Answer:
The angular displacement is [tex]\theta = 29.6 \ rad[/tex]
Explanation:
From the question we are told that
The initial angular speed is [tex]w = 5.35 \ rad/s[/tex]
The angular acceleration is [tex]\alpha = 0.331 rad /s^2[/tex]
The time take is [tex]t = 4.81 \ s[/tex]
Generally the angular displacement is mathematically represented as
[tex]\theta = w * t + \frac{1}{2} \alpha * t^2[/tex]
substituting values
[tex]\theta = 5.35 * 4.81 + \frac{1}{2} * 0.331 * (4.81)^2[/tex]
[tex]\theta = 29.6 \ rad[/tex]
A 1.70 m tall woman stands 5.00 m in front of a camera with a 50.00 cm focal
length lens. Calculate the size of the image formed on flim
Answer:
18.89cm
Explanation:
As we know that the person is standing 5m in front of the camera
[tex]d_0=5m=500cm[/tex]
The focal length of the lens =50cm
f=50 cm
By Lens formula we have:
[tex]\dfrac{1}{f} = \dfrac{1}{d_i} + \dfrac{1}{d_o}\\\dfrac{1}{50} = \dfrac{1}{d_i} + \dfrac{1}{500}\\\dfrac{1}{d_i} =\dfrac{1}{50}-\dfrac{1}{500}\\\dfrac{1}{d_i}=0.018\\d_i=55.56cm[/tex]
By the formula of magnification
[tex]\dfrac{h_i}{h_o} = \dfrac{55.56}{500}\\\\h_i = \dfrac{55.56}{500} \times h_o\\\\ h_o=1.70m=170cm\\\\Therefore: h_i=\dfrac{55.56}{500} \times$ 170 cm\\\\h_i =18.89 cm[/tex]
The height of the image formed is 18.89cm.
A population _____ follows a period of
Answer:
a population increase
Explanation:
During the 20th century, the world population increased from 1.65 billion to 6 billion. In 1970, the world's population was half that of today. In less than 15 years, 47% of the population will live in areas already under heavy water stress. In Africa, between 75 and 250 million people will face growing shortages in 2020 due to climate change. The scarcity of some arid and semi-arid regions will have a decisive impact on migration.
The motion of an object undergoing constant acceleration can be modeled by the kinematic equations. One such equation is xf=xi+vit+12at2 where xf is the final position, xi is the initial position, vi is the initial velocity, a is the acceleration, and t is the time. Let's say a car starts with an initial speed of 15 m/s, and moves between the 1000 m and 5000 m marks on a roadway in a time of 60 s. What is its acceleration?
Answer:
a = 1.72 m/s²
Explanation:
The given kinematic equation is the 2nd equation of motion. The equation is as follows:
xf = xi + (Vi)(t) + (1/2)(a)t²
where,
xf = the final position = 5000 m
xi = the initial position = 1000 m
Vi = the initial velocity = 15 m/s
t = the time taken = 60 s
a = acceleration = ?
Therefore,
5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²
5000 m = 1000 m + 900 m + a(1800 s²)
5000 m = 1900 m + a(1800 s²)
5000 m - 1900 m = a(1800 s²)
a(1800 s²) = 3100 m
a = 3100 m/1800 s²
a = 1.72 m/s²
An aluminium pot whose thermal conductivity is 237 W/m.K has a flat, circular bottom
with diameter 15 cm and thickness 0.4 cm. Heat is transferred steadily to boiling water in
the pot through its bottom at a rate of 1400 W. If the inner surface of the bottom of the pot
is at 105 °C, determine the temperature at the outer surface of the bottom of the pot
Answer:
T₁ = 378.33 k = 105.33°C
Explanation:
From Fourier's Law of heat conduction, we know that:
Q = - KAΔT/t
where,
Q = Heat Transfer Rate = 1400 W
K = Thermal Conductivity of Material (Aluminum) = 237 W/m.k
A =Surface Area through which heat transfer is taking place=circular bottom
A = π(radius)² = π(0.15 m)² = 0.0707 m²
ΔT = Difference in Temperature of both sides of surface = T₂ - T₁
T₁ = Temperature of outer surface = ?
T₂ = Temperature of inner surface = 105°C + 273 = 378 k
ΔT = 388 k - T₁
t = thickness of the surface (Bottom of Pot) = 0.4 cm = 0.004 m
Therefore,
1400 W = - (237 W/m.k)(0.0707 m²)(378 k - T₁)/0.004 m
(1400 W)/(4188.14 W/k) = - (378 k - T₁)
T₁ = 0.33 k + 378 k
T₁ = 378.33 k = 105.33°C
What is the period of a wave if the frequency is? 5 Hz
Answer: If the woodpecker drums upon a tree 5 times in one second, then the frequency is 5 Hz; each drum must endure for one-fifth a second, so the period is 0.2 s.
In Physics lab, a lab team places a cart on one of the horizontal, linear tracks with a fan attached to it. The cart is positioned at one end of the track, and the fan is turned on. Starting from rest, the cart takes 4.34 s to travel a distance of 1.62 m. The mass of the cart plus fan is 354 g. Assume that the cart travels with constant acceleration.
A) What is the net force exerted on the cart-fan combination?B) Mass is added to the cart until the total mass of the cart-fan combination is 762 g, and the experiment is repeated. How long does it take for the cart, starting from rest, to travel 1.62 m now?
Answer:
A. F = 0.06 N
B. t = 6.37 s
Explanation:
A)
First we need to find the constant acceleration of the cart. For this purpose, we use 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
where,
s = distance traveled = 1.62 m
Vi = 0 m/s (Since, it starts from rest)
t = Time Taken = 4.34 s
a = acceleration = ?
Therefore,
1.62 m = (0 m/s)(4.34 s) + (0.5)(a)(4.34 s)²
1.62 m/9.4178 s² = a
a = 0.172 m/s²
Now, from Newton's Second law, we know that:
F = ma
where,
F = Net Force of the combination = ?
m = Mass pf combination = 354 g = 0.354 kg
Therefore,
F = (0.354 kg)(0.172 m/s²)
F = 0.06 N
B)
Now, for the same force, but changed mass = 762 g = 0.762 kg, we have the acceleration to be:
F = ma
a = F/m
a = 0.06 N/0.762 kg
a = 0.08 m/s²
Now, using 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
1.62 m = (0 m/s)(t) + (0.5)(0.08 m/s²)t²
t² = 1.62 m/(0.04 m/s²)
t = √40.54 s²
t = 6.37 s
At a time when mining asteroids has become feasible, astronauts have connected a line between their 3220-kg space tug and a 6240-kg asteroid. They pull on the asteroid with a force of 362 N. Initially the tug and the asteroid are at rest, 311 m apart. How much time does it take for the ship and the asteroid to meet
-- F = m a ... ==> a = F/m
-- The tension in the rope is 362 N. That same force acts on the asteroid and on the tug, pulling them together.
-- The asteroid's acceleration is 362N / 6240 kg = 0.058 m/s², headed for a point on the rope somewhere between the asteroid and the tug.
-- The tug's acceleration is 362 N / 3220 kg = 0.112 m/s², also headed for a point on the rope somewhere between the tug and the asteroid.
-- So now we have a gap between them, initially 311 m long, closing with a speed that starts at zero and accelerates at 0.170 m/s² .
-- D = (1/2) a T²
311 m = (1/2) (0.170 m/s²) (T²)
T² = 311 m / 0.085 m/s²
T = √(311/0.085) seconds
T = 60.41 seconds
The answer I get is so durn near 60 seconds (1 minute) that it suggests two things to me: ==> That's where the weird numbers of 362N and 311m came from, and ==> there's a good chance that my answer is correct.
Note: It's important to me that you know that 5 points for this one is really cheap and chintzy, and the reason I decided to try it was only to see whether I could.
An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object
Answer:
v = 25.45 m/s
Explanation:
In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:
[tex]h_{max}=\frac{v_o^2}{g}[/tex] (1)
vo: initial speed of the object = 18 m/s
g: gravitational acceleration = 9.8 m/s²
Furthermore you use the following formula for the final speed of the object:
[tex]v^2=v_o^2-2gh[/tex] (2)
h: height
You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:
[tex]v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2[/tex]
Then, you solve the previous result for vo:
[tex]v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}[/tex]
The initial speed of the object was 25.45 m/s
A freight car moves along a frictionless level railroad track at constant speed. The freight car is open on top. A large load of coal is suddenly dumped into the car. What happens to the speed of the freight car
Answer:
The speed of the freight car decreases.
Explanation:
According to the law of conservation of momentum indicates that for colliding in an isolated system, the total momentum pre and post collision is same for the two objects this is done because the momentum that one item has lost is same for the momentum that the other received
In the given situation, the freight car travels at constant speed along a frictionless railroad line. The top floor freight car is open. Then a huge load of coal is dumped inside the car.
Therefore the speed of the freight car decreased by applying the law of conservation of momentum i
what statement is true according to newton’s first law of motion?
a. in the absence of unbalanced force an object at rest will stay at rest and an object in motion will come to a stop.
b. in the absence of an unbalanced force, an object will start moving and an object in motion will come to a stop.
c. in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.
d. in the absence of an unbalanced force, an object will start moving and an object in motion will stay in motion.
Answer:
c. in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.
Explanation:
First law: things keep doing what they are doing, unless force is applied.
When the play button is pressed, a CD accelerates uniformly from rest to 430 rev/min in 4.0 revolutions. If the CD has a radius of 7.0 cm and a mass of 17 g , what is the torque exerted on it?
Answer:
The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].
Explanation:
As CD is acceleration uniformly, the following equation of motion can be used to determine the angular acceleration:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot \Delta n[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]\ddot n[/tex] - Angular acceleration, measured in revolution per square minute.
[tex]\Delta n[/tex] - Change in angular position, measured in revolutions.
The angular acceleration is cleared and calculated:
[tex]\ddot n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \Delta n}[/tex]
Given that [tex]\dot n_{o} = 0\,\frac{rev}{min}[/tex], [tex]\dot n = 430\,\frac{rev}{min}[/tex] and [tex]\Delta n = 4\, rev[/tex], the angular acceleration is:
[tex]\ddot n = \frac{\left(430\,\frac{rev}{min} \right)^{2}-\left(0\,\frac{rev}{min} \right)^{2}}{2\cdot (4\,rev)}[/tex]
[tex]\ddot n = 23112.5\,\frac{rev}{min^{2}}[/tex]
The angular accelaration measured in radians per square second is:
[tex]\alpha = \left(23112.5\,\frac{rev}{min^{2}} \right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}} \right)[/tex]
[tex]\alpha \approx 40.339\,\frac{rad}{s^{2}}[/tex]
Net torque experimented by the CD during its accleration is equal to the product of its moment of inertia with respect to its axis of rotation and angular acceleration:
[tex]\tau = I \cdot \alpha[/tex]
Where:
[tex]I[/tex] - Moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].
[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.
In addition, a CD has a form of a uniform disk, whose moment of inertia is:
[tex]I = \frac{1}{2}\cdot m \cdot r^{2}[/tex]
Where:
[tex]m[/tex] - Mass of the CD, measured in kilograms.
[tex]r[/tex] - Radius of the CD, measured in meters.
If [tex]m = 0.017\,kg[/tex] and [tex]r = 0.07\,m[/tex], then:
[tex]I = \frac{1}{2}\cdot (0.017\,kg)\cdot (0.07\,m)^{2}[/tex]
[tex]I = 4.165\times 10^{-5}\,kg\cdot m^{2}[/tex]
Now, the net torque exerted on CD is:
[tex]\tau = (4.165\times 10^{-5}\,kg\cdot m^{2})\cdot \left(40.339\,\frac{rad}{s^{2}} \right)[/tex]
[tex]\tau = 1.680\times 10^{-3}\,N\cdot m[/tex]
The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].
Sr-90 has a half-life of T1/2 = 2.85 a (years). How much Sr-90 will remain in a 5.00 g sample after 5.00 a? Show all of your work. (2 marks)
Answer:
1.48 g
Explanation:
A = A₀ (½)^(t / T)
where A is the final amount,
A₀ is the initial amount,
t is time,
and T is the half life.
A = (5.00 g) (½)^(5.00 a / 2.85 a)
A = 1.48 g
EASY! WILL GIVE BRAINLIEST!
Find the conductivity of a conduit with a cross-sectional area of 0.50 cm2 and a length of 15 cm if its conductance G is 0.050 ohm-1.
σ = _____ ohm-1cm-l
3
75
1.5
0.0017
the answer is 1.5 hope this helps
Answer:
1.5
Explanation:
0.5=σ/(15/0.5)
σ=3/2 or 1.5
Zinc is added to a breaker containing hydrochloric acid and the beaker gets warm what type os reaction is this
Answer:
Exothermic
Explanation:
Depending on the unit you are in, the answer may vary.
This is an exothermic reaction because it produces heat (the beaker gets warm).
An arrow is shot from a height of 1.55 m toward a cliff of height H. It is shot with a velocity of 26 m/s at an angle of 60° above the horizontal. It lands on the top edge of the cliff 3.99 s later.
(a) Draw a sketch of the given example. Include the x-y coordinate system.
(b) What is the height of the cliff?
(c) What is the maximum height reached by the arrow along its trajectory?
(d) What is the arrow's impact speed just before hitting the cliff?
Answer:
Explanation:
vertical component of the velocity of arrow
= 26 sin 60 = 22.516 m
height reached by it after 3.99 s
h = ut - 1/2 g t²
= 22.516 x 3.99 - .5 x 9.8 x 3.99²
= 89.83 - 78
11.83 m
Total height of cliff = 1.55 + 11.83
= 13.38 m
c ) maximum height covered s
v² = u² - 2gs
0 = u² - 2gs
s = u² / 2g
= 22.516² / 2 x 9.8
= 25.86
maximum height reached
= 25.86 + 1.55
= 27.41 m
d )
vertical speed after 3.99 s
v = u - gt
= 22.516 - 9.8 x 3.99
= -16.586
Horizontal component will remain unchanged
Horizontal component = 26 cos 60
= 13 m /s
Resultant of two velocities
= √ 13²+ 16.568²
= 21 m /s
1. In 214 BC, Archimedes invented a large spherical-type mirror used to focus the sun's intense rays onto far away enemy boats, which would eventually light them on fire. If the boats were travelling in a nearby channel approximately 1,000 m from the river bank, what would the radius of curvature of his mirror need to be? Show your work.
Answer:
2000 m
Explanation:
since the boat is 1000 m from the river bank, the beam must be focused at this point. This indicates that the focal length is 1000 m
for a spherical mirror, the focal length is given by
f = R/2
where R is the radius of curvature
1000 = R/2
R = 2000 m
R = 2000 m
this means that the radius of curvature must be 2000 m
A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled has an acceleration of 2.10 m/s2. Part A By how much does the spring stretch if it pulls on the sled horizontally
Answer:
Stretch in the spring = 0.1643 (Approx)
Explanation:
Given:
Mass of the sled (m) = 9 kg
Acceleration of the sled (a) = 2.10 m/s ²
Spring constant (k) = 115 N/m
Computation:
Tension force in the spring (T) = ma
Tension force in the spring (T) = 9 × 2.10
Tension force in the spring (T) = 18.9 N
Tension force in the spring = Spring constant (k) × Stretch in the spring
18.9 N = 115 N × Stretch in the spring
Stretch in the spring = 18.9 / 115
Stretch in the spring = 0.1643 (Approx)
a steel ball is dropped from a diving platform use the approximate value of g as 10 m/s^2 to solve the following problem what is the velocity of the ball 0.9 seconds after its released
Answer:
The final speed of the ball is 9 m/s.
Explanation:
We have,
A steel ball is dropped from a diving platform. It is required to find the velocity of the ball 0.9 seconds after its released. It will move under the action of gravity. Using equation of motion to find it as :
[tex]v=u+at[/tex]
u = 0 (at rest), a = g
[tex]v=gt\\\\v=10\times 0.9\\\\v=9\ m/s[/tex]
So, the final speed of the ball is 9 m/s.
If radio waves are used to communicate with an alien spaceship approaching Earth at 10% of the speed of light c, the aliens would receive our signals at speed of:_______.
a. 0.99c
b. 1.10c
c. 1.00c
d. 0.90c
e. 0.10c
Answer:
3×10^7 m/s or 0.10c (e)
Explanation: If the actual value of the speed of light were to be put into consideration.
Given that the speed of light is c = 3.0×10^8m/s
The alien spaceship is approaching at the rate of 10% of the speed of light.
10% of 3.0×10^8m/s
10/100 × 3.0×10^8m/s
0.1 ×3.0×10^8m/s
3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c
Answer: 1.00c
Explanation: I got it correct on the homework
Why do some nucleus emit electrons?
Answer:
In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.
Explanation:
Hope this helps!
What is the answer for this question
Two hockey pucks, labeled A and B, are initially at rest on a smooth ice surface and are separated by a distance of 18.0 m . Simultaneously, each puck is given a quick push, and they begin to slide directly toward each other. Puck A moves with a speed of 3.90 m/s , and puck B moves with a speed of 4.30 m/s . What is the distance covered by puck A by the time the two pucks collide
Answer:
The distance covered by puck A before collision is [tex]z = 8.56 \ m[/tex]
Explanation:
From the question we are told that
The label on the two hockey pucks is A and B
The distance between the two hockey pucks is D 18.0 m
The speed of puck A is [tex]v_A = 3.90 \ m/s[/tex]
The speed of puck B is [tex]v_B = 4.30 \ m/s[/tex]
The distance covered by puck A is mathematically represented as
[tex]z = v_A * t[/tex]
=> [tex]t = \frac{z}{v_A}[/tex]
The distance covered by puck B is mathematically represented as
[tex]18 - z = v_B * t[/tex]
=> [tex]t = \frac{18 - z}{v_B}[/tex]
Since the time take before collision is the same
[tex]\frac{18 - z}{V_B} = \frac{z}{v_A}[/tex]
substituting values
[tex]\frac{18 -z }{4.3} = \frac{z}{3.90}[/tex]
=> [tex]70.2 - 3.90 z = 4.3 z[/tex]
=> [tex]z = 8.56 \ m[/tex]
The mass of a particular eagle is twice that of a hunted pigeon. Suppose the pigeon is flying north at ,2=17.1 m/s when the eagle swoops down, grabs the pigeon, and flies off. At the instant right before the attack, the eagle is flying toward the pigeon at an angle =52.7 ° below the horizontal and a speed of ,1=41.5 m/s.
Answer:
31.4 m/s
44.4°
Explanation:
Momentum is conserved in the horizontal direction:
pₓᵢ = pₓ
m vᵢ₂ + 2m vᵢ₁ cos θ = (m + 2m) vₓ
vᵢ₂ + 2 vᵢ₁ cos θ = 3 vₓ
17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vₓ
vₓ = 22.5 m/s
Momentum is conserved in the vertical direction:
pᵧᵢ = pᵧ
2m vᵢ₁ sin θ = (m + 2m) vᵧ
2 vᵢ₁ sin θ = 3 vᵧ
2 (41.5 m/s) (sin -52.7°) = 3 vᵧ
vᵧ = -22.0 m/s
The speed is:
v = √(vₓ² + vᵧ²)
v = √((22.5 m/s)² + (-22.0 m/s)²)
v = 31.4 m/s
The direction is:
θ = atan(vᵧ / vₓ)
θ = atan(-22.0 m/s / 22.5 m/s)
θ = -44.4°
The speed of the eagle at that instant is 31.4 m/s while it moves off in the direction of 44.4°.
Since momentum is conserved horizontally;
17.1 m/s + 2 (41.5 m/s) (cos -52.7°) = 3 vx
vx = 17.1 m/s + 2 (41.5 m/s) (cos -52.7°)/3
vx = 22.5 m/s
Also, momentum is conserved vertically hence;
2 (41.5 m/s) (sin -52.7°) = 3 vy
vy = 2 (41.5 m/s) (sin -52.7°) /3
vy = -22.0 m/s
The effective speed therefore, is;
v = √((22.5 m/s)² + (-22.0 m/s)²)
v = 31.4 m/s
The direction of this effective speed is;
θ = tan-1(22.0 m/s / 22.5 m/s)
θ = 44.4°
Learn more: https://brainly.com/question/13322477
A parallel-plate capacitor has square plates that are 7.20 cm on each side and 3.40 mm apart. The space between the plates is completely filled with two square slabs of dielectric, each 7.20 cm on a side and 1.70 mm thick. One slab is Pyrex glass and the other slab is polystyrene. If the potential difference between the plates is 96.0 V, find how much electrical energy (in nJ) can be stored in this capacitor.
Answer:
U = 218 nJ
Explanation:
We are given;
Spacing between the plates; d = 3.4 mm = 3.4 × 10^(-3) m
Voltage across the capacitor; V = 96 V
Dimension of the square plates is 7.2cm x 7.2cm.
So, Area = 7.2 × 7.2 = 51.84 cm² = 51.84 × 10^(-4) m²
Permittivity of free space; ε_o = 8.85 × 10^(-12) C²/N.m²
From relative permeability table;
Dielectric constant of Pyrex; k1 = 5.6
Dielectric constant of polystyrene; k2 = 2.56
Now, formula for capacitance of a capacitor with Dielectric is;
C = kC_o
Where, C_o = ε_o(A/d)
Since there are 2 capacitors, d will now be d/2 = (3.4 × 10^(-3))/2 m = 1.7 × 10^(-3)
Since we have 2 capacitor, thus ;
C1 = k1*ε_o*(A/d)
C1 = (5.6 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C1 = 1.51 × 10^(-10) F
Similarly;
C2 = (2.56 × 8.85 × 10^(-12) × (51.84 × 10^(-4))/(1.7 × 10^(-3))
C2 = 0.691 × 10^(-10) F
For capacitors in series, formula for total capacitance(Cs) is;
1/Cs = (1/C1) + (1/C2)
Simplifying this, we have;
Cs = (C1*C2)/(C1 + C2)
Plugging in the relevant values ;
Cs = (1.51 × 10^(-10)*0.691 × 10^(-10))/((1.51 × 10^(-10)) + (0.691 × 10^(-10)))
Cs = 0.474 × 10^(-10) F
The formula for energy stored in a capacitor with 2 Dielectrics is given as;
U = ½Cs*V²
So,
U = ½ × 0.474 × 10^(-10) × 96²
U = 2.18 × 10^(-7) J = 218 × 10^(-9) = 218 nJ
A string is stretched between fixed supports separated by 72.0 cm. It is observed to have resonant frequencies of 370 and 555 Hz, and no other resonant frequencies between these two.(a) What is the lowest resonant frequency for this string?(b) What is the wave speed for this string?
Answer:
(a) f = 185 Hz
(b) v = 266.4 m/s
Explanation:
(a) The lowest frequency can be calculated by using the following formula for the calculation of the modes (resonant frequencies) in a string:
[tex]f_n=\frac{nv}{2L}[/tex]
[tex]f_n=nf[/tex]
n: order of the mode
v: velocity of the waves in the string
L: length of the string = 72.0cm = 0.72m
fn: frequency of the n-th mode
With the information about two consecutive modes you can find the lowest resonant frequency. First you find the resonant mode n:
[tex]f_n=nf\\\\f_{n-1}=(n-1)f\\\\\frac{f_n}{f_{n-1}}=\frac{n}{n-1}[/tex]
you solve the previous equation for n:
[tex](n-1)f_n=nf_{n-1}\\\\555n-555=370n\\\\n=3[/tex]
With this information you can calculate the lowest resonant frequency:
[tex]f_n=nf\\\\f=\frac{f_n}{n}=\frac{555}{3}=185Hz[/tex]
b) You have information about two consecutive modes fn, fn-1. Then, you can calculate the velocity of the waves:
[tex]f_{n}-f_{n-1}=n\frac{v}{2L}-(n-1)\frac{v}{2L}\\\\f_n-f_{n-1}=\frac{v}{2L}\\\\v=2L(f_n-f_{n-1})[/tex]
fn = 555 Hz
fn-1: 370 Hz
[tex]v=2(0.72m)(555-370)Hz=266.4\frac{m}{s}[/tex]´
hence, the velocityof the waves in the string is 266.4 m/s
When an electromagnetic wave falls on a white, perfectly reflectingsurface, it exerts a force F on that surface. If the surfaceis now painted a perfectly absorbing black, the force that the samewave would exert on the surface is:___________.
A) F
B) F/2
C) F/4
D) 2F
E) 4F
Answer:
B. F/2
Explanation:
The radiation force per unit area (radiation pressure Prad) exerted by an electromagnetic wave on a perfectly absorbing body has been found by experiment to be equal to the energy density of the wave
i.e Prad = u
For a reflecting body, this force exerted per unit area has been found to be twice the energy density of the wave.
i.e Prad = 2u.
Therefore, if the force exerted on a perfectly reflective body is F, then the force exerted on a perfectly absorbing body will be F/2
The Great Lakes are all part of what? The Mississippi River The St. Lawrence Seaway A large body of salt lakes The Missouri River
Answer:
St Lawrence Sea way
Explanation:
The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.
1. Describe what must happen to an atom to make it
A. A cation
B. An anion
2. Describe why some acids are strong while other acids are weak
3. Compare protons, neutrons and electron, listing their similarities and differences
4. Explain why you breathe faster and deeper when exercising
Answer:
Explanation:
Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 × 10−23 g, and an electron ... The amu was originally defined based on hydrogen, the lightest element, ... but three-letter symbols have been used to describe some elements that have ...
Protons: Protons are positively charged particles that are also found in the nucleus. Like neutrons, protons give mass to the atom but do not participate in ... 3) Electrons: Electrons are negatively charged particles that are found in ... pair of electrons with 4 different hydrogen atoms, forming a molecule of CH4 (methane).Elements differ from each other in the number of protons they have, e.g. ... Atoms of an element that have differing numbers of neutrons (but a constant atomic ... Electrons, because they move so fast (approximately at the speed of light), ...toms are made up of particles called protons, neutrons, and electrons, which ... Therefore, they do not contribute much to an element's overall atomic mass. ... For instance, iron, Fe, can exist in its neutral state, or in the +2 and +3 ionic states. ... Isotopes of the same element will have the same atomic number but different ...
g A proton is held at rest in a uniform electric field. When it is released, the proton will lose... electrical potential energy. kinetic energy. both kinetic energy and electric potential energy. neither kinetic energy or electric potential energy.
Answer:
It will lose electrical potential energy.
Explanation:
A photon held at rest in a uniform electrical field will lose electrical potential energy when it is released this is because the electrical potential energy is the energy posses by the photon at rest or by virtue of the position is converted to kinetic energy which is energy posses by a body in motion.
Since the photon is released and set in motion , it now has kinetic energy and has lost the potential energy because it is set in motion.
1. For each of the following scenarios, describe the force providing the centripetal force for the motion: a. a car making a turn b. a child swinging around a pole c. a person sitting on a bench facing the center of a carousel d. a rock swinging on a string e. the Earth orbiting the Sun.
Complete Question
For each of the following scenarios, describe the force providing the centripetal force for the motion:
a. a car making a turn
b. a child swinging around a pole
c. a person sitting on a bench facing the center of a carousel
d. a rock swinging on a string
e. the Earth orbiting the Sun.
Answer:
Considering a
The force providing the centripetal force is the frictional force on the tires \
i.e [tex]\mu mg = \frac{mv^2}{r}[/tex]
where [tex]\mu[/tex] is the coefficient of static friction
Considering b
The force providing the centripetal force is the force experienced by the boys hand on the pole
Considering c
The force providing the centripetal force is the normal from the bench due to the boys weight
Considering d
The force providing the centripetal force is the tension on the string
Considering e
The force providing the centripetal force is the force of gravity between the earth and the sun
Explanation:
