and Nigel Scrutton at the University of Manchester (among others) has shown that enzymes can employ a weird quantum trick called tunnelling. Simply put, the enzyme encourages a process whereby electrons and protons vanish from one position in a biochemical and instantly rematerialise in another, without visiting any of the in-between places – a kind of teleportation.
Final answer:
Quantum tunneling is indeed a process that can be used by enzymes in our bodies to facilitate biochemical reactions by allowing particles to penetrate potential energy barriers, significantly affecting reaction rates.
Explanation:
Yes, it is possible that the enzymes in our bodies use quantum tunneling. Quantum tunneling is a phenomenon whereby particles penetrate a potential energy barrier despite having total energy less than the height of the barrier, which defies the principles of classical mechanics. This process was first analyzed by Friedrich Hund in 1927 and later used by George Gamow to explain alpha decay of atomic nuclei as a quantum-tunneling phenomenon.
In biological systems, quantum tunneling allows for the transfer of protons or electrons during enzyme-catalyzed reactions, significantly affecting the reaction rate by providing an alternate, low-energy pathway for the reaction to occur. Thus, enzymes can use quantum tunneling to speed up biochemical reactions that would otherwise occur much more slowly. This principle is vital in fields like biochemistry and molecular biology, providing a deeper understanding of enzyme kinetics and mechanism.
Grace rides her bike at a constant speed of 6 miles per hour. How far can she travel in 2 1/2 hours?
Answer:
15 miles
Explanation:
6 miles per hour
2 1/2 hours
6 x 2 = 12
6 x 1/2 = 3
12 + 3 = 15
A circular coil of radius 5.0 cm and resistance 0.20 Ω is placed in a uniform magnetic field perpendicular to the plane of the coil. The magnitude of the field changes with time according to B = 0.50e-20t T. What is the magnitude of the current induced in the coil at the time
Answer:
Explanation:
Given that,
Assume number of turn is
N= 1
Radius of coil is.
r = 5cm = 0.05m
Then, Area of the surface is given as
A = πr² = π × 0.05²
A = 7.85 × 10^-3 m²
Resistance of
R = 0.20 Ω
The magnetic field is a function of time
B = 0.50exp(-20t) T
Magnitude of induce current at
t = 2s
We need to find the induced emf
This induced voltage, ε can be quantified by:
ε = −NdΦ/dt
Φ = BAcosθ, but θ = 90°, they are perpendicular
So, Φ = BA
ε = −NdΦ/dt = −N d(BA) / dt
A is a constant
ε = −NA dB/dt
Then, B = 0.50exp(-20t)
So, dB/dt = 0.5 × -20 exp(-20t)
dB/dt = -10exp(-20t)
So,
ε = −NA dB/dt
ε = −NA × -10exp(-20t)
ε = 10 × NA exp(-20t)
Now from ohms law, ε = iR
So, I = ε / R
I = 10 × NA exp(-20t) / R
Substituting the values given
I = 10×1× 7.85 ×10^-3×exp(-20×2)/0.2
I = 1.67 × 10^-18 A
An observant fan at a baseball game notices that the radio commentators have lowered a microphone from their booth to just a few centimeters above the ground. (The microphone is used to pick up sounds from the field.) The fan also notices that the microphone is slowly swinging back and forth like a simple pendulum. Using her digital watch, she finds that 1010 complete oscillations take 20.2s20.2s. How high above the field is the radio booth?
Answer:
The radio booth is 0.993 meters above the field.
Explanation:
The pendulum covers 10 complete oscillations to take 20 s. We need to find the height above the radio booth. The time period of the pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{L}{g}} \\\\L=(\dfrac{T}{2\pi })^2g\\\\L=(\dfrac{(20/10)}{2\pi })^2\times 9.8\\\\L=0.993\ m[/tex]
So, the radio booth is 0.993 meters above the field.
The height of the radio booth is calculated to be approximately 9.9 cm by using the given oscillation period of 0.02 s and applying the pendulum formula.
A pendulum's period (T) is given by the formula:
T = 2π√(L/g)
Where:
T is the period of oscillationL is the length of the pendulum (distance from the pivot to the center of mass)g is the acceleration due to gravity (~9.8 m/s²)Given that 1010 complete oscillations take 20.2 seconds, we can find the period:
T = 20.2 s / 1010 = 0.02 sNow, using the period formula, we rearrange to solve for L:
T = 2π√(L/g)Square both sides to get:
(T / 2π)² = L / gThen solve for L:
L = g * (T / 2π)²Plugging in the values:
L = 9.8 m/s² * (0.02 s / 2π)²L ≈ 9.8 m/s² * (0.00318 s)²L ≈ 9.8 m/s² * 0.0000101 s²L ≈ 0.00009898 mSo, the height from which the microphone is hanging is approximately 9.9 cm.
Two rods are made of brass and have the same length. The cross section of one of the rods is circular, with a diameter of 2a. The other rod has a square cross section, where each side of the square is a length 2a. One end of the rods is attached to an immovable fixture which allows the rods to hang vertically. To the free end of each rod, a block of mass m is attached. Which rod, if either, will stretch more after the block is attached?
A. The one with the circular cross section will stretch more.
B. The one with the square cross section will stretch more.
C. Both will stretch by the same amount.
D. One cannot say which will stretch more without knowing the numerical values of a and m.
Answer:
A. The one with the circular cross section will stretch more.
Explanation:
According to the given data:
Two rods are made of brass and have the same length
Both rods having circular and square cross-section
Diameter of circular cross-section given is 2 a
therefore, Cross-section = [tex]A_c=\frac{\pi (2a)^2}{4}=\pi a^2[/tex]
If the length of square=2 a
then, Cross-section = [tex]A_{s}[/tex] = (2a)²=>4a²
Change in Length of rod = PL / AE
δL[tex]\alpha \frac{1}{A}[/tex]
Now, we are considering other factors same
the area of cross-section of square rod is more than Area of cross-section of circular rod
thus, the one with the circular cross section will stretch more
Answer:
The correct option is;
A. The one with the circular cross section will stretch more.
Explanation:
Here we have the cross section as being
1. Circular, with diameter, D = 2·a
2. Square cross section with each side length = 2·a
The area of the circular rod is then
Area of circle = π·D²/4 which is equal to
π×(2·a)²/4 = π·4·a²/4 = π·a²
The area of the rod with square cross section is
Area of square = Side² which gives
Area of cross section = (2·a)² = 4·a²
Therefore, since π = 3.142, the cross sectional area of the circular rod is less than that of the one with a square cross section
That is, π·a² = 3.142·a² < 4·a²
We note that the elongation or extension is directly proportional to the force applied as shown as follows
[tex]\frac{P}{A} = E\frac{\delta}{L}[/tex]
Where:
P/A = Force and
δ = Extension
The force is inversely proportional to the area, therefore a rod with less cross sectional area experiences more force and more elongation.
The motor in a toy car is powered by four batteries in series, which produce a total emf of 6.3 V. The motor draws 3.1 A and develops a 2.1 V back emf at normal speed. Each battery has a 0.18 Ω internal resistance.what is the resistance of the motor?
Answer:
0.635 Ω
Explanation:
Using
E-E' = IR.................... Equation 1
Where E = total Emf of the of the batteries, E' = back emf of the motor, I = current of the motor, R = combined resistance of the battery and the motor.
Make R the subject of the equation
R = E-E'/I.............. Equation 2
Given: E = 6.3 V, E' = 2.1 V, I = 3.1 A.
Substitute into equation 2
R = (6.3-2.1)/3.1
R = 4.2/3.1
R = 1.355 Ω
Since the motor and the batteries are connected in series,
R = R'+r'....................... Equation 3
Where R' = Resistance of the motor, r' = resistance of the batteries.
make R' the subject of the equation
R' = R-r'...................... Equation 4
Given: R = 1.355 Ω, r' = 0.18×4 (four batteries) = 0.72 Ω
Substitute into equation 4
R' = 1.355-0.72
R' = 0.635 Ω
Hence the resistance of the motor = 0.635 Ω
To find the motor's resistance, subtract the back emf from the total emf to get the effective voltage, then divide by the current and subtract the total internal resistance of the batteries. The resistance of the motor is found to be 1.35 Ω.
The question involves a toy car motor that is powered by four batteries producing a total electromotive force (emf) of 6.3 V, with each battery having an internal resistance of 0.18 Ω. Upon running, the motor develops a 2.1 V back emf and draws a current of 3.1 A. To find the resistance of the motor, one can use the net emf equation and Ohm's law.
The total emf minus the back emf is the effective voltage across the motor. Thus, the effective voltage (Veff) is 6.3 V - 2.1 V = 4.2 V. The total internal resistance (Rint) in series is 4 batteries × 0.18 Ω = 0.72 Ω. Applying Ohm's law, V = IR where I is the current and R is the resistance, we solve for the resistance (Rmotor) of the motor:
Veff = I(Rmotor + Rint)
Rmotor = Veff/I - Rint
Rmotor = 4.2 V / 3.1 A - 0.72 Ω
Rmotor = 1.35 Ω
Therefore, the resistance of the motor is 1.35 Ω.
You sit at the middle of a large turntable at an amusement park as it is set spinning on nearly frictionless bearings, and then allowed to spin freely. When you crawl toward the edge of the turntable, does the rate of the rotation increase, decrease, or remain unchanged, and why
Answer:
Decrease
Explanation:
If you crawl to the rim the rotational speed will decrease. The law of conservation of angular momentum supports this answer. And it states that :
"When the net external torque acting on a system about a given axis is. zero , the total angular momentum of the system about that axis remains constant."
Final answer:
When moving toward the edge of a spinning turntable, the rate of rotation decreases due to the principle of conservation of angular momentum, which necessitates a decrease in angular velocity to compensate for an increased moment of inertia.
Explanation:
When you crawl toward the edge of a large turntable at an amusement park while it is spinning, the rate of the rotation decreases. This phenomenon is explained by the principle of conservation of angular momentum, which states that if no external torque acts on a system, the total angular momentum of the system remains constant.
Angular momentum is given by the product of the moment of inertia (I) and the angular velocity (ω), represented by the equation L = Iω. As you move away from the axis of rotation, your moment of inertia increases because the moment of inertia is directly proportional to the square of the distance from the axis of rotation. In order to conserve angular momentum, if the moment of inertia increases, the angular velocity must decrease accordingly.
Match the following:Part A1. developed geocentric theory 2. developed heliocentric theory 3. founded nursing profession 4. invented barometer 5. considered "Father of Modern Science" 6. developed law of universal gravitation 7. examined the inner workings of the human body 8. developed metric temperature scale Part Ba. Galileo b. Copernicus c. Celsius d. Newton e. Nightingale f. Aristotle g. Torricelli h. Vesalius
Answer:
1 -f,2 -b,3-e,4 -g,5-a,6 - d, 7 - h, 8 - c
Explanation:
1. developed geocentric theory - Aristotle
2. developed heliocentric theory - Copernicus in 1543
3. founded nursing profession - Florence Nightingale
4. invented barometer - Evangelista Torricelli in 1643
5. "Father of Modern Science" - Galileo Galilei
6.law of universal gravitation -Sir Isaac Newton
7. examined the inner workings of the human body - Vesalius
8. developed metric temperature scale - Andres Celsius in 1742