Given that,
Central maximum = 1 cm
Distance from the window shade to the wall =4 m
We know that,
The visible range of the sun light is 400 nm to 700 nm.
(a). We need to calculate the average wavelength
Using formula of average wavelength
[tex]\lambda_{avg}=\dfrac{\lambda_{1}+\lambda_{2}}{2}[/tex]
Put the value into the formula
[tex]\lambda_{avg}=\dfrac{400+700}{2}[/tex]
[tex]\lambda_{avg}=550\ nm[/tex]
(b). We need to calculate the diameter of the pinhole
Using formula for diameter
[tex]w=\dfrac{2.44\lambda L}{D}[/tex]
[tex]D=\dfrac{2.44\lambda L}{w}[/tex]
Put the value into the formula
[tex]D=\dfrac{2.44\times550\times10^{-9}\times4}{1\times10^{-2}}[/tex]
[tex]D=0.537\ mm[/tex]
Hence, (a). The average wavelength 550 nm.
(b). The diameter of the pinhole is 0.537 mm.
A Cannonball is shot at an angle of 35.0 degrees and is in flight for 11.0 seconds before hitting the ground at the same height from which it was shot.
A. What is the magnitude of the inital velocity?B. What was the maximum height reached by the cannonball?C. How far, horizontally, did it travel?
Answer:
Explanation:
According to Equations of Projectile motion :
[tex]Time\ of\ Flight = \frac{2vsin(x)}{g}[/tex]
vsin(x) = 11 * 9.8 / 2 = 53.9 m/sec
(A) v (Initial velocity) = 11 * 9.8 / 2 * sin(35) = 94.56 m/sec
[tex]Maximum Height = \frac{(vsinx)^{2} }{2g}[/tex]
(B) Maximum Height = 53.9 * 53.9 / 2 * 9.8 = 142.2 m
[tex]Horizontal Range = vcosx * t[/tex]
(C) Horizontal Range = 94.56 * 0.81 * 11 = 842.52 m
A 500 nm wavelength light illuminates a soap film with an index of refraction 1.33 to make it look bright. If the beam of light is incident normal on the film, what is the minimum thickness of the film
Answer:
t(min) = 94nm
Explanation:
The wavelength of the light incident is 500 nm.
The refractive index of the film is 1.33.
The minimum thickness of the soap film required for constructive interference.
The thickness of the film for the constructive interference is given by:
2*t= (m + 1/2) λ′
Now, λ′ = λ/μ = 500/1.33 = 376nm
The minimum thickness of the film
′t′ will be at m=0 :
2*t(min) = (0 + 1/2) 376
t(min) = 94nm
You have two capacitors and want to connect them across a voltage source (battery) to store the maximum amount of energy. Should they be connected in series or in parallel?
Answer:
In parallel
Explanation:
Ctotal = C1 + C2 + ... + Cn
In the lab , you have an electric field with a strength of 1,860 N/C. If the force on a particle with an unknown charge is 0.02796 N, what is the value of the charge on this particle.
Answer:
The charge is [tex]q = 1.50 *10^{-5} \ C[/tex]
Explanation:
From the question we are told that
The electric field strength is [tex]E = 1860 \ N/C[/tex]
The force is [tex]F = 0.02796 \ N[/tex]
Generally the charge on this particle is mathematically represented as
[tex]q = \frac{F}{E}[/tex]
=> [tex]q = \frac{0.02796}{ 1860}[/tex]
=> [tex]q = 1.50 *10^{-5} \ C[/tex]
Nine tree lights are connected inparallel across 120-V potential difference. The cord to the wall socket carries a current of 0.43 A. Required:a. Detrmine the resistance of one of the bulbs.b. What would the current be if the bulbs were connected in series?
Answer:
a)3000ohm
b)4.44mA
Explanation:
a) we were given a Nine tree lights connected inparallel across 120-V potential difference, since the resistor are in parallel we use the expresion below
1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉
But according to ohm'law which can be expressed below
V=IR
R=V/I
R(total)= 120/0.36
= 333.33ohm
1/R(total)= 1/R₁ + 1/R₂ + 1/R₃ + 1/R₃ +.... 1/R₉
R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉
1/R(total)=9/R
1/333.33= 9/R
R= 3000ohm
Therefore, the resistance is 3000ohm
b)the bulbs were connected in series here, then for series connection we use below expression
R₁=R₂ =R₃ =R₄= R₅=R₆=R₇=R₈=R₉
R(total)=9R
= 9*3000
=27000ohm
I=VR
I=V/R
I= 120/27000
= 4.44*10⁻³A
4.44mA
Therefore, the current is 4.4mA
A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1.0 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).
Required:
a. Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.
b. Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm^2. How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?
Answer:
1. 6.99x 10^-6V/m
2. 18m
Explanation:
See attached file
A resistor and a capacitor are connected in series to an ideal battery of constant terminal voltage. At the moment contact is made with the battery, the voltage across the resistor is
Answer:
At the moment contact is made with the battery, the voltage across the resistor is equal to the batteries terminal voltage
Explanation;
Because at series connection the battery and resistor have equal voltage
An LR circuit consists of a 35-mH inductor, ac resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed
Answer:
I = 1.23 A
Explanation:
In an RL circuit current passing is described by
I = E / R (1 - [tex]e^{-Rt/L}[/tex])
Let's reduce the magnitudes to the SI system
L = 35 mH = 35 10⁻³ H
t = 5.0 ms = 5.0 10⁻³ s
let's calculate
I = 18/12 (1 - [tex]e^{-12 .. 5 {10}^{-3}/35 .. {10}^{-3} }[/tex]e (- 5 10-3 12/35 10-3))
I = 1.5 (1- [tex]e^{-1.715}[/tex])
I = 1.23 A
if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when the pipe flows full?
Answer:
0.35 m³/s
Explanation:
When the pipe's depth is 0.4 m, the area of the circular segment is:
A = ½ R² (θ − sin θ)
The depth of the water is:
h = R (1 − cos(θ/2))
Solving for θ:
0.4 = 0.5 (1 − cos(θ/2))
0.8 = 1 − cos(θ/2)
cos(θ/2) = 0.2
θ/2 = acos(0.2)
θ = 2 acos(0.2)
θ ≈ 2.74 rad
The area is therefore:
A = ½ (0.5 m)² (2.74 − sin 2.74)
A = 0.338 m²
The cross-sectional area when the pipe is full is:
A = π (0.5 m)²
A = 0.785 m²
The flow velocity is constant:
v = v
Q / A = Q / A
(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)
Q = 0.35 m³/s
Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
A. Which skater, if either, has the greater momentum after the push-off? Explain.
B. Which skater, if either, has the greater speed after the push-off? Explain.
Answer:
the two ice skater have the same momentum but the are in different directions.
Paula will have a greater speed than Ricardo after the push-off.
Explanation:
Given that:
Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.
A. Which skater, if either, has the greater momentum after the push-off? Explain.
The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off
The law of conservation of momentum states that the total momentum of two or more objects acting upon one another will not change, provided there are no external forces acting on them.
So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.
Momentum is the product of mass and velocity.
SO, from the information given:
Let represent the mass of Paula with [tex]m_{Pa}[/tex] and its initial velocity with [tex]u_{Pa}[/tex]
Let represent the mass of Ricardo with [tex]m_{Ri}[/tex] and its initial velocity with [tex]u_{Ri}[/tex]
At rest ;
their velocities will be zero, i.e
[tex]u_{Pa}[/tex] = [tex]u_{Ri}[/tex] = 0
The initial momentum for this process can be represented as :
[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = 0
after push off from each other then their final velocity will be [tex]v_{Pa}[/tex] and [tex]v_{Ri}[/tex]
The we can say their final momentum is:
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0
Using the law of conservation of momentum as states earlier.
Initial momentum = final momentum = 0
[tex]m_{Pa}[/tex][tex]u_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]u_{Ri}[/tex] = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
Since the initial velocities are stating at rest then ; u = 0
[tex]m_{Pa}[/tex](0) + [tex]m_{Pa}[/tex](0) = [tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] + [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex] = 0
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = - [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.
B. Which skater, if either, has the greater speed after the push-off? Explain.
Given that Ricardo weighs more than Paula
So [tex]m_{Ri} > m_{Pa}[/tex] ;
Then [tex]\mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}[/tex]
The magnitude of their momentum which is a product of mass and velocity can now be expressed as:
[tex]m_{Pa}[/tex][tex]v_{Pa}[/tex] = [tex]m_{Ri}[/tex][tex]v_{Ri}[/tex]
The ratio is
[tex]\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1[/tex]
[tex]v_{Pa} >v_{Ri}[/tex]
Therefore, Paula will have a greater speed than Ricardo after the push-off.
(A) Both the skaters have the same magnitude of momentum.
(B) Paula has greater speed after push-off.
Conservation of momentum:Given that two skaters Paula and Ricardo are initially at rest.
Ricardo weighs more than Paula.
Let us assume that the mass of Ricardo is M, and the mass of Paula is m.
Let their final velocities be V and v respectively.
(A) Initially, both are at rest.
So the initial momentum of Paula and Ricardo is zero.
According to the law of conservation of momentum, the final momentum of the system must be equal to the initial momentum of the system.
Initial momentum = final momentum
0 = MV + mv
MV = -mv
So, both of them have the same magnitude of momentum, but in opposite directions.
(B) If we compare the magnitude of the momentum of Paula and Ricardo, then:
MV = mv
M/m = v/V
Now, we know that M>m
so, M/m > 1
therefore:
v/V > 1
v > V
So, Paula has greater speed.
Learn more about conservation of momentum:
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What would happen in a State if its citizens lack relevant knowledge, skills
and positive attitude?
Answer:
If the older generation is lacking, the younger generation would likely have knowledge, skill, or a positive attitude in some combination, but it is relative to the culture.
The simple reason is the desirability for genetic variation using recessive genes.
In other words, if the older generation lacks something, it tends to be something they don’t need, but something that will look good on young people. But mostly relative to the culture and education system.
Hope this helps
Water is pumped with a 120 kPa compressor entering the lower pipe (1) and flows upward at a speed of 1 m/s. Acceleration due to gravity is 10 m/s and water density is1000 kg/m-3. What is the water pressure on the upper pipe (II).
Answer:
The water pressure on the upper pipe is 92.5 kPa.
Explanation:
Given that,
Pressure in lower pipe= 120 kPa
Speed of water in lower pipe= 1 m/s
Acceleration due to gravity = 10 m/s²
Density of water = 1000 kg/m³
Radius of lower pipe = 12 m
Radius of uppes pipe = 6 m
Height of upper pipe = 2 m
We need to calculate the velocity in upper pipe
Using continuity equation
[tex]A_{1}v_{1}=A_{2}v_{1}[/tex]
[tex]\pi r_{1}^2\times v_{1}=\pi r_{2}^2\times v_{2}[/tex]
[tex]v_{2}=\dfrac{r_{1}^2\times v_{1}}{r_{2}^2}[/tex]
Put the value into the formula
[tex]v_{2}=\dfrac{12^2\times1}{6^2}[/tex]
[tex]v_{2}=4\ m/s[/tex]
We need to calculate the water pressure on the upper pipe
Using bernoulli equation
[tex]P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}[/tex]
Put the value into the formula
[tex]120\times10^{3}+\dfrac{1}{2}\times1000\times1^2+1000\times10\times0=P_{2}+\dfrac{1}{2}\times1000\times(4)^2+1000\times10\times2[/tex]
[tex]120500=P_{2}+28000[/tex]
[tex]P_{2}=120500-28000[/tex]
[tex]P_{2}=92500\ Pa[/tex]
[tex]P_{2}=92.5\ kPa[/tex]
Hence, The water pressure on the upper pipe is 92.5 kPa.
A diver shines an underwater searchlight at the surface of a pond (n = 1.33). At what angle (relative to the surface) will the light be totally reflected?
Answer:
41.2°
Explanation:
Total internal reflection is the reflection of the incident ray at the interface between two media in which one of the media has a lower refractive index than the other. It occurs when the angle of incidence in the denser medium exceeds the critical angle.
The critical angle is the angle of incidence in the denser medium when the angle of incidence in the less dense medium is 90°.
Since
n= 1/sin C
C= sin^-(1/n)
C= sin^-(1/1.33)
C= 48.8°
Hence angle of incidence= 90-48.8 = 41.2°
A high school physics student claims her muscle car can achieve a constant acceleration of 10 ft/s/s. Her friend develops an accelerometer to confirm the feat. The accelerometer consists of a 1 ft long rod (mass=4 kg) with one end attached to the ceiling of the car, but free to rotate. During acceleration, the rod rotates. What will be the angle of rotation of the rod during this acceleration? Assume the road is flat and straight.
Answer: Ф = 17.2657 ≈ 17°
Explanation:
we simply apply ET =0 about the ending of the rod
so In.g.L/2sinФ - In.a.L/2cosФ = 0
g.sinФ - a.cosФ = 0
g.sinФ = a.cosФ
∴ tanФ = a/g
Ф = tan⁻¹ a / g
Ф = tan⁻¹ ( 10 / 32.17405)
Ф = tan⁻¹ 0.31080948777
Ф = 17.2657 ≈ 17°
Therefore the angle of rotation of the rod during this acceleration is 17.2657 ≈ 17°
A ceiling fan is spinning at 45 revolutions per minute when it is switched to a higher speed. It accelerates uniformly, and 2.0 seconds later it is spinning at 110 revolutions per minute. Through how many radians did it rotate during the transition of speeds
Answer:
θ = 16.2 rad
Explanation:
First we find the angular acceleration by using first equation of motion in angular form:
ωf = ωi + αt
where,
ωf =final angular speed = (110 rev/min)(2π rad/1 rev)(1 min/60 s) = 11.5 rad/s
ωi =initial angular speed = (45 rev/min)(2π rad/1 rev)(1 min/60 s) = 4.7 rad/s
α = angular acceleration = ?
t = time = 2 s
Therefore,
11.5 rad/s = 4.7 rad/s + α(2 s)
α = (6.8 rad/s)/(2 s)
α = 3.4 rad/s²
Now, we use 2nd equation of motion:
θ = ωi t + (1/2)αt²
where,
θ = rotation = ?
Therefore,
θ = (4.7 rad/s)(2 s) + (1/2)(3.4 rad/s²)(2 s)²
θ = 9.4 rad + 6.8 rad
θ = 16.2 rad
A 5.2 cm diameter circular loop of wire is in a 1.35 T magnetic field. The loop is removed from the field in 0.29 sec. Assume that the loop is perpendicular to the magnetic field. What is the average induced emf?
Answer:
9.88 milivolt
Explanation:
Given: diameter d = 5.2 cm
magnetic field B_1 = 1.35 T, final magnetic field B_2 =0 T
t = 0.29 sec.
we know emf = - dΦ/dt
and flux Φ = BA
A= area
therefore emf ε = -A(B_2-B_1)/Δt
[tex]=-\pi(d/2)^2\frac{B_2-B_1}{\Delta t} \\=-\pi(0.052/2)^2\frac{0-1.35}{0.29} \\=98.8\times10^4\\=9.88 mV[/tex]
wo 10-cm-diameter charged rings face each other, 25.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength
Complete question:
Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC. What is the electric field strength at:
a) the midpoint between the two rings?
b) the center of the left ring?
Answer:
a) the electric field strength at the midpoint between the two rings is 0
b) the electric field strength at the center of the left ring is 2712.44 N/C
Explanation:
Given;
distance between the two rings, d = 25 cm = 0.25 m
diameter of each ring, d = 10 cm = 0.1 m
radius of each ring, r = [tex]\frac{0.1}{2} = 0.05 \ m[/tex]
the charge on each ring, q = 20 nC
Electric field strength for a ring with radius r and distance x from the center of the ring is given as;
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}}[/tex]
The electric field strength at the midpoint;
the distance from the left ring to the mid point , x = 0.25 m / 2 = 0.125 m
[tex]E = \frac{kxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9}*0.125*20*10^{-9}}{(0.125^2 + 0.05^2)^{3/2}} \\\\E = 9210.5 \ N/C[/tex]
[tex]E_{left} = 9210.5 \ N/C[/tex]
The electric field strength due to right ring is equal in magnitude to left ring but opposite in direction;
[tex]E_{right} = -9210.5 \ N/C[/tex]
The electric field strength at the midpoint;
[tex]E_{mid} = E_{left} + E_{right}\\\\E_{mid} = 9210.5 \ N/C - 9210.5 \ N/C\\\\E_{mid} = 0[/tex]
(b)
The distance from the right ring to center of the left ring, x = 0.25 m.
[tex]E = \frac{KxQ}{(x^2 +r^2)^{3/2}} \\\\E = \frac{8.99*10^{9} *0.25*20*10^{-9}}{(0.25^2 + 0.05^2)^{3/2}} \\\\E = 2712.44 \ N/C[/tex]
The equivalent resistance of two resistors connected in series is always greater than the equivalent resistance of the same two resistors connected in parallel. True False
Answer:
True
Explanation:
Because the resistors in series is the sum of the two resistors given as
R= R1+R2
While that of resistors in parallel is the sum of the reciprocal of the resistance given as
1/R = 1/ R1+ 1/R2
So that of series connection will be greater
Two instruments produce a beat frequency of 5 Hz. If one has a frequency of 264 Hz, what could be the frequency of the other instrument
Answer:
259 Hz or 269 Hz
Explanation:
Beat: This is the phenomenon obtained when two notes of nearly equal frequency are sounded together. The S.I unit of beat is Hertz (Hz).
From the question,
Beat = f₂-f₁................ Equation 1
Note: The frequency of the other instrument is either f₁ or f₂.
If the unknown instrument's frequency is f₁,
Then,
f₁ = f₂-beat............ equation 2
Given: f₂ = 264 Hz, Beat = 5 Hz
Substitute into equation 2
f₁ = 264-5
f₁ = 259 Hz.
But if the unknown frequency is f₂,
Then,
f₂ = f₁+Beat................. Equation 3
f₂ = 264+5
f₂ = 269 Hz.
Hence the beat could be 259 Hz or 269 Hz
If we compare the force of gravity to strong nuclear force, we could conclude that
O gravity is the weaker force; it is related to mass
O gravity is the stronger force; it is related to distance
strong nuclear is the stronger force; it is related to mass
O strong nuclear is the weaker force; it is related to distance
Answer:
strong nuclear is the stronger force; it is related to mass
Explanation:
If we compare the force of gravity to strong nuclear force, we could conclude that strong nuclear is the stronger force; it is related to mass, therefore the correct answer is option C
What are nuclear forces?The nuclear force is the interaction between the subatomic particles that make up a nucleus. There are two types of nuclear forces: the strong nuclear force and the weak nuclear force. Depending on the separation between the proton neutron and proton pairs, these nuclear forces can be both attracting and positive.
Both types of nuclear forces come under the four fundamental forces of nature. There are mainly four fundamental forces of nature electromagnetic force, gravitational force, strong nuclear force, and weak nuclear force.
Thus, Option C is the appropriate response since, when compared to the force of gravity, the strong nuclear force is the greater force because it is tied to mass.
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Two hoops, staring from rest, roll down identical incline planes. The work done by nonconservative forces is zero. The hoops have the same mass, but the larger hoop has twice the radius. Which hoop will have the greater total kinetic energy at the bottom
Answer:
They both have the same total K.E at the bottom
Explanation:
This Is because If assuming no work is done by non conservative forces, total mechanical energy must be conserved
So
K1 + U1 = K2 + U2
But If both hoops start from rest, and and at the bottom of the incline the level for gravitational potential energy is zero for reference
thus
K1 = 0 , U2 = 0
ΔK = ΔU = m g. h
But if the two inclines have the same height, and both hoops have the same mass m,
So difference in kinetic energy, must be the same for both hoops.
3. El tambor de una lavadora que gira a 3 000 revoluciones por minuto (rpm) se acelera uniformemente hasta que alcanza las 6 000 rpm, completando un total de 12 revoluciones.
d. Determina la aceleración tangencial, centrípeta y la total en m.s-2 cuando el tambor a alcanzado los 60000 rpm
e. Explica lo que ocurre con la magnitud y dirección de los vectores aceleración tangencial, aceleración centrípeta, aceleración total, aceleración angular, velocidad angular cuando la lavadora ha girado desde 3000 rpm hasta 6000 rpm.
Answer:
d) α = 1693.5 rad / s² , a = 392.7 m / s² , a_total = α √(R² +1) ,
e) tan θ = a / α
Explanation:
This is an exercise in linear and angular kinematics.
We initialize reduction of all the magnitudes to the SI system
w₀ = 3000 rev / min (2π rad / 1rev) (1min / 60s) = 314.16 rad / s
w = 6000 rev / mi = 628.32 rad / s
θ = 12 rev = 12 rev (2π rad / 1 rev) = 75.398 rad
d) ask for centripetal, tangential and total acceleration.
Let's start by looking for centripetal acceleration, let's use the formula
w² = w₀² + 2 α θ
α = (w²- w₀²) / 2θ
we calculate
α = (628.32²2 - 314.16²) / 2 75.398
α = 1693.5 rad / s²
the quantity is linear and angular are related
the linear or tangential acceleration is
a = α R
where R is the radius of the drum
a = 1693.5 R
Unfortunately you do not give the radius of the drum for a complete calculation, but suppose it is a washing machine drum R = 20 cm = 0.20 m
a = 1693.5 0.20
a = 392.7 m / s²
the total acceleration is
a_total = √(a² + α²)
a_total = √ (α² R² + α²)
a_total = α √(R² +1)
e) The centripetal acceleration is directed towards the center of the movement is radial and its magnitude is constant
Tangential acceleration is tangency to radius and its value varies proportionally radius
the total accelracicon is the result of the vector sum of the two accelerations and their directions given by trigonometry
tan θ = a / α
the angular velocity increases linearly when with centripetal acceleration
Three resistors, each having a resistance, R, are connected in parallel to a 1.50 V battery. If the resistors dissipate a total power of 3.00 W, what is the value of R
Answer:
The value of resistance of each resistor, R is 2.25 Ω
Explanation:
Given;
voltage across the three resistor, V = 1.5 V
power dissipated by the resistors, P = 3.00 W
the resistance of each resistor, = R
The effective resistance of the three resistors is given by;
R(effective) = R/3
Apply ohms law to determine the current delivered by the source;
V = IR
I = V/R
I = 3V/R
Also, power is calculated as;
P = IV
P = (3V/R) x V
P = 3V²/R
R = 3V² / P
R = (3 x 1.5²) / 3
R = 2.25 Ω
Therefore, the value of resistance of each resistor, R is 2.25 Ω
This problem explores the behavior of charge on conductors. We take as an example a long conducting rod suspended by insulating strings. Assume that the rod is initially electrically neutral. For convenience we will refer to the left end of the rod as end A, and the right end of the rod as end B. In the answer options for this problem, "strongly attracted/repelled" means "attracted/repelled with a force of magnitude similar to that which would exist between two charged balls.A. A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?
What happens to end A of the rod when the ball approaches it closely this first time?a. It is strongly repelled.b. It is strongly attracted.c. It is weakly attracted.d. It is weakly repelled.e. It is neither attracted nor repelled.
Answer:
e. It is neither attracted nor repelled.
Explanation:
Electrostatic attraction or repulsion occurs between two or more charged particles or conductors. In this case, if the negatively charged ball is brought close to the neutral end A of the rod, there would be no attraction or repulsion between the rod end A and the negatively charged ball. This is because a charged particle or conductor has no attraction or repulsion to a neutral particle or conductor.
What is the maximum speed (in units of m/s) with which a car can round a
flat horizontal curve of radius (r=60 m), if the coefficient of static friction between tires and
road is (0.4)?
A) 5
B) 15.5
€) 240
D) 25.1
Answer:
B) 15.5 m/s
Explanation:
r = 60m
μs = 0.4
using the formula max V = √r*g*μs (flat roadway)
v = sqrt(60 * 10 * 0.4)
v = 15.5 m/s
The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresponds to a 1 % change in length.
A. Young's modulus for bone is about Y = 1.4 x 10 N/m². The tibia (shin bone) of a human is 0.35 m long and has an average cross-sectional area of 2.9 cm. What is the effective spring constant of the tibia?
B. If a man weighs 750 N, how much is the tibia compressed if it supports half his weight?
C. What is the maximum force that can be applied to a tibia with a cross-sectional area, A = 2.90 cm?
Answer:
a
[tex]k = 11600000 N/m[/tex]
b
[tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]
c
[tex]F = 3750.28 \ N[/tex]
Explanation:
From the question we are told that
The Young modulus is [tex]E = 1.4 *10^{10} \ N/m^2[/tex]
The length is [tex]L = 0.35 \ m[/tex]
The area is [tex]2.9 \ cm^2 = 2.9 *10^{-4} \ m ^2[/tex]
Generally the force acting on the tibia is mathematically represented as
[tex]F = \frac{E * A * \Delta L }{L}[/tex] derived from young modulus equation
Now this force can also be mathematically represented as
[tex]F = k * \Delta L[/tex]
So
[tex]k = \frac{E * A }{L}[/tex]
substituting values
[tex]k = \frac{1.4 *10^{10} * 2.9 *10^{-4} }{ 0.35}[/tex]
[tex]k = 11600000 N/m[/tex]
Since the tibia support half the weight then the force experienced by the tibia is
[tex]F_k = \frac{750 }{2} = 375 \ N[/tex]
From the above equation the extension (compression) is mathematically represented as
[tex]\Delta L = \frac{ F_k * L }{ A * E }[/tex]
substituting values
[tex]\Delta L = \frac{ 375 * 0.35 }{ (2.9 *10^{-4}) * 1.4*10^{10} }[/tex]
[tex]\Delta L = 3.2323 *10^{-5} \ m[/tex]
From the above equation the maximum force is
[tex]F = \frac{1.4*10^{10} * (2.9*10^{-4}) * 3.233*10^{-5} }{ 0.35}[/tex]
[tex]F = 3750.28 \ N[/tex]
What is the maximum wavelength of incident light that can produce photoelectrons from silver? The work function for silver is Φ=2.93 eV.
Answer:
The maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.
Explanation:
Given;
work function of silver, Φ = 2.93 eV = 2.93 x 1.602 x 10⁻¹⁹ J = 4.6939 x 10⁻¹⁹ J
Apply Einstein Photo electric effect;
E = K.E + Ф
Where;
E is the energy of the incident light
K.E is the kinetic of electron
Ф is the work function of silver surface
For the incident light to have maximum wavelength, the kinetic energy of the electron will be zero.
E = Ф
hf = Ф
[tex]h\frac{c}{\lambda} = \phi[/tex]
where;
c is speed of light = 3 x 10⁸ m/s
h is Planck's constant, = 6.626 x 10⁻³⁴ J/s
λ is the wavelength of the incident light
[tex]\lambda = \frac{hc}{\phi}\\\\\lambda =\frac{6.626*10^{-34} *3*10^8}{4.6939*10^{-19}} \\\\\lambda = 4.235 *10^{-7} \ m\\\\\lambda = 423.5 *10^{-9} \ m\\\\\lambda = 423.5 \ nm[/tex]
Therefore, the maximum wavelength of incident light that can produce photoelectrons from silver is 423.5 nm.
Physical properties of a mineral are a result of the arrangement of the atoms in the mineral. Use this fact to explain the following:_________
A. One mineral has a density of 2.7 g/ml while another has a density of 5.1 g/ml
B. The mineral mica cleaves into thin flat sheets while olivine does not show cleavage
Explanations:
a) The physical properties of a mineral is as a result of the arrangement of the atoms in the minerals. The reason behind one mineral having a density of 2.1 g/ml which is lower than that of another mineral with density of 5.1 g/ml is the packing density of the minerals. Minerals with high density have their atoms more closely packed together, leaving less space between the atoms. This characteristics means that they have more atomic mass per unit volume for a given molecular space, when compared to another mineral with its atoms less closely packed.
b) The property of cleavage is due to the crystalline structure of a mineral species. Cleavage is used to describe the ease with which minerals cleaves. Cleavage is due to a weak bonding strength between the molecules of the mineral, or a due to a greater lattice spacing across the the cleavage plane than in other planes within the crystal. The greater the lattice spacing, the weaker the strength of the bond across a plane.
From these, we can clearly see that the property of cleavage is due to the physical properties of the crystalline structure of the mineral species.
A chemist must dilute 55.6 ml of 1.48 M aqueous silver nitrate (AgNO3)solution until the concentration falls to 1.00 M. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to 3 significant digits.
Answer:
82.2 mL
Explanation:
The process of adding water to a solution to make it more dilute is known as dilution. The formula for dilution is;
C1V1=C2V2
Where;
C1= concentration of stock solution
V1= volume of stock solution
C2= concentration of dilute solution
V2= volume of dilute solution
V2= C1V1/C2
V2= 1.48 × 55.6/ 1.0
V2= 82.2 mL
An expensive vacuum system can achieve a pressure as low as 1.53 ✕ 10−7 N/m2 at 26°C. How many atoms are there in a cubic centimeter at this pressure and temperature?
Answer:
The value is [tex]N = 3.708*10^{7} \ \ atoms[/tex]
Explanation:
From the question we are told that
The pressure is [tex]P = 1.53 *10^{-7} \ N/m^2[/tex]
The temperature is [tex]T = 26 + 273 = 299 \ K[/tex]
The volume is 1 cubic cm = [tex]1 * 10^{-6} m^3[/tex]
Generally according to the ideal gas law we have that
[tex]PV = NkT[/tex]
here k is the Boltzmann constant with a value [tex]k = 1.38 *10^{-23} \ J/K[/tex]
=> [tex]N = \frac{PV}{ k T}[/tex]
=> [tex]N = \frac{ 1.53 *10^{-7} * (1* 10^{-6})}{ 1.38*10^{-23} * 299}[/tex]
=> [tex]N = 3.708*10^{7} \ \ atoms[/tex]
