A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from it at 21.8 m/s, what is the received sound frequency?
f= ? Hz
Answer:
f" = 40779.61 Hz
Explanation:
From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;
from the Doppler effect equation, we can calculate the initial observed frequency as:
f' = f(1 - (v_o/v))
We are given;
f = 46.2 kHz = 46200 Hz
v_o = 21.8 m/s
v is speed of sound = 343 m/s
Thus;
f' = 46200(1 - (21/343))
f' = 43371.4285 Hz
In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;
Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;
f" = f'/(1 + (v_o/v))
f" = 43371.4285/(1 + (21.8/343))
f" = 40779.61 Hz
1.How much work does it take to get a 2Kg ball moving 15m/s if it starts from rest?
2. If a force of 235N was added to the ball, through what distance would this force have to act to give the ball a velocity of 15m/s
which changes will increase the rate of reaction during combustion
Answer:
reducing temperature of the surrounding
Explanation:
combustion reactions are exothermic so they give off heat. reducing the temperature of the surrounding will enable more efficient energy transfer
The second law of thermodynamics imposes what limit on the efficiency of a heat engine?
A. The energy a heat engine must deposit in a cold reservoir is greater than or equal to the energy it extracts from a hot reservoir.
B. The energy a heat engine must deposit in a cold reservoir is greater than or equal to the energy extracted as useful work.
C. A heat engine must deposit some energy in a cold reservoir.
Answer:
C. A heat engine must deposit some energy in a cold reservoir.
Explanation:
The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir."
This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir.
Then we have the equation:
Q = W + q
From this we can conclude that the correct option is:
C. A heat engine must deposit some energy in a cold reservoir.
There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.
C. A heat engine must deposit some energy in a cold reservoir.
The second law of thermodynamics says that "It is impossible to extract an amount of heat Q from a hot reservoir and use it all to do work W. Some amount of heat q must be exhausted to a cold reservoir". This means that if we extract an amount of heat Q from the hot reservoir, the work W can never be exactly equal to Q, then there is a surplus of heat q that must be deposited in a cold reservoir. Then we have the equation: Q = W + q There will be always some energy that is not transformed into work, and is deposited in a cold reservoir.Therefore, option C is correct.
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A rectangular reflecting pool is 85.0 ft wide and 120 ft long. What is the area of the pool in square meters?
A car is traveling at a constant speed of 20 m/s for 3 seconds. Then the driver puts on the brakes. The total distance the car travels is 100 m. What is the total time the car was moving?
Answer:
15 seconds
Explanation:
If car was moving at 20m/s for 3 sec.
if car traveled 100m = 15 sec total
WHat does that mean?
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.69 times a second. A tack is stuck in the tire at a distance of 0.331 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed.
Answer:
the tack's tangential speed is 5.59 m/s
Explanation:
Given that;
R = 0.331 m
wheel rotates 2.69 times a second which means, the wheel complete 2.69 revolutions in a second, so
ω = 2.69 rev/s × 2π/1s = 16.9 rad/s
using the relation of angular speed with tangential speed
tangential speed v of the tack is expressed as;
v = R × ω
so we substitute
v = 0.331 m × 16.9 rad/s
v = 5.59 m/s
Therefore, the tack's tangential speed is 5.59 m/s
true or false A person's speed around the Earth is faster at the poles than it is at the equator.
Answer:False
Explanation:The Earth rotates faster at the equator than at the poles.
Plz help this is so confusing
Answer:
5 Km/h
Explanation:
From the question given above, the following data were obtained:
Distance travelled = 10 Km
Time = 2 hours
Speed =?
Speed is simply defined as the distance travelled per unit time. Mathematically, it can be represented as:
Speed = distance travelled /time.
With the above formula, we can obtain the speed at which the duck is travelling as follow:
Distance travelled = 10 Km
Time = 2 hours
Speed =?
Speed = distance travelled /time.
Speed = 10 / 2
Speed = 5 Km/h
Thus, the duck is travelling at a speed of 5 Km/h
A pingpong ball has 2 kg/s of momentum when
thrown 8 m/s. Find the mass of the ball.
Answer:
0.25 kg
Explanation:
p = mv
2 = m(8)
2/8 = m(8)/8 *cancels
m = 1/4 OR 0.25 kg
Two spherical objects have masses of 100 kg and 200 kg. Their centers are
separated by a distance of 40 cm. Find the gravitational attraction between
them.
Answer:
8.34 x 10⁻⁶N
Explanation:
Given parameters:
Mass 1 = 100kg
Mass 2 = 200kg
Distance of separation = 40cm = 0.4m
Unknown:
Gravitational force of attraction between them = ?
Solution:
To solve this problem, we use the expression below which is derived from the Newton's law of universal gravitation:
Fg = [tex]\frac{G x mass 1 x mass 2}{d^{2} }[/tex]
G is the universal gravitation constant = 6.67 x 10⁻¹¹
d is the separation
Now;
Fg = [tex]\frac{6.67 x 10^{-11} x 100 x 200}{0.4^{2} }[/tex] = 8.34 x 10⁻⁶N
Our Sun’s mass is 1.0 and our Earth’s mass is 2.0. The distance is standard as given on the simulation. Describe the path of the Earth.
Answer:
Earth orbits the Sun at an average distance of 149.60 million km (92.96 million mi), and one complete orbit takes 365.256 days (1 sidereal year), during which time Earth has traveled 940 million km (584 million mi).
Explanation:
John runs 3 km north then walks 2 km south. What is his total distance traveled and displacement?
Answer:
the total distance is 5km and the displacement is 1km
Explanation:
The total distance would be the addition of John running both ways so 3 km, 2 km.
However since he only walked back from a distance of 3 km to 2 km, he would be displaced 1 km because displacement is more like the position from the original point.
Think about 2 km as a positive value for the first part of the question and a negative value for the second part.
A racecar makes 24 revolutions around a circular track of radius 2 meters in
162 seconds. Find the racecar's frequency
Answer:
[tex]0.15\: \mathrm{Hz}[/tex]
Explanation:
The frequency is of an object is given by [tex]f=\frac{1}{T}[/tex], where [tex]T[/tex] is the orbital period of the object.
Since the racecar makes 24 revolutions around a circular track in 162 seconds, it will take the racecar [tex]\frac{162}{24}=6.75\:\mathrm{s}[/tex] per revolution.
Therefore, the frequency of the racecar is [tex]\frac{1}{6.75}=\fbox{$0.15\:\mathrm{Hz}$}[/tex] (two significant figures).
The radius of the track is irrelevant in this problem.
The speed limit on some segments of interstate 4 is 70 mph. What is this in km/h?
Answer:
112.63km/hr
Explanation:
The given dimension is :
70mph
We are to convert this to km/hr
1 mile = 1.609km
so;
70mph x 1.609 = 112.63km/hr
So,
The solution is 112.63km/hr
To understand and apply the formula τ=Iα to rigid objects rotating about a fixed axis. To find the acceleration a of a particle of mass m, we use Newton's second law: F net=ma , where F net is the net force acting on the particle.To find the angular acceleration α of a rigid object rotating about a fixed axis, we can use a similar formula: τnet=Iα, where τnet=∑τ is the net torque acting on the object and I is its moment of inertia.
Part A:
Assume that the mass of the swing bar, is negligible. Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, l, as well as the acceleration due to gravity g.
Part B:
Now consider a similar situation, except that now the swing bar itself has mass mbar.Find the magnitude of the angular acceleration α of the seesaw.
Express your answer in terms of some or all of the quantities m1, m2, mbar, l, as well as the acceleration due to gravity g.
Answer:
Hello your question is incomplete attached below is the missing part of the question
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these situations, two objects of masses m1 and m2 are attached to a seesaw. The seesaw is made of a bar that has length l and is pivoted so that it is free to rotate in the vertical plane without friction. Assume that the pivot is attached tot he center of the bar.
You are to find the angular acceleration of the seesaw when it is set in motion from the horizontal position. In all cases, assume that m1>m2.I
answer : part A = 2*[(M1 - M2)/(M1 + M2)]*g/L
part A = attached below
Explanation:
Part A :
Assuming that mass of swing is negligible
α = T/I
where ; T = torque, I = inertia,
hence T = L/2*9*(M1 - M2)
also; I = [tex]M1*(L/2)^2 + M2*(L/2)^2[/tex]= ( M1 + M2) * (L/2)^2
Finally the magnitude of the angular acceleration α
α = 2*[(M1 - M2)/(M1 + M2)]*g/L
Part B attached below
Two 0.60-kilogram objects are connected by a thread that passes over a light, frictionless pulley. The objects are initially held at rest. If a third object with a mass of 0.30 kilogram is added on top of one of the 0.60-kilogram objects and the objects are released, the magnitude of the acceleration of the 0.30-kilogram object is most nearly:______
Answer:
2 m/s²
Explanation:
From the given information:
The first mass m_1 = 0.6 kg
The second mass m_2 = 0.3 kg
The magnitude for the acceleration of 0.3 kg is:
a = net force/ effective mass
Mathematically, it can be computed as follows:
[tex]a = \dfrac{F}{m}[/tex]
[tex]a = \dfrac{(m_2 +m_1 -m_1) }{(m_2+m_1+m_1)}(g)[/tex]
[tex]a = \dfrac{0.3 +0.6 -0.6}{(0.3 +0.6+0.6)}(9.8)[/tex]
a ≅ 2 m/s²
There are two different isotopes; X and Y, both contain the same number of radioactive substances. If sample X
has a longer half-life than Y, compare their rate of radioactive decay.
O A. Rate does not depend on half-life
B. Both of their rates are equal
O C. X has a smaller rate than Y
O D. X has a greater rate than Y
Answer:
Half life refers to the time for 1/2 of the radioactive atoms to decay.
Suppose that X has a half life of 10 days and Y has a half life of 20 days
If both start out with 1000 radioactive atoms then after 20 days
X would have 250 radioactive atoms and Y would have 500 atoms
The rate of decay is greater for the shorter half life:
In the example given X must have the smaller rate of decay because it has a longer half life.
A receiver catches a football on the 50.0 yard line and is tackled 5.42 seconds later on the 12 yard line. What
was the runner's average speed?
Answer:
7.01yard/sec
Explanation:
Given parameters:
Initial position = 50yard
Final position = 12yard
Time = 5.42s
Unknown:
Average speed of runner = ?
Solution:
To solve this problem;
Speed = [tex]\frac{distance}{time}[/tex]
Distance covered = Initial position - final position = 50 - 12 = 38yards
So;
Speed = [tex]\frac{38}{5.42}[/tex] = 7.01yard/sec