Explanation:
(a) The acceleration is 9.8 m/s² downwards.
(b) At the maximum height, the velocity is 0 m/s.
(c) v = at + v₀
0 m/s = (-9.8 m/s²) (1.91 s) + v₀
v₀ = 18.7 m/s
(d) Δy = vt − ½ at²
Δy = (0 m/s) (1.91 s) − ½ (-9.8 m/s²) (1.91 s)²
Δy = 17.9 m
g beats can be heard when Question 5 options: when sound waves interfere constructively when sound waves of same frequency interfere when sound waves of slightly different frequencies interfere when sound waves of different amplitudes interfere
Answer:
Explanation:
Beats are produced when two sound waves from two different sources having slightly different frequencies interfere . Due to both constructive and destructive interference , high and low intensity sound is heard at some regular interval . This is called rate of beat formation . Amplitudes of sound may be same or different . High intensity sound heard per unit time is called beat . It is equal to difference of frequencies of the sound waves interfering
each other .
If an object is determined to have a negative charge of 1.6 micro Coulomb, you can conclude that the object has an excess of
Answer:
The object has an excess of [tex]10^{13}[/tex] electrons.
Explanation:
When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:
[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]
The object has an excess of [tex]10^{13}[/tex] electrons.
symbol of science hhshsjsiwtwwisjzhJava
Answer:
is this a company name.? java is a computer software right..
a jet fighter accelerates at 17.7 m/s^2 , increasing its velocity from 119 m/s to 233 m/s. how much time does that take?
Answer:
6.44 s
Explanation:
Given:
v₀ = 119 m/s
v = 233 m/s
a = 17.7 m/s²
Find: t
v = at + v₀
(233 m/s) = (17.7 m/s²) t + (119 m/s)
t = 6.44 s
59-kg girl weighs herself by standing on a scale in an elevator. What does the scale read when: the elevator is descending at a constant rate of 8 m/s
Answer:
578.2 N
Explanation:
The computation of reading on the scale is shown below:-
Data provided in the question
Weight of a girl = 59 kg
Constant rate = 8 m/s
Since the elevator is descended so the acceleration is zero
As we know that
Reading on the scale is
[tex]F = m\times g[/tex]
where, m = 59 kg
g [tex]= 59 \times 9.8 m/s^2[/tex]
So, the reading on the scale is
= 578.2 N
Therefore for computing the reading on the scale we simply applied the above formula.
A 2.0-m wire carrying a current of 0.60 A is oriented parallel to a uniform magnetic field of 0.50 T. What is the magnitude of the force it experiences
Answer:
The force experienced is 0.6 N
Explanation:
Given data
length of wire L= 2 m
current in wire I= 0.6 A
magnetic field B= 0.5
The force experienced can be represented as
[tex]F= BIL[/tex]
[tex]F= 0.5*0.6*2\\\F= 0.6 N[/tex]
A piano tuner hears a beat every 2.20 s when listening to a 266.0 Hz tuning fork and a single piano string. What are the two possible frequencies (in Hz) of the string? (Give your answers to at least one decimal place.)
Answer:
The lower frequency is [tex]f_1 = 265.55 \ Hz[/tex]
The higher frequency is [tex]f_2 = 266.4546 \ Hz[/tex]
Explanation:
From the question we are told that
The period is [tex]T = 2.20 \ s[/tex]
The frequency of the tuning fork is [tex]f = 266.0 \ Hz[/tex]
Generally the beat frequency is mathematically represented as
[tex]f_b = \frac{1}{T}[/tex]
substituting values
[tex]f_b = \frac{1}{2.20}[/tex]
[tex]f_b = 0.4546 \ Hz[/tex]
Since the beat frequency is gotten from the beat produced by the tuning fork and and the string then
The possible frequency of the string ranges from
[tex]f_1 = f- f _b[/tex]
to
[tex]f_2 = f + f_b[/tex]
Now substituting values
[tex]f_1 = 266.0 - 0.4546[/tex]
[tex]f_1 = 265.55 \ Hz[/tex]
For [tex]f_2[/tex]
[tex]f_2 = 266 + 0.4546[/tex]
[tex]f_2 = 266.4546 \ Hz[/tex]
Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal to which of the following quantities?
a. The flux of the magnetic field through a surface which has the loop as its boundary.
b. The negative of the time rate of change of the flux of the magnetic field through a surface which has the loop as its boundary.
c. The line integral of the magnetic field around the closed loop.
d. The flux of the electric field through a surface which has the loop as its boundary.
Answer:
(C). The line integral of the magnetic field around a closed loop
Explanation:
Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.
This can be written mathematically as;
[tex]EMF = -\frac{\delta \phi _B}{\delta t}[/tex]
[tex](\frac{\delta \phi _B}{\delta t} )[/tex] is the rate of change of the magnetic flux through a surface bounded by the loop.
ΔФ = BA
where;
ΔФ is change in flux
B is the magnetic field
A is the area of the loop
Thus, according to Faraday's law of electric generators
∫BdL = [tex]\frac{\delta \phi _B}{\delta t}[/tex] = EMF
Therefore, the line integral of the magnetic field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.
The correct option is "C"
(C). The line integral of the magnetic field around a closed loop
Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal to: D. The flux of the electric field through a surface which has the loop as its boundary.
In Physics, the surface integral with respect to the normal component of a magnetic field over a surface is the magnetic flux through that surface and it is typically denoted by the symbol [tex]\phi[/tex].
Faraday's Law states that the negative of the time rate of change ([tex]\Delta t)[/tex] of the flux of the magnetic field ([tex]\phi[/tex]) through a surface is directly proportional to the flux ([tex]\phi[/tex]) of the electric field through a surface which has the loop as its boundary.
Mathematically, Faraday's Law is given by the formula:
[tex]E.m.f = -N\frac{\Delta \phi}{\Delta t}[/tex]
Where:
N is the number of turns.Read more: https://brainly.com/question/15121836
9. A 30 cm ruler is found to have a center of mass of 15.6 cm. The percent error of the center of mass is _____, if the ruler is assumed to have uniform mass.
Answer:
3.85 percent
Explanation:
From the question,
Percentage error = (error/actual)×100................ Equation 1
Given: actual center of mass = 15 cm, error = 15.6-15 = 0.6 cm
Substitute these values into equation 1
Percentage error = (0.6/15.6)×100
Percentage error = 3.85 percent
Hence the percentage error of the uniform mass = 3.85 percent
A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 15.2 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. (a) Find the net charge on the shell.
Answer:
The net charge on the shell is 30x10^-9C
Explanation:
Pls see attached file
Determine the maximum height and range of a projectile fired at a height of 6 feet above the ground with an initial velocity of 100 feet per second at an angle of 40 degrees above the horizontal.Maximum heightRange Question 20 options:a) 70.56 feet183.38 feet b) 92.75 feet310.59 feet c) 92.75 feet183.38 feet d) 70.56 feet314.74 feet e)
Answer:
C is the correct answer
Explanation:
If a sound with frequency fs is produced by a source traveling along a line with speed vs. If an observer is traveling with speed vo along the same line from the opposite direction toward the source, then the frequency of the sound heard by the observer is fo = c + vo c − vs fs where c is the speed of sound, about 332 m/s. (This is the Doppler effect.) Suppose that, at a particular moment, you are in a train traveling at 32 m/s and accelerating at 1.3 m/s2. A train is approaching you from the opposite direction on the other track at 48 m/s, accelerating at 1.9 m/s2, and sounds its whistle, which has a frequency of 439 Hz. At that instant, what is the perceived frequency that you hear? (Round your answer to one decimal place.) Hz
Answer: The frequency heard is 562.7 Hz.
Explanation: Doppler Effect happens when there is shift in frequency during a realtive motion between a source and the observer of that source.
It can be calculated as:
[tex]f_{o} = f_{s}(\frac{c+v_{o}}{c-v_{s}} )[/tex]
where:
c is the speed of light (c = 332m/s)
all the subscripted s is related to the Source (frequency, velocity);
all the subscripted o is related to the Observer (frequency, velocity);
As the source is moving towards the observer and the observer is moving towards the source, the velocities of each are opposite related to direction.
So, the frequency perceived by the observer:
[tex]f_{o} = 439(\frac{332+32}{332-48} )[/tex]
[tex]f_{o} = 439(\frac{364}{284} )[/tex]
[tex]f_{o} = 439(1.282 )[/tex]
[tex]f_{o}[/tex] = 562.7 Hz
At this condition, the observer hears the train's horn in a perceived frequency of 562.7 Hz
The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the bal
Complete Question
The complete question is gotten from OpenStax
A soccer player starts at rest and accelerates forward, reaching a velocity of 8.00 m/s in 2.50 s ? The player’s mass is 70.0 kg, and air resistance is negligible.
The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball
Answer:
The time it will take is [tex]t = 1.4907 \ s[/tex]
Explanation:
From the question we are told that
The force experienced by the player is [tex]F = 126 \ N[/tex]
The distance of the ball from the player is [tex]d = 2.00 \ m[/tex]
The initial velocity is u = 0 m/s because the player stopped
From the Newton law the acceleration of the player is mathematically evaluated as
[tex]a = \frac{F}{m }[/tex] [i,e F = ma ]
substituting values
[tex]a = \frac{126}{70}[/tex]
[tex]a = 1.8 \ m/s^2[/tex]
Now from the equation of motion we have that
[tex]s = ut + \frac{1}{2} at^2[/tex]
substituting values
[tex]2.0 = 0 + \frac{1}{2} * 1.8 * t^2[/tex]
[tex]t = \sqrt{ \frac{2.0}{0.9} }[/tex]
[tex]t = 1.4907 \ s[/tex]
A centrifuge has an angular velocity of 3,000 rpm, what is the acceleration (in unit of the earth gravity) at a point with a radius of 10 cm
Answer:
[tex]a_{r} = 1006.382g \,\frac{m}{s^{2}}[/tex]
Explanation:
Let suppose that centrifuge is rotating at constant angular speed, which means that resultant acceleration is equal to radial acceleration at given radius, whose formula is:
[tex]a_{r} = \omega^{2}\cdot R[/tex]
Where:
[tex]\omega[/tex] - Angular speed, measured in radians per second.
[tex]R[/tex] - Radius of rotation, measured in meters.
The angular speed is first determined:
[tex]\omega = \frac{\pi}{30}\cdot \dot n[/tex]
Where [tex]\dot n[/tex] is the angular speed, measured in revolutions per minute.
If [tex]\dot n = 3000\,rpm[/tex], the angular speed measured in radians per second is:
[tex]\omega = \frac{\pi}{30}\cdot (3000\,rpm)[/tex]
[tex]\omega \approx 314.159\,\frac{rad}{s}[/tex]
Now, if [tex]\omega = 314.159\,\frac{rad}{s}[/tex] and [tex]R = 0.1\,m[/tex], the resultant acceleration is then:
[tex]a_{r} = \left(314.159\,\frac{rad}{s} \right)^{2}\cdot (0.1\,m)[/tex]
[tex]a_{r} = 9869.588\,\frac{m}{s^{2}}[/tex]
If gravitational acceleration is equal to 9.807 meters per square second, then the radial acceleration is equivalent to 1006.382 times the gravitational acceleration. That is:
[tex]a_{r} = 1006.382g \,\frac{m}{s^{2}}[/tex]
Write the equations of motion of a pendulum consisting of a thin, 2 kg stick of length suspended from a pivot. How long should the rod be in order for the period to be exactly 1 sec
Answer:
3g/(8π²) ≈ 0.372 m
Explanation:
Draw a free body diagram. There is a weight force at the center of the pendulum.
Sum of the torques about the pivot:
∑τ = Iα
mg (½ L sin θ) = (⅓ mL²) α
3g sin θ = 2L α
α = 3g/(2L) sin θ
For small θ, sin θ ≈ θ.
α = 3g/(2L) θ
θ" = 3g/(2L) θ
The solution of this differential equation is:
θ = θ₀ cos(√(3g/(2L)) t)
So the period is:
T = 2π / √(3g/(2L))
If the period is 1 second:
1 = 2π / √(3g/(2L))
√(3g/(2L)) = 2π
3g/(2L) = 4π²
L = 3g/(8π²)
L ≈ 0.372 m
The length of the pendulum rod is 0.37 m.
What is meant by time period of a pendulum ?The time period of a pendulum is defined as the time taken by the pendulum to complete one oscillation.
Here,
The mass of the pendulum, m = 2 kg
Time period of the pendulum, T = 1 s
Since, the pendulum is suspended from the pivot and is oscillating, at the position of the pendulum when it makes an angle θ with the pivot, there is a force of weight acting at the center of the pendulum.
The length of pendulum at that point = L/2
The perpendicular distance at that point, r = (L/2) sinθ
Therefore, the torque acting on the pendulum at that point,
τ = Iα
where I is the moment of inertia of the pendulum and α is the angular acceleration.
mg (L/2 sinθ) = (mL²/3)α
1/2 gsinθ = 1/3 Lα
Therefore,
α = (3g/2L) sinθ
For smaller values of θ, we can take sinθ = θ
So, α = (3g/2L) θ
We know that α = θ''
where θ is the angular displacement.
Therefore,
θ'' = (3g/2L) θ
So, ω = √3g/2L
Therefore, the equation of motion of the pendulum can be written as,
θ = θ₀ cos(ωt)
θ = θ₀ cos [(√3g/2L) t]
So, time period of the pendulum,
T = 2[tex]\pi[/tex]/ω
T = 2[tex]\pi[/tex]/(√3g/2L)
2[tex]\pi[/tex]/(√3g/2L) = 1
(√3g/2L) = 2[tex]\pi[/tex]
Therefore, length of the rod,
L = 3g/8[tex]\pi[/tex]²
L = 0.37 m
Hence,
The length of the pendulum rod is 0.37 m.
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Attaching the image file here.
A 6.50-m-long iron wire is 1.50 mm in diameter and carries a uniform current density of 4.07 MA/m^2. Find the voltage between the two ends of the wire.
Answer:
V = 0.45 Volts
Explanation:
First we need to find the total current passing through the wire. That can be given by:
Total Current = I = (Current Density)(Surface Area of Wire)
I = (Current Density)(2πrL)
where,
r = radius = 1.5/2 mm = 0.75 mm = 0.75 x 10⁻³ m
L = Length of Wire = 6.5 m
Therefore,
I = (4.07 x 10⁻³ A/m²)[2π(0.75 x 10⁻³ m)(6.5 m)]
I = 1.25 x 10⁻⁴ A
Now, we need to find resistance of wire:
R = ρL/A
where,
ρ = resistivity of iron = 9.71 x 10⁻⁸ Ωm
A = Cross-sectional Area = πr² = π(0.75 x 10⁻³ m)² = 1.77 x 10⁻⁶ m²
Therefore,
R = (9.71 x 10⁻⁸ Ωm)(6.5 m)/(1.77 x 10⁻⁶ m²)
R = 0.36 Ω
From Ohm's Law:
Voltage = V = IR
V = (1.25 x 10⁻⁴ A)(0.36 Ω)
V = 0.45 Volts
A stretched string is 120 cm long and has a linear density of 0.022 g/cm. What tension in the string will result in a second harmonic of 590 Hz
Answer:
T = 15,576 N
Explanation:
The speed of a wave on a string is given by
v = √ T /ρ rho
also the speed of the wave is given by the relationship
v = λ f
we substitute
λ f = √ T /ρ
T = (lam f)² ρ
let's find the wavelength in a string, fixed at the ends, the relation that gives the wavelength is
L= λ/2 n
λ= 2L / n
we substitute
T = (2L / n f)²ρ rho
let's calculate
T = (2 1.20 / 2 590) 0.022
T = 15,576 N
A spherical balloon is inflated with helium at a rate of 140pift^3/min How fast is the balloon's radius increasing when the radius is 7ft
Answer:
dr = 0.71 ft/min
the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.
Explanation:
Given;
Rate of inflation dV = 140pift^3/min
Radius r = 7 ft
Change in radius = dr
Volume of a spherical balloon is;
V = (4/3)πr^3
The change in volume can be derived by differentiating both sides;
dV = (4πr^2)dr
Making dr the subject of formula;
dr = dV/(4πr^2)
Substituting the given values;
dr = 140π/(4π×7^2)
dr = 0.714285714285 ft/min
dr = 0.71 ft/min
the balloon's radius is increasing at 0.71 ft/min when the radius is 7ft.
Find the change in the force of gravity between two planets when the distance between them becomes 10 times smaller. g
Answer:
F' = F/16
So, the force of gravity has become 16 times less than initial value.
Explanation:
The force of gravity between two planets, is given by the following formula:
F = Gm₁m₂/r² ----------- equation 1
where,
F = Force of gravity between two planets
G = Gravitational Constant
m₁ = Mass of one planet
m₂ = Mass of other plant
r = Distance between two planets
Now, if the distance between the planets (r) is 10 times smaller, then Force of gravity will become:
F' = Gm₁m₂/(4r)²
F' = (1/16) (Gm₁m₂/r²)
using equation 1:
F' = F/16
So, the force of gravity has become 16 times less than initial value.
The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?
Answer:
The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.
Explanation:
The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.
The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.
Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.
I hope it helps you!
Defination coulomb's law.
Answer:
a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.
What length (in m) should an oboe have to produce a fundamental frequency of 294 Hz on a day when the speed of sound is 350 m/s? It is open at both ends.
Answer:
L = 0.60 m
The length in metres should be 0.60 m
Explanation:
A pipe open at both ends can have a standing wave pattern with resonant frequency;
f = nv/2L ........1
Where;
v = velocity of sound
L = length of pipe
n = 1 for the fundamental frequency f1
Given;
Fundamental frequency f1 = 294 Hz
Velocity v = 350 m/s
n = 1
From equation 1;
Making L the subject of formula;
L = nv/2f1
Substituting the given values;
L = 1×350/(2×294)
L = 0.595238095238 m
L = 0.60 m
The length in metres should be 0.60 m
You have a circuit of three resistors in series connected to a battery. You add a fourth resistor, also in series, to the combination. As a result:_______.
A. Power delivered from the battery to combination increases.
B. It is impossible to give the answer without knowing the actual resistances and voltage of the battery.
C. Power delivered from the battery to combination is unchanged.
D. Power delivered from the battery to combination decreases.
Answer:
D
Explanation:
The power equation is P= V^2/R
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As a result of the given scenario, power delivered from the battery to combination decreases. The correct option is D.
What is a resistors?A resistor is a two-terminal passive electrical component that uses electrical resistance as a circuit element.
Resistors are used in electronic circuits to reduce current flow, adjust signal levels, divide voltages, and bias active elements.
A resistor is a component of an electronic circuit that limits or regulates the flow of electrical current. Resistors can also be used to supply a fixed voltage to an active device such as a transistor.
The current through resistors is the same when they are connected in series. The battery voltage is divided among resistors.
Adding more resistors to a series circuit increases total resistance and thus lowers current. However, in a parallel circuit, adding more resistors in parallel creates more options while decreasing total resistance.
Thus, the correct option is D.
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is tantalum least reactive or more
Answer:
it is more reactive in high temperature than in low temperature.
At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)
Answer:
The temperature of silver at this given resistivity is 2971.1 ⁰C
Explanation:
The resistivity of silver is calculated as follows;
[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]
where;
Rt is the resistivity of silver at the given temperature
Ro is the resistivity of silver at room temperature
α is the temperature coefficient of resistance
To is the room temperature
T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature
[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]
Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m
When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;
[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]
Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C
Near the surface of Earth, an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (magnitude and sign) would have to be placed on a penny that has a mass of 3 g to cause it to rise into the air with an upward acceleration of 0.19 m/s2?
Answer:
q = 2.997*10^-4C
Explanation:
In order to find the required charge that the penny have to have, to acquire an upward acceleration, you take into account that the electric force on the penny must be higher than the weight of the penny.
You use the second Newton law to sum both electrical and gravitational forces:
[tex]F_e-W=ma\\\\qE-mg=ma[/tex] (1)
Fe: electric force
W: weight of the penny
q: required charge = ?
m: mass of the penny = 3g = 0.003kg
E: magnitude of the electric field = 100N/C
g: gravitational acceleration = 9.8m/s^2
a: acceleration of the penny = 0.19m/s^2
You solve the equation (1) for q, and replace the values of the other parameters:
[tex]q=\frac{ma+mg}{E}=\frac{m(a+g)}{E}\\\\q=\frac{(0.003kg)(0.19m/s^2+9.8m/s^2)}{100N/C}\\\\q=2.997*10^{-4}C[/tex]
It is necessary that the penny has a charge of 2.997*10^-4 C, in order to acquire an upward acceleration of 0.19m/s^2
Saturn has a diameter of 74,900 miles. As the planet rotates on its axis, a point on its equator travels at 527,787 miles per day.
a. Find the angular speed of a point on its equator in radians per day.
b. Find the number of rotations the planet makes per day.
Answer:
a. ω = 14.1 radians per day
b. ω = 2.24 rotations per day
Explanation:
a.
The relationship between linear and angular velocity is given as:
v = rω
where,
v = linear speed of point = 527787 miles/day
r = radius of Saturn = diameter/2 = 74900 miles/2 = 37450 miles
ω = angular velocity of point = ?
Therefore,
527787 miles/day = (37450 miles)ω
ω = (527787 miles/day)/(37450 miles)
ω = 14.1 radians per day
b.
The number of rotations per day can be found by converting the units of angular acceleration:
ω = (14.1 radians per day)(1 rotation/2π radians)
ω = 2.24 rotations per day
PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!! In California, Clay is surfing on a wave that propels him toward the beach with a speed of 5.0 m/s. The wave crests are each 20 m apart. What is the frequency of the water wave? (please show your work and equation used)
Answer:
Frequency= 0.25m
Period= 4.0 secs
Explanation:
Clay is surfing on a wave with a speed of 5.0m/s
The wave crests are 20m apart
Therefore, the frequency of the wave can be calculated as follows
Frequency= wave speed/distance
= 5.0/20
= 0.25m
The period (T) can be calculated as follows
T= 1/frequency
T = 1/0.25
T= 4.0secs
Hence the frequency is 0.25m and the period is 4.0 secs
Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle. The horizontal surface on which block A (mass 2.10 kg) moves is frictionless. The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s. a)What is the tension force that the rope exerts on block B? b)What is the tension force that the rope exerts on block A? c)What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?
Answer:
(a) 62.3 N
(b) 1.89 N
(c) 0.430 kg m²
Explanation:
(a) Find the acceleration of block B.
Δy = v₀ t + ½ at²
1.80 m = (0 m/s) (2.00 s) + ½ a (2.00 s)²
a = 0.90 m/s²
Draw a free body diagram of block B. There are two forces:
Weight force mg pulling down,
and tension force Tb pulling up.
Sum of forces in the -y direction:
∑F = ma
mg − Tb = ma
Tb = m (g − a)
Tb = (7.00 kg) (9.8 m/s² − 0.90 m/s²)
Tb = 62.3 N
(b) Draw a free body diagram of block A. There are three forces:
Weight force mg pulling down,
Normal force N pushing up,
and tension force Ta pulling right.
Sum of forces in the +x direction:
∑F = ma
Ta = ma
Ta = (2.10 kg) (0.90 m/s²)
Ta = 1.89 N
(c) Draw a free body diagram of the pulley. There are two forces:
Tension force Tb pulling down,
and tension force Ta pulling left.
Sum of torques in the clockwise direction:
∑τ = Iα
Tb r − Ta r = Iα
(Tb − Ta) r = I (a/r)
I = (Tb − Ta) r² / a
I = (62.3 N − 1.89 N) (0.080 m)² / (0.90 m/s²)
I = 0.430 kg m²
The tension force that the rope exerts on block B is 62.3 N, the tension force that the rope exerts on block A is 1.89 N, and the moment of inertia of the pulley for rotation about the axle on which it is mounted is [tex]\rm 0.430 \; kg\;m^2[/tex].
Given :
Block A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block B is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius 0.080 m) because the pulley rotates on a frictionless axle.The horizontal surface on which block A (mass 2.10 kg) moves is frictionless.The system is released from rest, and block B (mass 7.00 kg) moves downward 1.80 m in 2.00 s.a) First, determine the acceleration of the B block.
[tex]\rm s = ut + \dfrac{1}{2}at^2[/tex]
[tex]\rm 1.8 = \dfrac{1}{2}\times a\times (2)^2[/tex]
[tex]\rm a = 0.9\; m/sec^2[/tex]
Now, apply Newton's second law of motion in order to determine the tension force that the rope exerts on block B.
[tex]\rm \sum F=ma[/tex]
[tex]\rm mg-T_b=ma[/tex]
[tex]\rm T_b = m(g-a)[/tex]
[tex]\rm T_b = 7\times (9.8-0.9)[/tex]
[tex]\rm T_b = 62.3\;N[/tex]
b) Now, again apply Newton's second law of motion in order to determine the tension force that the rope exerts on block A.
[tex]\rm \sum F=ma[/tex]
[tex]\rm T_a=ma[/tex]
[tex]\rm T_a = 2.1\times 0.9[/tex]
[tex]\rm T_a = 1.89\;N[/tex]
c) The sum of the torque in order to determine the moment of inertia of the pulley for rotation about the axle on which it is mounted.
[tex]\rm \sum \tau = I\alpha[/tex]
[tex]\rm T_br-T_ar = I\alpha[/tex]
[tex]\rm I = \dfrac{(T_b-T_a)r^2}{a}[/tex]
Now, substitute the values of the known terms in the above expression.
[tex]\rm I = \dfrac{(62.3-1.89)(0.080)^2}{0.90}[/tex]
[tex]\rm I = 0.430 \; kg\;m^2[/tex]
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A charge of 4.4 10-6 C is located in a uniform electric field of intensity 3.9 105 N/C. How much work (in Joules) is required to move this charge 50 cm along a path making an angle of 40
Answer:
The work required to move this charge is 0.657 J
Explanation:
Given;
magnitude of charge, q = 4.4 x 10⁻⁶ C
Electric field strength, E = 3.9 x 10⁵ N/C
distance moved by the charge, d = 50 cm = 0.5m
angle of the path, θ = 40°
Work done is given as;
W = Fd
W = FdCosθ
where;
F is the force on the charge;
According the coulomb's law;
F = Eq
F = 3.9 x 10⁵ x 4.4 x 10⁻⁶ = 1.716 N
W = FdCosθ
W = 1.716 x 0.5 x Cos40
W = 0.657 J
Therefore, the work required to move this charge is 0.657 J