Answer:
(a) 1470 N/m
(b) 48.2 m
Explanation:
Applying,
(a) F = ke.................... Equation 1
Where F = force applied to the spring, k = spring constant, e = extension
make k the subject of the equation
k = F/e............... Equation 2
But,
F = mg............. Equation 3
Where m = mass, g = acceleration due to gravity
Substitute equation 3 into equation 2
k = mg/e.............. Equation 4
From the question,
Given: m = 6 kg, e = 42-38 = 4 cm = 0.04 m
Constant: g = 9.8 m/s²
Substitute these values into equation 4
k = (6×9.8)/0.04
k = 1470 N/m
(b) Consider the end of the spring to the left which exert a force to the right
Then,
e = F/k............. Equation 5
Given: F = 150 N, k = 1470 N/m
Substitute these values into equation 5
e = 150/1470
e = 0.102 m
Hence the length of the spring is
L = 0.38+0.102 = 0.482 cm = 48.2 m
A system can only achieve a lower energy state by:_______
Record the lengths of the sides of ABC and ADE.
what is effort arm
don't say the answer of gogle
Answer:
effort arm mean the use of any work by using your hand force motion or by hand power
Calculate the terminal velocity of a rain drop of radius 0.12cm
Explanation:
Given that,
The radius of rain drop, r = 0.12 cm = 0.0012 m
The viscocity of air is, [tex]\eta=18\times 10^{-5}\ poise[/tex]
Let the viscous force is, [tex]F = 0.010173\ N[/tex]
The viscous force is given by :
[tex]F=6\pi \eta rv\\\\v=\dfrac{F}{6\pi \eta r}[/tex]
Put all the values,
[tex]v=\dfrac{0.010173}{6\pi 18\times 10^{-5}\times 0.0012 }\\\\v=2498.58\ m/s[/tex]
How does the magnitude of the normal force exerted by the ramp in the figure compare to the weight of the static block? The normal force is:______ a. greater than the weight of the block. b. possibly greater than or less than the weight of the block, depending on whether or not the ramp surface is smooth. c. equal to the weight of the block. d. possibly greater than or equal to the weight of the block, depending on whether or not the ramp surface is smooth. less than the weight of the block.
Answer:
less than the weight of the block.
Explanation:
From the free body diagram, we get.
The normal force is N = Mg cosθ
The tension in the string is T = Mg sinθ
Wight of the block when the block is static, W = Mg
Now since the magnitude of cosθ is in the range of : 0 < cosθ < 1,
therefore, the normal force is less than the weight of the static block.
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h, and g : Are any of the answers changed if the initial angle is changed?
Complete question is;
A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h and g:
(a) the work done by the force of gravity on the projectile,
(b) the change in kinetic energy of the projectile since it was fired, and
(c) the final kinetic energy of the projectile.
(d) Are any of the answers changed if the initial angle is changed?
Answer:
A) W = mgh
B) ΔKE = mgh
C) K2 = mgh + ½mv_o²
D) No they wouldn't change
Explanation:
We are expressing in terms of m, v0, h, and g. They are;
m is mass
v0 is initial velocity
h is height of projectile fired
g is acceleration due to gravity
A) Now, the formula for workdone by force of gravity on projectile is;
W = F × h
Now, Force(F) can be expressed as mg since it is force of gravity.
Thus; W = mgh
Now, there is no mention of any angles of being fired because we are just told it was fired horizontally.
Therefore, even if the angle is changed, workdone will not change because the equation doesn't depend on the angle.
B) Change in kinetic energy is simply;
ΔKE = K2 - K1
Where K2 is final kinetic energy and K1 is initial kinetic energy.
However, from conservation of energy, we now that change in kinetic energy = change in potential energy.
Thus;
ΔKE = ΔPE
ΔPE = U2 - U1
U2 is final potential energy = mgh
U1 is initial potential energy = mg(0) = 0. 0 was used as h because at initial point no height had been covered.
Thus;
ΔKE = ΔPE = mgh
Again like a above, the change in kinetic energy will not change because the equation doesn't depend on the angle.
C) As seen in B above,
ΔKE = ΔPE
Thus;
½mv² - ½mv_o² = mgh
Where final kinetic energy, K2 = ½mv²
And initial kinetic energy = ½mv_o²
Thus;
K2 = mgh + ½mv_o²
Similar to a and B above, this will not change even if initial angle is changed
D) All of the answers wouldn't change because their equations don't depend on the angle.
A given wave has a wavelength of 1.4 m and a frequency of 2.0 Hz. How fast
is the wave moving?
O A. 0.6 m/s
O B. 3.4 m/s
O C. 0.7 m/s
D. 2.8 m/s
Answer:
The wave speed is 2.8 m.
Explanation:
Wavelength = 1.4 m
frequency, f = 2 Hz
the wave speed is given by
wave speed = wavelength x frequency
wave speed = 1.4 x 2 = 2.8 m
option (D) is correct.
help plzzzzzzzzzzzz ?
Explanation:
1. First, let's find the total resistance of the circuit. We begin by combining [tex]R_{4}[/tex], [tex]R_{5}[/tex] and [tex]R_{6}[/tex]:
[tex]R_{456}=R_{4} + \dfrac{R_{5}R_{6}}{R_{5} + R_{6}}[/tex]
[tex]= 6\:Ω + \dfrac{(3\:Ω)(5\:Ω)}{3\:Ω+5\:Ω} = 7.9\:Ω[/tex]
Now time to combine [tex]R_{2}[/tex] and [tex]R_{3}[/tex] and they are connected in series so
[tex]R_{23} =R_{2} + R_{3} = 17\:Ω[/tex]
Note that [tex]R_{23}[/tex] and [tex]R_{456}[/tex] are connected in parallel so
[tex]R_{23456} = \dfrac{R_{23}R_{456}}{R_{23}+R_{456}}=5.4\:Ω[/tex]
Finally, [tex]R_{23456}[/tex] is connected in series with [tex]R_{1}[/tex] so the total resistance [tex]R_{T}[/tex] is
[tex]R_{T} = R_{1} + R_{23456} = 10\:Ω + 5.4\:Ω = 15.4\:Ω[/tex]
2. The total current in the circuit is
[tex]I_{T} = \dfrac{V}{R_{T}} = \dfrac{20\:V}{15.4\:Ω} = 1.3\:A[/tex]
3. The voltage drop across [tex]R_{1},\:V_{1}[/tex] is
[tex]V_{1} = I_{T}R_{1} = (1.3\:A)(10\:Ω) = 13\:V[/tex]
4. We can see that [tex]I_{T} = I_{1} + I_{2}[/tex]. To solve for [tex]I_{1}[/tex], we need [tex]V_{23}[/tex], which is just [tex]V_{T} - V_{1} = 20\:V - 13\:V = 7\:V[/tex] , which gives us
[tex]I_{1} = \dfrac{V_{23}}{R_{23}} = \dfrac{7\:V}{17\:Ω} = 0.4\:A[/tex]
5. From #2 & #4, [tex]I_{2} = 1.3\:A - 0.4\:A = 0.9\:A[/tex] and we also know that the voltage drop across [tex]R_{456}[/tex] is 7 V, the same as that of [tex]R_{23}[/tex]. The voltage drop across [tex]R_{4}[/tex] is
[tex]V_{4} = I_{2}R_{4} =(0.9\:A)(6\:Ω) = 5.4\:V[/tex]
This means that the voltage drop across [tex]R_{6}[/tex] is 7 V - 5.4 V = 1.6 V. Knowing this, the current through [tex]R_{6}[/tex] is
[tex]I_{6} = \dfrac{1.6\:V}{5\:Ω} = 0.3\:A[/tex]
A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 33 N. Starting from rest, the sled attains a speed of 1.6 m/s in 9.8 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.
Answer:
[tex]\mu=0.185[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=17kg[/tex]
Force [tex]F=33N[/tex]
Velocity [tex]v=1.6m/s[/tex]
Distance [tex]d= 9.8m[/tex]
Generally the equation for Work done is mathematically given by
[tex]W=\triangle K.E+\triangle P.E[/tex]
Where
[tex]\triangle K.E=(F-F_f)*2[/tex]
[tex]F_f=F+\frac{\triangle K.E}{d}[/tex]
[tex]F_f=33+\frac{0.5*17*1.6^2}{9.8}[/tex]
[tex]F_f=30.8N[/tex]
Since
[tex]f = \mu*m*g[/tex]
[tex]\mu= 30.8/(m*g)[/tex]
[tex]\mu= 30.8/(17*9.81)[/tex]
[tex]\mu=0.185[/tex]
A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 m/s.
Required:
a. Compute the magnitude and direction of the velocity of the stone after it is struck.
b. Is the collision perfectly elastic?
Answer:
(a)Magnitude=28.81 m/s
Direction=33.3 degree below the horizontal
(b) No, it is not perfectly elastic collision
Explanation:
We are given that
Mass of stone, M=0.150 kg
Mass of bullet, m=9.50 g=[tex]9.50\times 10^{3} kg[/tex]
Initial speed of bullet, u=380 m/s
Initial speed of stone, U=0
Final speed of bullet, v=250m/s
a. We have to find the magnitude and direction of the velocity of the stone after it is struck.
Using conservation of momentum
[tex]mu+ MU=mv+ MV[/tex]
Substitute the values
[tex]9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V[/tex]
[tex]3.61i=2.375j+0.150V[/tex]
[tex]3.61 i-2.375j=0.150V[/tex]
[tex]V=\frac{1}{0.150}(3.61 i-2.375j)[/tex]
[tex]V=24.07i-15.83j[/tex]
Magnitude of velocity of stone
=[tex]\sqrt{(24.07)^2+(-15.83)^2}[/tex]
|V|=28.81 m/s
Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s
Direction
[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
=[tex]tan^{-1}(\frac{-15.83}{24.07})[/tex]
[tex]\theta=tan^{-1}(-0.657)[/tex]
=33.3 degree below the horizontal
(b)
Initial kinetic energy
[tex]K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2[/tex]
[tex]K_i=685.9 J[/tex]
Final kinetic energy
[tex]K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2[/tex]
=[tex]\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2[/tex]
[tex]K_f=359.12 J[/tex]
Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.
A boy observes a fish at a depth of 4 metres under the surface of a lake. If the refractive index of the water is 4/3. what is the apparent depth of the fish as seen by the boy? Apparent depth Real depth
Answer:
[tex]{ \tt{n _{w} = \frac{real \: depth}{aparent \: depth} }} \\ \\ \frac{4}{3} = \frac{4}{d} \\ { \tt{d = 3 \: meters}}[/tex]
A boy observes a fish at the depth of 4 meters under the water's surface. The refractive index of the water is 4/3 then the apparent depth of the fish as seen by the boy is 3 meters.
What is the Refractive index?Refractive index is also known as the refraction, is a measure of the way a light beam bends when it goes through different media. The refractive index n is computed by dividing the sine of the angle of incidence by the sine of the angle of refraction, i.e., n = sin i/sin r, if I will be the angle of incidence of a light source in a void (angle between the inbound ray and the normal, or perpendicular to the surface of a medium).
The Index of refraction is also equal to c/v, or the ratio of a wavelength of light's velocity in a substance to its velocity in a void.
As per the given data in the question,
n(w) = real depth/apparent depth
4/3 = 4/d
d = 3 meters.
Therefore, the apparent depth will be equal to 3 meters.
To know more about the Refractive index:
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HELP ME PLS!!!!
Find the location of beryllium (Be) on the periodic table. What type of ion will
beryllium form?
A. An ion with a -2 charge
B. An ion with a +6 charge
C. An ion with a +2 charge
D. An ion with a -6 charge
Answer:
the answer is c which is a+2 charge
Explanation:
Beryllium is in group 2A. It's nearest noble gas is Helium, which is 2 elements behind Beryllium. ThBeryllium wants to lose two electrons. When it does that, Beryllium will have a positive chargeof two, and it will be stated as B-e two plus.
The Beryllium (Be) has an atomic number of 4 and belongs to Group-2 elements. The Beryllium will form a divalent cation (+2). Thus, option C is correct.
What are cations and anions?In an atom, the number of electrons equals the number of protons. If the electrons are removed from the atom or the electrons are added to the atom, the atom has an excessive positive or negative charge.
This excessive of electrons or lack of electrons forms Ions. The excess of electrons has a negative charge or anions and the lack of electrons has a positive charge or cations.
Beryllium has 4 electrons. Two electrons are occupied in the valence shell of beryllium. Group 2 elements always form the positive ions or cations, to become stable ions.
The outermost shell of beryllium has two electrons. In order to form a stable ion, beryllium should lose its two electrons or gain six electrons. Beryllium belongs to the Group-2 element, it always loses two electrons and forms Be²⁺, to form a stable ion.
Hence, Beryllium forms an ion with a +2 charge. Thus, the correct option is C.
To learn more about the Cations and Anions:
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Malar is playing with a toy car and track set. She has made a hill to let the car roll down, followed by a series of small hills and then a short flat section on the floor. The car starts at the top of the hill which is 137 cm above the short flat section. How much energy was lost due to friction (between the car and the sides of the track, and the car’s axels) during the entire trip? You don’t have any details about the time while the car is going down the hill or through the loop, so you don’t know how fast it is going at the top of the loop.
Mass is 150 g
Answer:
E = 2.01 J
Explanation:
The energy lost by the car due to friction through the entire trip must be equal to the potential energy of the car at the starting point.
Energy Lost = E = Potential Energy
E = mgh
where,
m = mass of car = 150 g = 0.15 kg
g = acceleration due to gravity = 9.81 m/s²
h = height of the track = 137 cm = 1.37 m
Therefore,
E = (0.15 kg)(9.81 m/s²)(1.37 m)
E = 2.01 J
You are a member of an alpine rescue team and must get a box of supplies, with mass 2.50 kg, up an incline of constant slope angle 30.0° so that it reaches a stranded skier who is a vertical distance 3.50 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 6.00x102. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s
Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier. Express your answer numerically, in meters per second.
1. How to approach the problem
2. Find the total work done on the box
3. Initial kinetic energy
4. What is the final kinetic energy?
Answer:
v₀ = 2.67 m / s
Explanation:
This problem can be solved using the Kinetic Enemy Work Theorem
W = ΔK
Work is defined by the relation
W = fr. d
The bold letters indicate vectors, in this case the blow is in the direction of the slope of the ramp and the displacement is also in the direction of the ramp, therefore the angle between the force and the displacement is zero.
the friction force opposes the displacement therefore its angle is 180º
W = - fr d
Let's use Newton's second law, we define a reference frame with the horizontal axis parallel to the plane
Y axis
N- Wy = 0
N - W cos tea = 0
the friction force has the expression
fr = μ N
fr = μ W cos θ
we substitute
W = - μ W cos θ d
let's look for kinetic energy
the minimum velocity at the highest point is zero
K_f = 0
the initial kinetic energy is
K₀ = ½ m v₀²
we substitute energy in the work relationship
- μ W cos θ d = 0 - ½ m v₀²
v₀² = - μ W cos θ 2d / m
Let's use trigonometry to find distance d
sin θ= y / d
d = y /sin θ
d = 3.50 / sin 30
d = 7 m
let's calculate
v₀² = (6 10⁻² 2.50 9.8 cos 30) 2 7 / 2.50
v₀ = √7.129
v₀ = 2.67 m / s
What is the resistance of a bulb of 4ow
connected in a line of 220v?
2
Answer:
1210 ohm
Explanation:
Given :
P=40 W
V=220 V
Now,
[tex]P=\frac{V^{2} }{R} \\40=\frac{(220)^{2} }{R} \\40R=48400\\R=\frac{48400}{40} \\R=1210 ohm[/tex]
Therefore, resistance of bulb will be 1210 ohm
which statement summarized the difference between mass and weight?
Answer:
The second statement.
A car accelerates from 0-30m/s in 5 s. It has a mass of 1200kg. What force does the engine produce? *
A) 36000N
B) 7200N
C) 60000N
D) 2000N
include explanation please
Answer:
a = ?
u = 0
v = 30
by using v = u + at equation we can find " a "
30 = 0 + 5a
6 m/s = a
by using f = ma equation we can find force produce by engine ,
f = ?
a = 6
m = 1200
f = 1200 × 6
f = 7200 N
so the answer is "B"
Matthew throws a ball straight up into the air. It rises for a period of time and then begins to drop. At which points in the ball’s journey will gravity be the greatest force acting on the ball?
Answer:
If air resistance is taken as negligible, then the ball is in freefall the moment it is thrown so gravity is the only force acting on the object. If air resistance is not negligible then gravity will be the greatest force acting on the ball while it is going up and coming down, because Fair has to be less than gravity at all times otherwise the atmosphere would wither away.
Question 18/55 (2 p.)
A vibrating object produces ripples on the surface of a liquid. The object completes 20 vibrations
every second. The spacing of the ripples, from one crest to the next, is 3.0 cm.
What is the speed of the ripples?
D
C 60 cm/s
120 cm/s
A 0.15cm/s
B 6.7 cm/s
Answer:
the correct answer is C v = 60 cm / s
Explanation:
The speed of a wave is related to the frequency and the wavelength
v = λ f
They indicate that the object performs 20 oscillations every second, this is the frequency
f = 20 Hz
the wavelength is the distance until the wave repeats, the distance between two consecutive peaks corresponds to the wavelength
λ = 3 cm = 0.03 m
let's calculate
v = 20 0.03
v = 0.6 m / s
v = 60 cm / s
the correct answer is C
How many atoms of carbon, C, are in 0.020 g of carbon?
Answer:
9.6352× 10²⁰ C atoms
Explanation:
From the given information,
The molar mass of Carbon = 12 g/mol
number of moles = 0.020g/ 12 g/mol
number of moles = 0.0016 mol
If 1 mole of C = 6.022 × 10²³ C atoms
∴
0.0016 mol of C = (6.022 × 10²³ C atoms/ 1 mol of C)×0.0016 mol of C
= 9.6352× 10²⁰ C atoms
Hence, the number of carbon atoms present in 0.020 g of carbon = 9.6352× 10²⁰ C atoms
A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. What will the horizontal speed be right before it hits the ground?
A. 15 m/s
B. 0 m/s
C. 30 m/s
D. 40 m/s
Answer:
C
Explanation:
horizintal speed stays same
only vertical speed changes
so H speed will stay 30 m/s
A 16 kg science book is dropped of a 120 meter high cliff. Assuming a closed system:
a) how fast is book traveling the instant before it impacts the ground below the cliff?
b) how far above the bottom of the cliff is the object moving at 12 m/s?
Answer:
Explanation:
The mass of that science book...wow. In pounds that would be 35.2! Yikes!
Anyway, we need final velocity here, and the mass of the book has nothing to do with how fast it falls. Everything is pulled by the same gravity. A feather falls at 9.8 m/s/s and so does an elephant. Mass is useless information. The equation we will use is
[tex]v^2=v_0^2+2a[/tex]Δx where
v is the final velocity, our unknown,
v₀ is the initial velocity which is 0 since someone had to be holding the book before dropping it,
a is the pull of gravity which is always -9.8 m/s/s, and
Δx = -120 which is the displacement (it's negative because the book falls below the point from which it was dropped). Filling in:
[tex]v^2=0^2+2(-9.8)(-120)[/tex] so
[tex]v=\sqrt{2(-9.8)(-120)}[/tex] and
v = 48 m/s
As far as how far above the bottom of the cliff the object is when it is moving at 12 m/s we will use the same equation, but the velocity will be 12:
[tex]12^2=0^2+2(-9.8)[/tex]Δx and
144 = -19.6Δx so
Δx = -7.3 m. That's how far from the top of the cliff it is. We subtract then t find out how far it is from the bottom:
120 - 7.3 = 112.7 m off the ground.
A baby leaves a bowl of food on the floor and crawls westwards to fetch a toy placed 5 m away.At the same time a dog walks eastwards towards the baby. it takes the baby 30 s to reach the toy. The dog walks past the toy to eat the baby's food in the bowl
Determine the position of the dog relative to the baby before they both moved?
They create a heat engine where the hot reservoir is filled with water and steam at equilibrium, and the cold reservoir is filled with ice and water at equilibrium. What is the Carnot efficiency for their heat engine if the pressure is constant at 1.0 atmospheres?
Answer:
The efficiency of Carnot's heat engine is 26.8 %.
Explanation:
Temperature of hot reservoir, TH = 100 degree C = 373 K
temperature of cold reservoir, Tc = 0 degree C = 273 K
The efficiency of Carnot's heat engine is
[tex]\eta = 1-\frac{Tc}{T_H}\\\\\eta = 1 -\frac{273}{373}\\\\\eta = 0.268 =26.8 %[/tex]
The efficiency of Carnot's heat engine is 26.8 %.
A student wishes to construct a mass-spring system that will oscillate with the same frequency as a swinging pendulum with a period of 3.45 S. The student has a spring with a spring constant of 72.0 N/m. What mass should the student use to construct the mass-spring system?
Answer:
21.73 kg
Explanation:
Applying,
T = 2π√(m/k)............... Equation 1
Where T = period, m = mass on the spring, k = spring constant, π = pie.
make m the subject of the equation
m = T²k/4π²................. Equation 2
From the question,
Given: T = 3.45 s, k = 72.0 N/m, π = 3.14
Substitute these values into equation 2
m = (3.45²×72)/(4×3.14²)
m = 21.73 kg.
Hence the mass should be 21.73 kg
What is the answer can you explain it to me
Answer:
C) 300 Ohm.
Explanation:
In a series circuit, total resistance is just adding all the resistance together. So R (total) = 75 +75 +75+ 75 = 300 ohms
Parallel circuit are different because you add the inverses of resistance and you flip the final answer.
You can confirm your answers using the tools below:
https://www.omnicalculator.com/physics/series-resistor
https://www.omnicalculator.com/physics/parallel-resistor
if an object is moving with a velocity of 24 and has an acceleration of -4 square root how long will it take it to stop
Answer:
Time, t = 8 seconds.
Explanation:
Given the following data;
Initial velocity = 24 m/s
Final velocity = 0 m/s (since the object is coming to a stop).
Acceleration = -4 m/s²
To find how long it will take the object to stop, we would use the first equation of motion;
[tex] V = U + at[/tex]
Where;
V is the final velocity. U is the initial velocity. a is the acceleration. t is the time measured in seconds.Making time, t the subject of formula, we have;
[tex] t = \frac{V - U}{a}[/tex]
Substituting into the equation, we have;
[tex] t = \frac{0 - 24}{-4}[/tex]
[tex] t = \frac{-24}{-4}[/tex]
Time, t = 8 seconds.
Một ô tô khối lượng một tấn chuyển động trên một đường nằm ngang. Hệ số ma sát giữa bánh ô tô và mặt đường là 0,07. Gia tốc trọng trường g=9,8m/s2
a) vẽ và xác định tên các lực tác động lên vật. Viết phương trình chuyển động của vật.
b) nếu ô tô chuyển động đều, xuống dốc có độ dốc 5%. Tính lực kéo của ô tô.
c) nếu ô tô chuyển động đều. Lên dốc có độ dốc 5%. Tính lực kéo của động cơ ô tô
You decide to impress Grandpa by showing him how fast sound travels. You have a piece of plastic pipe with an adjustable closed end, and a 312 Hz tuning fork. The piece of pipe resonates in the 2nd resonant length when it is adjusted to a length of 81.0 cm. What is the speed of sound on that day?
Answer:
336.96m/s
Explanation:
answer is in photo above
The speed of sound on that day is 336.96 m/s.
What is speed?Speed is distance travelled by the object per unit time. Due to having no direction and only having magnitude, speed is a scalar quantity With SI unit meter/second.
given that:
frequency of the sound = 312 Hz.
2nd resonant length = 81.0 cm.
If the wavelength of sound is λ; the 2nd resonant length of plastic pipe with an adjustable closed end = 3λ/4
Hence, wavelength of sound is = 81×(4/3) cm = 0.81 ×(4/3) m.
So, the speed of sound on that day is = frequency × wavelength
= 312 Hz × 0.81 ×(4/3) m.
= 336.96 m/s.
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A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.2 m/s2. A green car arrives at the position of the stop-light 7.5 s after the light had turned green. If t = 0 when the light turns green, at what time does the green car catch the blue car if the green car maintains the slowest constant speed necessary to catch up to the blue car?
Answer:
After 15 seconds, the green car will catch up with the blue car
Explanation:
Let the time for the green car to catch up with the blue car be T
When the green car catches up to the blue car, the distances covered by each car after time T will be equal. Also, their velocities at that instant will be equal
Distance covered by blue car after time T is given by: s = ut + 0.5 at²
Where u = 0, a = 0.2 m/s², t = T
S = 0.5 × 0.2 × T² = 0.1 T²
Velocity of blue car, v = u+ at
v = 0.2T
Distance covered by green car at T is given as: S = Velocity × time
Where v = 0.2T, t = T - 7.5 (since the blue car started 7.5 seconds earlier)
S = 0.2T (T - 7.5)
S = 0.2 T² - 1.5T
Equating the distance covered by the two cars
0.2T² - 1.5T = 0.1T²
0.1T² - 1.5T = 0
T(0.1T - 1.5) = 0
T = 0 or
T = 1.5/0.1 = 15 secs
Therefore, after 15 seconds, the green car will catch up with the blue car