Answer:
6 bits
Explanation:
The quality of digitized signal can be improved by reducing quantizing error. This is done by increasing the number of amplitude levels, thereby minimizing the difference between the levels and hence producing a smoother signal.
Also, Sampling frequently (also known as oversampling) can help in improving signal quality.
To get the number of bits, we use:
2ⁿ = amplitude level
where n is the number of bits.
Given an amplitude level of 64, hence:
2ⁿ = 64
2ⁿ = 2⁶
n = 6 bits
A person jumps out of an airplane above the surface of the Earth, and falls a distance h before opening their parachute. Once the prachute is open the person coasts to the ground a distance d at constant velocity.
a. The work done on the person by the Earth is:
b. The change in gravitational potential energy of the person + Earch system is:
Answer:
a) W_total = mg (2h + d) , b) E_total = - mg (h + d)
Explanation:
a) We must solve this problem in two parts, the first for the accelerated movement and the second for the movement with constant speed
Let's look for work for the part that is in free fall
y = y₀ + v₀ t - ½ g t²
when he jumps out of a plane his vertical speed is zero
y =y₀ - ½ g t²
dy = 0 - ½ g 2t dt
the work in this first part is
W₁ = ∫ F dy
W₁ = mg ∫ g t dt
W₁ = m g² t² / 2
the time it takes to travel the distance y₀-y = h is
y₀-y = ½ g t²
t =[tex]\sqrt{2h/g}[/tex]
we substitute
W₁ = m g² 2h / g
W₁ = m g 2h
now we look for the work for the part with constant speed
since the velocity is constant let's use the uniform motion ratio
W₂ = F d
W₂ = mg d
the total work is
W_total = W₁ + W₂
W_total = 2mgh + m gd
W_total = mg (2h + d)
b) The change in gravitational potential energy
U = mg Δy
in the part with accelerated movement
U₁ = mg h
in the part with uniform movement
U₂ = mg d
the total potential energy is
E_total = U₁ + U₂
E_total = - mg (h + d)
The atoms in your body are mostly empty space . And so are the atoms in any wall. Why then is your body unable to pass through walls ?
First of all, both are not a single sheet of atom. There are many layers of atoms, so the empty part gets beside each other, so there are less empty part. Secondly, there are so many atoms that the probability that they will have empty space at the same place necessary, is negligible.
This was something from logic.
The reason I was taught in my class was that only a limited number of electrons can be in a given orbit, so atoms cannot overlap each other.
A 0.033-kg bullet is fired vertically at 222 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball
Answer:
The maximum height risen by the bullet-baseball system after the collision is 81.76 m.
Explanation:
Given;
mass of the bullet, m₁ = 0.033 kg
mass of the baseball, m₂ = 0.15 kg
initial velocity of the bullet, u₁ = 222 m/s
initial velocity of the baseball, u₂ = 0
let the common final velocity of the system after collision = v
Apply the principle of conservation of linear momentum to determine the common final velocity.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
0.033 x 222 + 0.15 x 0 = v(0.033 + 0.15)
7.326 = v(0.183)
v = 7.326 / 0.183
v = 40.03 m/s
Let the height risen by the system after collision = h
Initial velocity of the system after collision = Vi = 40.03 m/s
At maximum height, the final velocity, Vf = 0
acceleration due to gravity for upward motion, g = -9.8 m/s²
[tex]v_f^2 = v_i^2 +2gh\\\\0 = 40.03^2 - (2\times 9.8)h\\\\19.6h = 1602.4\\\\h = \frac{1602.4}{19.6} \\\\h = 81.76 \ m[/tex]
Therefore, the maximum height risen by the bullet-baseball system after the collision is 81.76 m.
You are helping your friend move a new refrigerator into his kitchen. You apply a horizontal force of 275 N in the positive x direction to try and move the 61 kg refrigerator. The coefficient of static friction is 0.58. (a) How much static frictional force does the floor exert on the refrigerator
Answer:
f = 347.08 N
Explanation:
The frictional force exerted by the floor on the refrigerator is given as follows:
[tex]f = \mu R = \mu W[/tex]
where,
f = frictional force = ?
μ = coefficient of static friction = 0.58
W = Weight of refrigerator = mg
m = mass of refrigerator = 61 kg
g = acceleration due to gravity = 9.81 m/s²
Therefore,
[tex]f = \mu mg\\f = (0.58)(61\ kg)(9.81\ m/s^2)\\[/tex]
f = 347.08 N
Question 4(Multiple Choice Worth 4 points)
(02.04 MC)
Which explanation justifies why the theory of evolution is a theory and not a law?
Predicts an organism's ability to adapt to its environment
It can be expressed as a simple mathematical statement
Explains the existence of diverse forms of life on Earth
O Additional evidence will change the theory into a law
Answer:
A(predicts an organisms ability to adapt to its enviroment, it is not a fact that each organization can adapt)
Explanation:
The density of blood is 1055 kg/m3 . If the blood at the very top of your head exerts a minimum gauge pressure of 45 mm Hg (6000 Pa), estimate the gauge pressure at your heart in pascals.
Answer:
P = 10135.6 Pa
Explanation:
For this exercise we use that the pressure varies with the height
P = P₀ + ρ g h
where h is the height from the head to the heart, which is approximately
h = 40 cm = 0.40m and P₀ is the head pressure P₀ = 6000 Pa
P = 6000 + 1055 9.8 0.40
P = 6000 + 4135.6
P = 10135.6 Pa
What would the radius (in mm) of the Earth have to be in order for the escape speed of the Earth to equal (1/21) times the speed of light (300000000 m/s)? You may ignore all other gravitational interactions for the rocket and assume that the Earth-rocket system is isolated. Hint: the mass of the Earth is 5.94 x 1024kg and G=6.67×10−11Jmkg2G=6.67\times10^{-11}\frac{Jm}{kg^2}G=6.67×10−11kg2Jm
Answer:
The expected radius of the Earth is 3.883 meters.
Explanation:
The formula for the escape speed is derived from Principle of Energy Conservation and knowing that rocket is initially at rest on the surface of the Earth and final energy is entirely translational kinetic, that is:
[tex]U = K[/tex] (1)
Where:
[tex]U[/tex] - Gravitational potential energy, in joules.
[tex]K[/tex] - Translational kinetic energy, in joules.
Then, we expand the formula by definitions of potential and kinetic energy:
[tex]\frac{G\cdot M\cdot m}{r} = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (2)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of the Earth. in kilograms.
[tex]m[/tex] - Mass of the rocket, in kilograms.
[tex]r[/tex] - Radius of the Earth, in meters.
[tex]v[/tex] - Escape velocity, in meters per second.
Then, we derive an expression for the escape velocity by clearing it within (2):
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]v = \sqrt{\frac{2\cdot G \cdot M}{r} }[/tex] (3)
If we know that [tex]v = \frac{1}{21}\cdot c[/tex], [tex]c = 3\times 10^{8}\,\frac{m}{s}[/tex], [tex]M = 5.94\times 10^{24}\,kg[/tex], [tex]G = 6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex] and [tex]M = 5.94\times 10^{24}\,kg[/tex], then the expected radius of the Earth is:
[tex]\frac{GM}{r} = \frac{1}{2}\cdot v^{2}[/tex]
[tex]r = \frac{2\cdot G \cdot M}{v^{2}}[/tex]
[tex]r = \frac{2\cdot \left(6.67\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.94\times 10^{24}\,kg)}{\left[\frac{1}{21}\cdot \left(3\times 10^{8}\,\frac{m}{s} \right) \right]^{2}}[/tex]
[tex]r = 3.883\,m[/tex]
The expected radius of the Earth is 3.883 meters.
I need help with this problem can anybody help me please , it’s physics 2 course
Answer:
ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|ਹੈਲੋ, ਇੰਡੀਆ ਦਾ ਆਪਣਾ ਵੀਡੀਓ ਐਪ - ਰੋਪੋਸੋ ਤੇ Manjeet Warval ਦਾ ਵੀਡੀਓ ਦੇਖੋ | ਨਾਲ ਹੀ PM ਮੋਦੀ ਦੇ 'ਵੋਕਲ ਫ਼ਾਰ ਲੋਕਲ' ਮੋਹਿਮ ਨੂੰ ਸਫ਼ਲ ਬਣਾਉਣ ਲਈ ਰੋਪੋਸੋ ਤੇ 5 ਕਰੋੜ ਤੋਂ ਜ਼ਿਆਦਾ ਭਾਰਤੀਆਂ ਦੇ ਨਾਲ ਜੋੜੋ| ਹੁਣੇ ਰੋਪੋਸੋ ਐਪ ਡਾਊਨਲੋਡ ਕਰੋ ਅਤੇ 100 ਕੋਇਨਜ਼ ਪਾਣ ਲਈ 24 ਘੰਟੇ ਦੇ ਅੰਦਰ ਸਾਇਨ ਅੱਪ ਕਰੋ|
Explanation:
The 2 capacitors in the middle are connected in parallel so simply add their capacitance together:
[tex]5.0\:\mu\text{F} + 8.0\:\mu\text{F} = 13.0\:\mu \text{F}[/tex]
Now we have 3 capacitors connected in series so their equivalent capacitance [tex]C_{eq}[/tex] is
[tex]\dfrac{1}{C_{eq}} = \dfrac{1}{10.0\:\mu \text{F}} + \dfrac{1}{13.0\:\mu \text{F}} + \dfrac{1}{9.0\:\ mu \text{F}} [/tex]
or
[tex]C_{eq} = 3.5\:\mu \text{F}[/tex]
Two electrons are passing 20.0 mm apart. What is the electric repulsive force that they exert on each other
Answer:
0.5766422350752*10^-24 N
Explanation:
Couloumb's law states that states that there is an electrical force acting on 2 static charges. The magnitude is directly proportional to the product of the 2 charges.
Strength of electrons : q1 = q2 = 1.602 x 10-19. C
Substitute and solve:
F = (9*10^9)(1.602 x 10-19)(1.602 x 10-19) / (0.02)^2
Done.
When air expands adiabatically (without gaining or losing heat), its pressure P and volume V are related by the equation PV1.4=C where C is a constant. Suppose that at a certain instant the volume is 420 cubic centimeters and the pressure is 99 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centimeters per minute is the volume increasing at this instant?
Answer:
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Explanation:
We are given that
[tex]PV^{1.4}=C[/tex]
Where C=Constant
[tex]\frac{dP}{dt}=-7KPa/minute[/tex]
V=420 cubic cm and P=99KPa
We have to find the rate at which the volume increasing at this instant.
Differentiate w.r.t t
[tex]V^{1.4}\frac{dP}{dt}+1.4V^{0.4}P\frac{dV}{dt}=0[/tex]
Substitute the values
[tex](420)^{1.4}\times (-7)+1.4(420)^{0.4}(99)\frac{dV}{dt}=0[/tex]
[tex]1.4(420)^{0.4}(99)\frac{dV}{dt}=(420)^{1.4}\times (7)[/tex]
[tex]\frac{dV}{dt}=\frac{(420)^{1.4}\times (7)}{1.4(420)^{0.4}(99)}[/tex]
[tex]\frac{dV}{dt}=21.21cm^3/min[/tex]
Answer:
[tex]\dot V=2786.52~cm^3/min[/tex]
Explanation:
Given:
initial pressure during adiabatic expansion of air, [tex]P_1=99~kPa[/tex]
initial volume during the process, [tex]V_1=420~cm^3[/tex]
The adiabatic process is governed by the relation [tex]PV^{1.4}=C[/tex] ; where C is a constant.
Rate of decrease in pressure, [tex]\dot P=7~kPa/min[/tex]
Then the rate of change in volume, [tex]\dot V[/tex] can be determined as:
[tex]P_1.V_1^{1.4}=\dot P.\dot V^{1.4}[/tex]
[tex]99\times 420^{1.4}=7\times V^{1.4}[/tex]
[tex]\dot V=2786.52~cm^3/min[/tex]
[tex]\because P\propto\frac{1}{V}[/tex]
[tex]\therefore[/tex] The rate of change in volume will be increasing.
any one tell me about the earth rotation it rotatining or not with any proof?
A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from the Sun are r1 and r2, respectively. Find the relation between the time period of the planet in terms of r1 and r2.
Answer:
the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
Explanation:
Given the data i the question;
mass of sun = M
minimum and maximum distance = r1 and r2 respectively
Now, using Kepler's third law,
" the square of period T of any planet is proportional to the cube of average distance "
T² ∝ R³
average distance a = ( r1 + r2 ) / 2
we know that
T² = 4π²a³ / GM
T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]
T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]
T² = 4π² [( r1 + r2 )³ / 8GM ]
T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]
T = 2π √[( r1 + r2 )³ / 8GM ]
Therefore, the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
A pilot flies her route in two straight-line segments. The displacement vector A for the first segment has a magnitude of 243 km and a direction 30.0o north of east. The displacement vector for the second segment has a magnitude of 178 km and a direction due west. The resultant displacement vector is R = A + B and makes an angle ? with the direction due east. Using the component method, find (a) the magnitude of R and (b) the directional angle ?.
(a) R = km
(b) ? = degrees
Answer:
a) [tex]R=126Km[/tex]
b) [tex]\theta=74.6\textdegree[/tex]
Explanation:
From the question we are told that:
1st segment
243km at Angle=30
2nd segment
178km West
Resolving to the X axis
[tex]F_x=243cos30+178[/tex]
[tex]F_x=33.44Km[/tex]
Resolving to the Y axis
[tex]F_y=243sin30+178sin0[/tex]
[tex]R=\sqrt{F_y^2+F_x^2}[/tex]
[tex]F_y=121.5Km[/tex]
Therefore
Generally the equation for Directional Angle is mathematically given by
[tex]\theta=tan^{-1}\frac{F_y}{F_x}[/tex]
[tex]\theta=tan^{-1}\frac{121.5}{33.44}[/tex]
[tex]\theta=74.6\textdegree[/tex]
Generally the equation for Magnitude is mathematically given by
[tex]R=\sqrt{F_y^2+F_x^2}[/tex]
[tex]R=\sqrt{33.44^2+121.5^2}[/tex]
[tex]R=126Km[/tex]
A single-turn square loop of wire, 2.00 cm on each edge, carries a clockwise current of 0.240 A. The loop is inside a solenoid, with the plane of the loop perpendicular to the magnetic field of the solenoid. The solenoid has 30.0 turns/cm and carries a clockwise current of 15.0 A. Find:
a. the force on each side of the loop
b. the torque acting on the loop.
Answer:
Explanation:
a )
Magnetic field inside solenoid B = μ₀ NI ,
μ₀ = 4π x 10⁻⁷ ; N is no of turns per meter length in solenoid and I is current B= 4π x 10⁻⁷ x 30 x 10² x 15
= .0565 T .
Force on each side of square loop = B i L
B is external magnetic field , i is current in loop and L is length of side
Force on each side of square loop = .0565 x .24 x 2 x 10⁻²
= 2.7 x 10⁻⁴ N .
b )
Torque on the loop = F x d
F is force on one side , d is distance between two sides , that is side of the square loop
= 2.7 x 10⁻⁴ x 2 x 10⁻² N.m
= 5.4 x 10⁻⁶ N.m .
A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels, from left to right along a long, horizontal stretched string with a speed of 36.0 m s. I Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement,
(a) What is the frequency of the wave?
(b) What is the angular frequency of the wave?
(c) What is the wave number of the wave?
(d) What is the function y(x,t) that describes the wave?
(e) What is y(t) for a particle at the left end of the string?
(f) What is y(t) for a particle 1.35 m to the right of the origin?
(g) What is the maximum magnitude of transverse velocity of any particle of the string?
(h) Find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
(i) Find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
Explanation:
Given that,
Amplitude, A = 2.5 nm
Wavelength,[tex]\lambda=1.8\ m[/tex]
The speed of the wave, v = 36 m/s
At time t = 0 the left end of the string has its maximum upward displacement.
(a) Let f is the frequency. So,
[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]
(b) Angular frequency of the wave,
[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]
(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]
[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]
The gravitational force Asteroid A experiences is the gravitational force Asteroid C experiences
Answer:
The gravitational force Asteroid A experiences is greater than the gravitational force Asteroid C experiences
A resident of a lunar colony needs to have her blood pressure checked in one of her legs. Assume that we express the systemic blood pressure as we do on earth and that the density of blood does not change. Suppose also that normal blood pressure on the moon is still 120/80 (which may not actually be true).
Required:
If a lunar colonizer has her blood pressure taken at a point on her ankle that is 1.5 m below her heart, what will be her systemic blood-pressure reading, expressed in the standard way, if she has normal blood pressure? The acceleration due to gravity on the moon is 1.67 m/s^2
Answer:
The pressure is 2505 Pa.
Explanation:
Height, h = 1.5 m
density of blood, d = 1000 kg/cubic meter
Gravity, g = 1.67 m/s^2
let the pressure is P.
The pressure due to the fluid is given by
P = h d g
P = 1.5 x 1000 x 1.67
P = 2505 Pa
a baseball is thrown vertically upward with an initial velocity of 20m/s.
A,what maximum height will it attain? B,what time will elapse before it strike the ground?
C,what is the velocity just before it strike the ground?
Answer:
Look at explanation
Explanation:
a)Only force acting on the object is gravity, so a=-g (consider up to be positive)
use: v^2=v0^2+2a(y-y0)
plug in givens, at max height v=0
0=400-19.6(H)
Solve for H
H= 20.41m
b) Use: y=y0+v0t+1/2at^2
Plug in givens
0=0+20t-4.9t^2
solve for t
t=4.08 seconds
c) v=v0+at
v=20-39.984= -19.984m/s
A flag pole 18m high casts a shadow 9.6m long . Find the distance of top of pole from the far of end of Shadow.
Answer:
[tex]{ \bf{pythogras \: theorem :}} \\ \\ { \tt{ = \sqrt{ {9.6}^{2} + {18}^{2} } }} \\ = 20.4 \: cm[/tex]
what is conservation energy?
Explanation:
Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant
hope it is helpful to you
If the sum of the external forces on an object is zero, then the sum of the external torques on it
must also be zero.
A) True
B) False
Answer:
True.
Explanation:
If the sum of the external forces on an object is zero, then the sum of the external torques on it must also be zero.
The net external force and the net external torque acting on the object have to be zero for an object to be in mechanical equilibrium.
Hence, the given statement is true.
The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part to 80 dB during a loud part. Approximately how many times louder is the latter sound
Answer:
[tex]\alpha=-3.01dB[/tex]
Explanation:
From the question we are told that:
Sound level intensity
[tex]\triangle I=40dB-80dB[/tex]
Generally the equation for intensity level is mathematically given by
[tex]\alpha=10log_{10}(I/I_x)dB[/tex]
Where
I= Intensity measured
[tex]I_x=Threshold\ of\ audibility[/tex]
[tex]I_x= 10-12 W / m2[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_x} - 10 log10 \frac{}I_2{I_x}[/tex]
[tex]\alpha= 10 log10 \frac{I_1}{I_2}[/tex]
[tex]\alpha=10 log10\frac{40}{80}[/tex]
[tex]\alpha=-3.01dB[/tex]
A baseball of mass 0.145 kg is thrown at a speed of 40.0 m/s. The batter strikes the ball with a force of 15,000 N; the bat and ball are in contact for 0.500 ms. The force is exactly opposite to the original direction of the ball. Determine the final speed of the ball.
The final speed of the ball is 91.72 m/s.
Given the following data:
Mass of baseball = 0.145 kgInitial speed = 40.0 m/sForce = 15,000 NewtonTime = 0.500 milliseconds (ms) to seconds = 0.0005 seconds.To find the final speed of the ball, we would use the following formula:
[tex]F = \frac{M(V - U)}{t}[/tex]
Where:
F is the force applied. u is the initial speed. v is the final speed. t is the time measured in seconds.Substituting the parameters into the formula, we have;
[tex]15000 = \frac{0.145(V \;- \;40)}{0.0005}\\\\15000(0.0005) = 0.145(V \;- \;40)\\\\7.5 = 0.145V - 5.8\\\\0.145V = 7.5 + 5.8\\\\0.145V = 13.3\\\\V = \frac{13.3}{0.145}[/tex]
Final speed, V = 91.72 m/s
Therefore, the final speed of the ball is 91.72 m/s.
Read more here: https://brainly.com/question/24029674
Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2 , đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g, mang điện tích q = 10-8 C treo gần vào mặt phẳng, sao cho dây treo lúc đầu song song với mặt phẳng. Lấy g = 9,8m/s2 . Khi cân bằng, dây treo quả cầu hợp với mặt phẳng 1 góc bằng bao nhiêu?
Answer:
The angle is 16 degree.
Explanation:
A uniformly charged infinite plane, density σ = 4.10-9 C/cm2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2 . In equilibrium, by what angle does the string hanging from the ball make an angle with the plane?
Surface charge density, σ = 4 x 10^-5 C/m^2
charge, q = 10^-8 C
mass, m = 0.008 kg
The electric field due to the plate is
[tex]E= \frac{\sigma }{2\varepsilon 0}[/tex]
Let the angle make with the vertical is A and T is the tension in the string.
[tex]T sin A = q E....(1)\\\\T cos A = m g .... (2)\\\\Divie (1) by (2)\\\\tan A =\frac{q E}{m g}\\\\tan A = \frac{10^{-8}\times 4\times 10^{-5}}{2\times 8.85\times 10^{-12}\times 0.008\times9.8}\\\\tan A = 0.288\\\\A = 16 degree\\[/tex]
Question 9 of 10
According to the law of conservation of momentum, the total initial
momentum equals the total final momentum in a(n)
A. Interacting system
B. System interacting with one other system
C. Isolated system
D. System of balanced forces
Answer:
The answer is C. Isolated System
Answer:
C. Isolated system
Explanation :
∵According to law of conservation of momentum ,In an isolated system ,the total momentum remains conserved.
Test your prediction through calculation for the situations of the clay bob and the bouncy ball. Assume each has a mass of 100 grams, and each has an initial velocity of 20 m/s straight at the door. Ignore the effects of gravity. Calculate the change in momentum of
Answer:
a) Δp = -2.0 kgm / s, b) Δp = -4 kg m / s
Explanation:
In this exercise the change in moment of a ball is asked in two different cases
a) clay ball, in this case the ball sticks to the door and we have an inelastic collision where the final velocity of the ball is zero
Δp = p_f - p₀
Δp = 0 - m v₀
Δp = - 0.100 20
Δp = -2.0 kgm / s
b) in this case we have a bouncing ball, this is an elastic collision, as the gate is fixed it can be considered an object of infinite mass, therefore the final speed of the ball has the same modulus of the initial velocity, but address would count
v_f = - v₀
Δp = p_f -p₀
Δp = m v_f - m v₀
Δp = m (v_f -v₀)
Δp = 0.100 (-20 - 20)
Δp = -4 kg m / s
You walk into a room and you see 4 chickens on a bed 2 cows on the floor and 2 cats in a chair. How many legs are on the ground? (I know this answer just a riddle to see who knows it) (:
Answer:
18
Explanation:
I'm pretty sure I got it right
The period of a pendulum is the time it takes the pendulum to swing back and forth once. If the only dimensional quantities that the period depends on are the acceleration of gravity, g, and the length of the pendulum, l, what combination of g and l must the period be proportional to
Explanation:
Let T is the period of a pendulum. The SI unit of time is seconds (s).
It depends on the acceleration of gravity, g, and the length of the pendulum, l.
The SI unit of acceleration of gravity, g and the length of the pendulum, l are m/s² and m respectively.
If we divide m and m/s², we left with s². If the square root of s² is taken, we get s only i.e. the SI unit of period of a pendulum.
So,
[tex]T\propto \sqrt{\dfrac{l}{g}}[/tex]
Hence, this is the required solution.
Una pelota se lanza verticalmente hacia arriba desde la azotea de un edificio con una velocidad inicial de 35 m/s. Si se detiene en el aire a 200 m del suelo, ¿Cuál es la altura del edificio?
a. 138,8 m
b. 51.2 m
c. 71,2 m
d. 45,0 m
If Katie swims from one end of the pool, to the other side, and then swims back to her original spot, her average velocity is half of her average speed when she swam to the other side.a) trueb) false
Answer:
false.
Explanation:
Ok, we define average velocity as the sum of the initial and final velocity divided by two.
Remember that the velocity is a vector, so it has a direction.
Then when she goes from the 1st end to the other, the velocity is positive
When she goes back, the velocity is negative
if both cases the magnitude of the velocity, the speed, is the same, then the average velocity is:
AV = (V + (-V))/2 = 0
While the average speed is the quotient between the total distance traveled (twice the length of the pool) and the time it took to travel it.
So we already can see that the average velocity will not be equal to half of the average speed.
The statement is false