Answer:
plz give the question... floor the proper answer
Explanation:
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a sound wave has a frequency of 85 Hz and a wavelength of 4.2 m. what is the wave speed of the sound wave?
A. 357 seconds
B. 89.2 m/s
C. 357 m/s
D. 20 m/s
Answer:C
Explanation:
I did the test
a sound wave has a frequency of 85 Hz and a wavelength of 4.2 m. Then the wave speed of the sound wave is 20.23 m/s. Hence option D is correct.
What is wave ?Wave is is a disturbance in a medium that carries energy as well as momentum . wave is characterized by amplitude, wavelength and phase. Amplitude is the greatest distance that the particles are vibrating. especially a sound or radio wave, moves up and down. Amplitude is a measure of loudness of a sound wave. More amplitude means more loud is the sound wave.
Wavelength is the distance between two points on the wave which are in same phase. Phase is the position of a wave at a point at time t on a waveform. There are two types of the wave longitudinal wave and transverse wave.
The wave speed is given by,
c = νλ
where λ is wavelength, ν is frequency.
Given,
frequency ν = 85 Hz
wavelength λ = 4.2 m
The speed of the wave,
c = 85 × 4.2 = 20.23 m/s
Hence option D is correct.
To know more about Speed :
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A column of soldiers, marching at 100 steps per minute, keep in step with the beat of a drummer at the head of the column. It is observed that the soldiers in the rear end of the column are striding forward with the left foot when the drummer is advancing with the right. What is the approximate length of the column? (Take the speed of sound to be 343 m/s.)
Answer:
The value is [tex]D = 205.8 \ m [/tex]
Explanation:
The time taken for the column to take a step mathematically represented as
100 steps => 1 minutes => 60 seconds
1 step => t
=> [tex]t = 0.6 \ s [/tex]
Generally the length of the column is mathematically represented as
[tex]D = v * t[/tex]
substituting 343 m/s for v we have
[tex]D = 343 * 0.6 [/tex]
=> [tex]D = 205.8 \ m [/tex]
True or false: points that lie on the same plane are Collinear
If two tug boats are towing a ship with force of 5 tons each and the angle between the two ropes is 60 degrees, what is the resultant force on the ship? Explain how to use a force table to verify answer.
Answer:
8.6602 tons
Explanation:
We first draw the known vector forces.
2fcos30⁰
We have f to be equal to 5tons
Inserting into formula
Σfx = 2(5)cos30⁰
= 8.6602 tons
Σfy is equal to 0, this is because in the y direction, the forces cancel themselves out.
Therefore the resultant force on the ship is equal to 8.6602 tons
I hope this helps!
Please check attachment for diagram.
Oppositely charged parallel plates are separated by 4.49 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? N/C (b) What is the magnitude of the force on an electron between the plates? N (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 3.14 mm from the positive plate?
Answer:
A. Using
E=V/d
= 600/4.49*10^-3
= 1.336 x10^5 N/C
b) F = E*q = 1.33610^5 x 1.6*10^-19
= 2.17 x 10^-14 N
c) Work = Fs distance = 2.17 x 10^-14 N (4.49-3.14)*.001= 1.35 x 10^-17 J
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
Answer:
what's the question ,so I can answer it right ?
Newtons Second Law of motion states that for acceleration to stay the same...
Answer:
newton force is the state of gravity force they abstract when they came closer
A basketball leaves a player's hands at a height of 2.20 m above the floor. The basket is 2.70 m above the floor. The player likes to shoot the ball at a 36.0 ∘ angle. Of the shot is made from a horizontal distance of 9.10 m and must be accurate to ±0.23m (horizontally), what is the range of initial speeds allowed to make the basket
Answer:
The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].
Explanation:
We must notice that basketball depicts a parabolic motion, which consists of combining a constant speed motion in x-direction and free fall motion in the y-direction. The motion is described by the following kinematic formulas:
x-Direction
[tex]x = x_{o} + v_{o}\cdot t \cdot \cos \alpha[/tex]
y-Direction
[tex]y = y_{o} + v_{o}\cdot t\cdot \sin \alpha +\frac{1}{2}\cdot g\cdot t^{2}[/tex]
Where:
[tex]x_{o}[/tex], [tex]y_{o}[/tex] - Initial position of the basketball, measured in meters.
[tex]x[/tex], [tex]y[/tex] - Final position of the basketball, measured in meters.
[tex]v_{o}[/tex] - Initial speed of the basketball, measured in meters per second.
[tex]t[/tex] - Time, measured in seconds.
[tex]\alpha[/tex] - Tilt angle, measured in sexagesimal degrees.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
If we know that [tex]x_{o} = 0\,m[/tex], [tex]y_{o} = 2.20\,m[/tex], [tex]\alpha = 36^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]x = (9.10\pm0.23)\,m[/tex] and [tex]y = 2.70\,m[/tex], the system of equation is reduce to this:
[tex](9.10\pm 0.23)\,m = 0\,m + v_{o}\cdot t \cdot \cos 36^{\circ}[/tex]
[tex]9.10\pm 0.23 = 0.809\cdot v_{o}\cdot t[/tex] (Ec. 1)
[tex]2.70\,m = 2.20\,m + v_{o}\cdot t \cdot \sin 36^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]
[tex]0.50 = 0.588\cdot v_{o}\cdot t-4.904\cdot t^{2}[/tex] (Ec. 2)
At first we clear [tex]v_{o}\cdot t[/tex] in (Ec. 1):
[tex]v_{o}\cdot t = \frac{9.10\pm 0.23}{0.809}[/tex]
[tex]v_{o}\cdot t = 11.248\pm 0.284[/tex]
(Ec. 1) in (Ec. 2):
[tex]0.5 = 0.588\cdot (11.248\pm 0.284)-4.904\cdot t^{2}[/tex]
Now we clear the time in the resulting expression:
[tex]4.904\cdot t^{2} = 0.588\cdot (11.248\pm 0.284)-0.5[/tex]
[tex]t = \sqrt{\frac{0.588\cdot (11.248\pm 0.284)-0.5}{4.904} }[/tex]
There are two solutions:
[tex]t_{1} = \sqrt{\frac{0.588\cdot (11.248- 0.284)-0.5}{4.904} }[/tex]
[tex]t_{1} \approx 1.101\,s[/tex]
[tex]t_{2} = \sqrt{\frac{0.588\cdot (11.248+ 0.284)-0.5}{4.904} }[/tex]
[tex]t_{2}\approx 1.131\,s[/tex]
The initial velocity is cleared within (Ec. 2):
[tex]v_{o}=\frac{0.50+4.904\cdot t^{2}}{0.588\cdot t}[/tex]
The bounds of the range of initial speed is determined hereafter:
[tex]t_{1} \approx 1.101\,s[/tex]
[tex]v_{o} = \frac{0.50+4.904\cdot (1.101)^{2}}{0.588\cdot (1.101)}[/tex]
[tex]v_{o} = 9.954\,\frac{m}{s}[/tex]
[tex]t_{2}\approx 1.131\,s[/tex]
[tex]v_{o} = \frac{0.50+4.904\cdot (1.131)^{2}}{0.588\cdot (1.131)}[/tex]
[tex]v_{o} = 10.185\,\frac{m}{s}[/tex]
The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].
When the sun provides energy for photosynthesis, an interaction with the __________ takes place.
A bird flies 3.7 meters in 46 seconds, what is its speed?
Answer:
Speed is 0.08 m/s.
Explanation:
Given the distance that the bird flies = 3.7 meters
The time is taken by the bird to fly the 3.7 meters = 46 seconds
We have given distance and time. Now we have to find the speed at which the bird flies. So, to calculate the speed of the bird we have to divide the distance by the time.
Below is the formula to find the speed.
Speed = Distance / Time
Now insert the given value in the formula.
Speed = 3.7 / 46 = 0.08 m/s
You are trying to get to class on time using the UCF Shuttle. You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. In 40.9 m you will reach a barrier and you must catch the shuttle before that point. The shuttle has a constant acceleration of 4.5 m/s2. What is the minimum velocity you have to run at to catch the bus before it reaches the barrier
Answer:
20.1 m/s
Explanation:
Since You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. And In 40.9 m you will reach a barrier and you must catch the shuttle before that point.
Given that the shuttle has a constant acceleration of 4.5 m/s2.
The total distance to cover is:
Total distance = 40.9 + 3.9 = 44.8 m
Assuming you are starting from rest. Then initial velocity U = 0
Using the 3rd equation of motion to calculate the minimum velocity.
V^2 = U^2 + 2as
V^2 = 0 + 2 × 4.5 × 44.8
V^2 = 403.2
V = sqrt (403.2)
V = 20.1 m/s
Therefore, the minimum velocity you have to run at to catch the bus before it reaches the barrier is 20.1 m/s
A watermelon is dropped from the edge of the roof of a build- ing and falls to the ground. You are standing on the sidewalk and see the watermelon falling when it is 30.0 m above the ground. Then 1.50 s after you first spot it, the watermelon lands at your feet. What is the height of the building
Answer:
The hight of the building is 38.16 m
Explanation:
These two pieces of information given, first, the watermelon is 30 m above the ground and after 1.50 s the watermelon has been spotted. Now we are required to find the height of the building.
Use the below formula to find the height of buildings.
S = ut + ½ gt^2
30 =1.5u + (1/2) × 9.8 (1.5)^2
u = 12.65 m/sec
v^2 – u^2 = 2gs
(12.65)^2 = 2×9.8 s’
S’ = 8.16 m
h = s + s’
h = 30 + 8.16 = 38.16 m
The hight of the building is 38.16 m.
The height of the building is 38.16 m.
Given data:
The height above the ground is, h = 30.0 m.
The time interval after observation of first spot is, t = 1.50 s.
We need to find the height of building. And since two pieces of information given, first, the watermelon is 30 m above the ground and after 1.50 s the watermelon has been spotted. So, using the second kinematic equation of motion as,
[tex]h = ut + \dfrac{1}{2}gt^{2}[/tex]
Here, u is the initial speed. Solving as,
[tex]30 = (u \times 1.50) + \dfrac{1}{2} \times 9.8 \times (1.50)^{2}\\\\u =12.65 \;\rm m/s[/tex]
Now landing distance (s') is calculated using the third kinematic equation of motion as,
[tex]v^{2} =u^{2}+2(-g)s\\\\0^{2} =12.65^{2}+2(-9.8)s\\\\s = 8.16 \;\rm m[/tex]
Then the height of building is given as,
H = h + s
H = 30 m + 8.16 m
H = 38.16 m
Thus, we can conclude that the height of the building is 38.16 m.
Learn more about the kinematic equations of motion here:
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Let's start by calculating what acceleration the rocket must produce to launch into earth orbit. In order to attain orbit around earth, the ATLAS V rocket must accelerate up to a speed of about 7700 meters per second in about 4.2 minutes. What average acceleration is required to accomplish this
Answer:
30.56 m/s^2
Explanation:
Given that In order to attain orbit around earth, the ATLAS V rocket must accelerate up to a speed of about 7700 meters per second in about 4.2 minutes.
The average acceleration that is required to accomplish this will be
Average acceleration = change in velocity / time
Average acceleration = 7700/ 4.2 × 60
Average acceleration = 7700/252
Average acceleration = 30.56 m/s^2
The chilled water system for a 27-story building has a pump located at ground level. The lost head in a vertical riser from the pump to an equipment room on the twenty-seventhfloor is 40ftof water, and the pump produces 270ft of head. What is the pressure on the suction side of the pump for a pressure of 8 psig to exist in the riser on the twenty-fifth floor
This question is incomplete, the complete question is;
The chilled water system for a 27-story building has a pump located at ground level. The lost head in a vertical riser from the pump to an equipment room on the twenty-seventh floor is 40ft of water, and the pump produces 270ft of head. What is the pressure on the suction side of the pump for a pressure of 8 psig to exist in the riser on the twenty-fifth floor
Assume 12ft of elevation per floor
Answer: 48.68 psig
Explanation:
First we calculate the elevation of the building
hb = 27 story * 12ft per floor/story
hb = 324 ft
given that the head lost in the vertical riser hL = 40 ft
now the delivery head required in the riser on he 27th floor;
hd = 8 psig * (2.31 ft / 1 psig)
hd = 18.46 ft
Now calculate the suction head required by balancing the energy per unit weight of water, considering pump as the control volume
hp = (hb + hL + hd) - hs
hs = hb + hL + hd - hp
where hp is the head developed by the pump (270 ft)
hb is the elevation of the 27th floor of the building ( 324 ft)
hL is the head lost in the vertical riser ( 40 ft)
hd is the head required to exist in the riser on the 27th floor (18.46 ft)
so we substitute
hs = 324 ft + 40 ft + 18.46 ft - 270 ft
hs = 112.46
so 112.46ft * (1 psig / 2.31 ft)
= 48.68 psig
An airplane with a hot-wire anemometer mounted on its wing tip is to fly through the turbulent boundary layer of the atmosphere at a speed of 50 m/sec. The velocity fluctuations in the atmosphere are of order 0.5 m/sec, the length scale of the large eddies is about 100 m. The hot-wire anemometer is to be designed so that it will register the motion of the smallest eddies.What is the highest frequency the anemometer will encounter
Answer:
0.55 hz
Explanation:
Given that the plane fly through the turbulent boundary layer of the atmosphere at a speed of 50 m/sec. And the velocity fluctuations in the atmosphere are of order 0.5 m/sec, the length scale of the large eddies is about 100 m.
The maximum speed attained will be
Maximum speed = 50 + 0.5 = 5.5 m/s
The Length = 100m
Speed = FL
Where F = frequency
Substitute speed and distance length into the formula
55 = 100F
F = 55/100
F = 0.55 Hz
Therefore, the highest frequency the anemometer will encounter will be 0.55 Hz
Answer:
40079 Hz
Explanation:
1. The first step is to calculate energy dissipation ∈=u^3/l and here u is fluctuating velocity
∈=(0.5^3)/100 = 0.00125 m^2/s^3
2. Find out the length scale of the small eddies
η=(viscosity/∈)^1/4
η=(1.470e-5/0.00125)^1/4 = 0.00126 m
3. The frequency associated with these small-scale eddies will be the greatest frequency the anemometer will encounter, thus:
u_max=f_max * η
u_max = u + u' = 50+0.5=50.5 m/s
f_max = u_max/η = 50.5/0.00126 = 40079 Hz
This is the heighest frequency the anemometer will encounter.
What is the velocity of a plane to travel 3000 miles from New York to California in 5.0 hours
Answer:
10 miles per minute.
Drive-reduction theory states that motivation comes from a combination of both reinforcement and drive.
ОА.
True
OB. False
(7%) Problem 14: A robot cheetah can jump over obstacles. Suppose the launch speed is vo = 4.74 m/s, and the launch angle is 0 = 25.5
degrees above horizontal.
What is the maximum height in meters?
Air that initially occupies 0.22 m3 at a gauge pressure of 86 kPa is expanded isothermally to a pressure of 101.3 kPa and then cooled at constant pressure until it reaches its initial volume. Compute the work done by the air. (Gauge pressure is the difference between the actual pressure and atmospheric pressure.)
Answer:
total work done = -5960.8 J
Explanation:
given data
initial volume v1 = 0.22 m³
initial pressure p1 = 86 kPa
final pressue p2 = 101.3 kPa
solution
we apply here isothermal expansion that is express as
p1 × v1 = p2 × v2 ......................1
put here value
86 × 0.22 = 101.3 × v2
v2 = 0.1867 m³
and
work done will be here
w1 = p1 × v1 × ln([tex]\frac{p1}{p2}[/tex]) ....................2
w1 = 86 × 10³ × 0.22 × [tex]ln(\frac{86}{101.3})[/tex]
w1 = -3.097 × 10³ J
and
it is cooled to initial volume at constant pressure so here work done will be
w2 = p(v2 - v1) .................3
w2 = 86 × 10³ × ( 0.1867 - 0.22 )
w2 = -2863.8 J
so
total work done is
total work done = w1 + w2
total work done = -3097 + -2863.8
total work done = -5960.8 J
Two cylindrical resistors are made from the same material. The shorter one has a length LL and diameter DD . The longer one has a length 16L16L and diameter 4D4D . How do their resistances compare? The resistance of the longer resistor is four times the resistance of the shorter resistor. The resistance of the longer resistor is twice the resistance of the shorter resistor. The resistance of the longer resistor is the same as the resistance of the shorter resistor. The resistance of the longer resistor is half the resistance of the shorter resistor. The resistance of the longer resistor is a quarter of the resistance of the shorter resistor.
Answer:
The resistance of the longer resistor is a quarter of the resistance of the shorter resistor.Explanation:
If Two cylindrical resistors are made from the same material, then their resistivity will be the same. Formula for calculating resistivity of a material is expressed as;
[tex]\rho = \frac{RA}{L} \ where \ A = \frac{\pi d^2}{4}[/tex] where;
R is the resistance
A is the cross sectional area of the material
L is the length of the material
For the shorter cylinder:
Length = L
diameter = D
[tex]\rho = \dfrac{R_s(\frac{\pi D^2}{4})}{L} \\\\\rho = \dfrac{R_s{\pi D^2}}{4L}[/tex]
For the longer cylinder:
Length = 16L
diameter = 4D
[tex]\rho = \dfrac{R_l(\frac{\pi (4D)^2}{4})}{16L} \\\\\\\rho = \dfrac{R_l(\frac{\pi (16D^2)}{4})}{16L} \\\\\rho = \dfrac{R_l{16\pi D^2}}{16L}\\\\\rho = \dfrac{R_l{\pi D^2}}{L}[/tex]
Since their resistivity are the same then;
[tex]\dfrac{R_s{\pi D^2}}{4L} = \dfrac{R_l{\pi D^2}}{L} \\\\ \dfrac{R_s}{4} = {R_l} \\\\R_s = 4R_l\\\\R_l = \frac{R_s}{4}[/tex]
Hence the resistance of the longer resistor is a quarter of the shorter resistor.
A pendulum is 0.760 m long, and the bob has a mass of 1,00 kg. At the bottom of its swing, the bob's speed is 1.60 m/s. The tension is greater than the weight of the bob because Multiple Choice 0 its horizontal component is increasing from zero. O the bob has a downward acceleration, so the net Fy must be downward O O the bob has zero acceleration. O ( the bob has an upward acceleration, so the net Fy must be upward Required information A pendulum is 0.760 m long, and the bob has a mass of 1.00 kg. At the bottom of its swing, the bob's speed is 1.60 m/s. What is the tension in the string at the bottom of the swing?
Answer:
T = 13.17 N
the correct one is there is a digested centripetal acceleration towards upward
Explanation:
For this exercise we can use Newton's second law at the bottom of the pendulum's path
we will assume that the upward direction is positive
T - W = m a
in this case it is describing a circle the acceleration is central
a = v² / R
where the radius of the trajectories equals the length of the pendulum
R = L
we substitute
T = mg + m v² / L
T = m (g + v² / L)
When reviewing the different statements, the correct one is there is a digested centripetal acceleration towards upward
let's calculate the tension of the rope
T = 1 (9.8 + 1.6 2 / 0.760)
T = 13.17 N
An object, initially at rest, moves 250 m in 17 s. What is its acceleration?
Answer:
1.73 m/s²
Explanation:
Given:
Δx = 250 m
v₀ = 0 m/s
t = 17 s
Find: a
Δx = v₀ t + ½ at²
250 m = (0 m/s) (17 s) + ½ a (17 s)²
a = 1.73 m/s²
The acceleration of this object is 1.730 meter per seconds square.
Given the following data:
Initial velocity = 2.5 m/s (since the object is starting from rest).Time = 17 seconds.To find the acceleration of this object, we would use the second equation of motion.
Mathematically, the second equation of motion is given by the formula;
[tex]S = ut + \frac{1}{2} at^2[/tex]
Where:
S is the displacement or distance covered.u is the initial velocity.a is the acceleration.t is the time measured in seconds.Substituting the given values into the formula, we have;
[tex]250 = 0(17) + \frac{1}{2} (a)(17^2)\\\\250 = \frac{1}{2} (289)a\\\\250 = 144.5a\\\\a = \frac{250}{144.5}[/tex]
Acceleration, a = 1.730 [tex]m/s^2[/tex]
Therefore, the acceleration of this object is 1.730 meter per seconds square.
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You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but you ask a friend down on the ground to throw another ball upward at speed v. Your friend throws the ball upward at the same moment that you drop yours from the window. At some location, the balls pass each other. Is this location.
Answer:
y = y₀ (1 - ½ g y₀ / v²)
Explanation:
This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i
y = y₀ + v₀ t - ½ g t²
y = y₀ - ½ g t²
for the ball thrown from the ground with initial velocity v₀₂ = v
y₂ = y₀₂ + v₀₂ t - ½ g t²
in this case y₀ = 0
y₂2 = v t - ½ g t²
at the point where the two balls meet, they have the same height
y = y₂
y₀ - ½ g t² = vt - ½ g t²
y₀i = v t
t = y₀ / v
since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height
y = y₀ - ½ g t²
y = y₀ - ½ g (y₀ / v)²
y = y₀ - ½ g y₀² / v²
y = y₀ (1 - ½ g y₀ / v²)
with this expression we can find the meeting point of the two balls
Whenever an object is moving at a constant rate, the value(s) that equal zero is (are):
a) Speed
b) Acceleration
c) Velocity
d) All of the above
Whenever an object is moving at a constant rate . . .
a). its speed is that constant rate.
b). Its acceleration MAY BE zero, if it's also moving in a straight line.
c). Its velocity is that constant rate if it's moving in a straight line. Otherwise, its velocity could have many different values, depending on the path it's following.
The hot reservoir for a Carnot engine has a temperature of 889 K, while the cold reservoir has a temperature of 657 K. The heat input for this engine is 4710 J. The 657-K reservoir also serves as the hot reservoir for a second Carnot engine. This second engine uses the rejected heat of the first engine as input and extracts additional work from it. The rejected heat from the second engine goes into a reservoir that has a temperature of 406 K. Find the total work delivered by the two engines.
Answer:
Explanation:
Efficiency of first engine
= T₁ - T₂ / T₁ where T₁ is temperature of hot reservoir , T₂ is temperature of cold reservoir
= (889 - 657 ) / 889
= ,261 or 26.1 %
output of work = .261 x 4710 = 1229.15 J .
Efficiency of second engine
= (657 - 406 ) / 657
= .382
Heat rejected in engine one is heat input of second engine
heat input of second engine = 4710 - 1229.15 = 3480.85
output of work of second engine
= .382 x 3480.85 = 1329.68 J
Total work delivered by two engine
= 1229.15 + 1329.68 J
= 2558.83 J
A hockey puck initially travelling to the right at 34 m/s. It moves for 7 before
coming to a stop. How far did it move in 7 seconds?
You can use kinematic equations
Answer:
[tex]x=119m[/tex]
Explanation:
Hello,
In this case, since the hockey puck was moving at 34 m/s and suddenly stopped (final velocity is zero) in 7 seconds, we can first compute the acceleration via:
[tex]a=\frac{v_f-v_o}{t}=\frac{0m/s-34m/s}{7s}\\ \\a=-4.86m/s^2[/tex]
In such a way, we can compute the displacement via:
[tex]x=\frac{v_f^2-v_o^2}{2a}\\ \\x=\frac{0^2-(34m/s)^2}{2*-4.86m/s^2}\\ \\x=119m[/tex]
Best regards.
When a piano tuner strikes both the A above middle C on the piano and a 440 Hz tuning fork, he hears 4 beats each second. The frequency of the piano's:____________.
A) 444 Hz
B) 880 Hz
C) 436 Hz
D) either 436 Hz or 444 Hz
Answer:
D) either 436 Hz or 444 Hz
Explanation:
frequency of the tuning fork, F₁ = 440 Hz
frequency of the piano, F₂ = ?
Beat frequency, F = 4 Hz
Beat frequency is given as the difference between the frequency of the two instruments and it is given by;
F = F₂ - F₁ or F = F₁ - F₂
F₂ = F + F₁ or F - F₁ = - F₂
F₂ = 4 Hz + 440 Hz or 4 - 440 = - F₂
F₂ = 444 Hz or - 436 = - F₂
F₂ = 444 Hz or F₂ = 436 Hz
Therefore, the frequency of the piano is 444 Hz or 436 Hz
ou are walking down a straight path in a park and notice there is another person walking some distance ahead of you. The distance between the two of you remains the same, so you deduce that you are walking at the same speed of 1.05 m/s. Suddenly, you notice a wallet on the ground. You pick it up and realize it belongs to the person in front of you. To catch up, you start running at a speed of 2.75 m/s. It takes you 18.5 s to catch up and deliver the lost wallet. How far ahead of you was this person when you started running
Answer:
The value is [tex]d = 31.45 \ m [/tex]
Explanation:
Generally the relative speed at which you are moving with respect to the person ahead of you is mathematically represented as
[tex]v_r = v_s - v_c[/tex]
substituting 1.05 m/s for [tex] v_c [/tex] and 2.75 m/s for [tex]v_s[/tex]
So
[tex]v_r = 2.75 - 1.05[/tex]
=> [tex]v_r = 1.7 \ m/s [/tex]
Generally the distance by which the person is ahead of you is mathematically represented as
[tex]d = v_r * t[/tex]
substituting 18.5 s for [tex] t [/tex]
[tex]d = 1.7 * 18.5[/tex]
=> [tex]d = 31.45 \ m [/tex]
A bucket is being lowered by a very light rope with a constant downward velocity. The tension in the rope must be
Answer:
The tension in the light rope must be equal to the weight of the bucket
Explanation:
Given that,
Constant velocity of bucket and direction of bucket in downward
We need to find the tension in the rope
Using given data,
When a bucket moves downward with a constant velocity then the net force does not applied on the bucket.
So, The weight of the bucket will be equal to the tension in the light rope
In mathematically,
[tex]T=mg[/tex]
Where, T = tension
m = mass of bucket
g = acceleration due to gravity
Hence, The tension in the light rope must be equal to the weight of the bucket.
Using the equation for Impact, can you explain the following:
Why are car steering rods designed to collapse?
Why are highway guard rails designed to crumple up on impact?
Why are traffic saftey barrels filled with water or sand?
Explanation:
Equation for Impact
FΔt = ΔP,
F = force
Δt = Impact of time
ΔP = Change in momentum
Car steering is engineered to fail in order to maximize the time of contact and hence reduce the initial impact and mitigate the damage incurred.
Road guard railing crumple on contact to maximize impact time and hence reduce impact intensity and mitigate damage.
Road safety containers are loaded with liquid or sand as they improve the period of impact.