Answer:
0.372 kg
Explanation:
The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:
[tex]mu=(M+m)v[/tex] (1)
where
[tex]m=11.9 g = 11.9\cdot 10^{-3}kg[/tex] is the mass of the bullet
[tex]u=261 m/s[/tex] is the initial velocity of the bullet
[tex]M[/tex] is the mass of the block
[tex]v[/tex] is the velocity at which the bullet and the block travels after the collision
We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write
[tex]\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2[/tex] (2)
where
k = 205 N/m is the spring constant
x = 35.0 cm = 0.35 m is the compression of the spring
From eq(1) we get
[tex]v=\frac{mu}{M+m}[/tex]
And substituting into eq(2), we can solve to find the mass of the block:
[tex](M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg[/tex]
The mass of the wooden block is approximately 0.404 kg.
The initial kinetic energy of the bullet is given by:
KE_initial = 0.5 * mb * vb^2
where:
`m_b` is the mass of the bullet (11.9 g = 0.0119 kg)
`v_b` is the velocity of the bullet (261 m/s)
The potential energy stored in the compressed spring is given by:
PE_spring = 0.5 * k * x^2
where:
`k` is the spring constant (205 N/m)
`x` is the compression of the spring (0.35 m)
Since the system comes to a stop, the initial kinetic energy is converted to potential energy stored in the spring:
KE_initial = PE_spring
Solving for the mass of the wooden block, we get:
m_w = (2 * KE_initial) / (v_b^2 - (k * x) / m_b)
Plugging in the values, we get,
m_w = (2 * 0.5 * 0.0119 kg * (261 m/s)^2) / ((261 m/s)^2 - (205 N/m * 0.35 m) / 0.0119 kg)
m_w ≈ 0.404 kg
Therefore, the mass of the wooden block is approximately 0.404 kg.
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Suppose that the coefficient of kinetic friction between Zak's feet and the floor, while wearing socks, is 0.250. Knowing this, Zak decides to get a running start and then slide across the floor.
a) If Zak's speed is 3.00 when he starts to slide, what distance will he slide before stopping? d=1.84
b) Now, suppose that Zak's younger cousin, Greta, sees him sliding and takes off her shoes so that she can slide as well (assume her socks have the same coefficient of kinetic friction as Zak's). Instead of getting a running start, she asks Zak to give her a push. So, Zak pushes her with a force of 125 over a distance of 1.00 . If her mass is 20.0 , what distance does she slide after Zak's push ends? Remember that the frictional force acts on Greta during Zak's push and while she is sliding after the push.
a) 1.84 m
b) 1.55 m
Explanation:
a)
In this problem, the only force acting on Zak along the direction of motion (horizontal direction) is the force of friction, which is
[tex]F_f=-\mu mg[/tex]
where
[tex]\mu=0.250[/tex] is the coefficient of friction
m is Zak's mass
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
According to Newton's second law of motion, the net force acting on Zak is equal to the product between its mass (m) and its acceleration (a), so we have
[tex]F=ma[/tex]
Here the only force acting is the force of friction, so this is also the net force:
[tex]-\mu mg = ma[/tex]
Therefore we can find Zak's acceleration:
[tex]a=-\mu g=-(0.250)(9.8)=-2.45 m/s^2[/tex]
Since Zak's motion is a uniformly accelerated motion, we can now use the following suvat equation:
[tex]v^2-u^2=2as[/tex]
where
v = 0 is the final velocity (he comes to a stop)
u = 3.00 m/s is the initial velocity
[tex]a=-2.45 m/s^2[/tex] is the acceleration
s is the distance covered before stopping
Solving for s,
[tex]s=\frac{v^2-u^2}{2a}=\frac{0^2-3.0^2}{2(-2.45)}=1.84 m[/tex]
b)
In this second part, Zak gives a push to Greta.
We can find Greta's velocity after the push by using the work-energy theorem, which states that the work done on her is equal to her change in kinetic energy:
[tex](F-F_f)d =\frac{1}{2}mv^2-\frac{1}{2}mu^2[/tex]
where
F = 125 N is the force applied by Zak
d = 1.00 m is the distance
[tex]F_f=\mu mg[/tex] is the force of friction, where
[tex]\mu=0.250[/tex]
m = 20.0 kg is Greta's mass
[tex]g=9.8 m/s^2[/tex]
v is Greta's velocity after the push
u = 0 is Greta's initial velocity
Solving for v, we find:
[tex]v=\sqrt{\frac{2(F-\mu mg)d}{m}}=\sqrt{\frac{2(125-(0.250)(20.0)(9.8))(1.00)}{20.0}}=2.76 m/s[/tex]
After that, Zak stops pushing, so Greta will slide and the only force acting on her will be the force of friction; so the acceleration will be:
[tex]a=-\mu g = -(0.250)(9.8)=-2.45 m/s^2[/tex]
And so using again the suvat equation, we can find the distance she slides after Zak's push ends:
[tex]s=\frac{v'^2-v^2}{2a}[/tex]
where
v = 2.76 m/s is her initial velocity
v' = 0 when she stops
Solving for s,
[tex]s=\frac{0-(2.76)^2}{2(-2.45)}=1.55 m[/tex]
(a) The distance traveled by Zack before stopping is 1.84 m.
(b) The distance traveled by Greta after Zack's push ends is 1.56 m.
The given parameters;
coefficient of kinetic friction, [tex]\mu_k[/tex] = 0.25initial speed of Zack, u = 3 m/sThe distance traveled by Zack before stopping is calculated as follows;
The acceleration of Zack;
[tex]-F_k = ma\\\\-\mu_k mg = ma\\\\-\mu_k g = a\\\\-(0.25 \times 9.8) = a\\\\- 2.45 \ m/s^2 = a[/tex]
The distance traveled by Zack;
[tex]v^2 = u^2 + 2as\\\\when \ Zack \ stops \ v = 0\\\\0 = u^2 + 2as\\\\0 = (3)^2 +2(-2.45)s\\\\0 = 9 - 4.9s\\\\4.9s = 9\\\\\s = \frac{9}{4.9} \\\\s = 1.84 \ m[/tex]
The distance traveled by Greta is calculated as follows;
Apply law of conservation of energy to determine the velocity of Greta after the push.
[tex]Fd - F_kd = \frac{1}{2} mv^2\\\\125\times 1 - (0.25 \times 20 \times 9.8 \times 1) = (0.5 \times 20)v^2\\\\76 = 10v^2\\\\v^2 = \frac{76}{10} \\\\v ^2 = 7.6\\\\v = \sqrt{7.6} \\\\v = 2.76 \ m/s[/tex]
The acceleration of Greta;
[tex]a = -\mu_k g\\\\a = 0.25 \times 9.8\\\\a = -2.45 \ m/s^2[/tex]
The distance traveled by Greta;
[tex]v_f^2 = v^2 + 2as\\\\when \ Greta \ stops \ v_f = 0\\\\0 = v^2 + 2as\\\\-2as = v^2\\\\-2(-2.45)s = (2.76)^2\\\\4.9s = 7.62\\\\s = \frac{7.62}{4.9} \\\\s = 1.56 \ m[/tex]
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An electron has a velocity of 1.50 km/s (in the positive x direction) and an acceleration of 2.00 ✕ 1012 m/s2 (in the positive z direction) in uniform electric and magnetic fields. If the electric field has a magnitude of strength of 18.0 N/C (in the positive z direction), determine the following components of the magnetic field. If a component cannot be determined, enter 'undetermined'.
Answer:
see explanation
Explanation:
Given that,
velocity of 1.50 km/s = 1.50 × 10³m/s
acceleration of 2.00 ✕ 1012 m/s2
electric field has a magnitude of strength of 18.0 N/C
[tex]\bar F= q[\bar E + \bar V \times \bar B]\\\\\bar F = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )]\\\\\\m \bar a = [\bar E + \bar V \times ( B_x \hat i +B_y \hat j +B_z \hat z )][/tex]
[tex]9.1 \times 10^-^3^1 \times 2\times 10^1^2 \hat k=-1.6\times10^-^1^9 \hat k [18\hat k+ 1.5\times 10^3 \hat i \times (B_x \hat i +B_y \hat j +B_z \hat k)][/tex][tex]42.2 \times 10^-^1^9 \hat k = -2.4 \times 10^1^6B_y \hat k + 2.4 \times 10 ^1^6 \hat j B_z\\[/tex]
[tex]B_x = undetermined[/tex]
[tex]B_y = \frac{42.2 \times 10^-^1^9}{-2.4 \times 10^-^1^6} \\\\= - 0.0176 T[/tex]
[tex]B_z = 0T[/tex]
A police officer's radar gun emits radio waves at a frequency of 10.7 GHz10.7 GHz . What is the difference in the frequencies reflected back to the radar gun by a car traveling at the 75 mph75 mph speed limit and a car travelling 9 mph9 mph faster? Assume that the cars are both headed directly away from the stationary officer.
Answer:
Explanation:
A police car's radar gun emits microwaves with frequency f. 1. =36 GHz. The beam reflects from a car that speeds away from the cruiser with 43 m/s. The receiver ...
Moving from boron to carbon, the intensity of the bulb Moving from boron to carbon, the intensity of the bulb blank because Z increases from blank to blank. The thickness of the frosting blank because the core electron configuration is the same for both atoms. because Z increases from Moving from boron to carbon, the intensity of the bulb blank because Z increases from blank to blank. The thickness of the frosting blank because the core electron configuration is the same for both atoms. to Moving from boron to carbon, the intensity of the bulb blank because Z increases from blank to blank. The thickness of the frosting blank because the core electron configuration is the same for both atoms.. The thickness of the frosting Moving from boron to carbon, the intensity of the bulb blank because Z increases from blank to blank. The thickness of the frosting blank because the core electron configuration is the same for both atoms. because the core electron configuration is the same for both atoms.
Complete Question
The complete Question is shown on the first and second uploaded image
Answer:
The underlined words are the answers
Part A
Moving from boron to carbon, the intensity of the bulb Increases because Z increases from 5 to 6 , The thickness of the frosting stays the same because the core electron configuration is the same for both atoms
Part B
Moving from boron to aluminum the intensity of the bulb Increases because Z increases from 5 to 13 . The thickness of the frosting also increases because Al has the core configuration of Ne, while B has the core configuration of He
Explanation:
Here Z denotes the atomic number
Ne denoted the element called Neon and its electronic configuration is
[tex]1s^2 \ 2s^2 \ 2p^6[/tex]
He denoted the element called Helium and its electronic configuration is
[tex]1s^2[/tex]
B denoted the element called Boron and its electronic configuration is
[tex]1s^2 \ 2s^2\ 2p^1[/tex]
Looking at its electronic configuration we can see that the core is He
I,e [tex][He]\ 2s^2 2p^1[/tex]
Al denoted the element called Aluminium and its electronic configuration is
[tex]1s^2 \ 2s^2 \ 2p^2 \ 3s^2 \ 3p^1[/tex]
Looking at its electronic configuration we can see that the core is Ne
I,e [tex][Ne]\ 3s^2 \ 3p^1[/tex]
A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of 0.390.
How far along the rough ice does she go before stopping?
Answer:
18.8m
Explanation:
Let the distance traveled before stopping be 'd' m.
Given:
Mass of the skater (m) = 90 kg
Initial velocity of the skater (u) = 12.0 m/s
Final velocity of the skater (v) = 0 m/s (Stops finally)
Coefficient of kinetic friction (μ) = 0.390
Acceleration due to gravity (g) = 9.8 m/s²
Now, we know from work-energy theorem, that the work done by the net force on a body is equal to the change in its kinetic energy.
Hence, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.
Frictional force is given as:
[tex]f=\mu N[/tex]
Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg
[tex]f=\mu mg\\=0.390\times 90\times 9.8\\=343.98\ N[/tex]
work done by friction is a negative work as friction and displacement are in opposite direction and is given as
[tex]W=-fd=-343.98d[/tex]
Now, change in kinetic energy is given as:
[tex]\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J[/tex]
Therefore, from work-energy theorem,
[tex]W=\Delta K\\\\-343.98d=6480\\\\d=\frac{6480}{343.98}\\\\d=18.8m[/tex]
Hence, the skater covers a distance of 18.8 m before stopping.
Answer:
18.84m
Explanation:
We are given;
Mass; m = 90 Kg
Initial velocity; u = 12 m/s
Coefficient of friction; μ = 0.390
Let,the combined mass of the ice skater and the skate be M
Thus, So,if the coefficient of frictional force is μ,then frictional force acting is; μN. N= Mg. Thus F_friction = μMg
Now, the deceleration due to friction will be, F_friction/M
Thus, deceleration = μMg/M
M will cancel out and we have; μg
Now, deceleration means negative acceleration. Thus acceleration (a) =
-μg
Now, to find the distance, let's use equation of motion which is;
v² = u² + 2as
Putting -μg for a, we have;
v² = u² + 2(-μg)s
v² = u² — 2μgs
We want to know the distance covered before coming to rest. Thus, v = 0m/s
So,
0² = 12² - (2 x 0.39 x 9.8 x s)
0 = 144 - 7.644s
Thus,
7.644s = 144
Thus, s = 144/7.644 = 18.84m
If the opening to the harbor acts just like a single-slit which diffracts the ocean waves entering it, what is the largest angle, in degrees relative to the incident direction, that a boat in the harbor would be protected from the wave action
Answer:
The angle that the wave would be [tex]\theta = sin ^{-1}\frac{2 \lambda}{D}[/tex]
Explanation:
From the question we are told that the opening to the harbor acts just like a single-slit so a boat in the harbor that at angle equal to the second diffraction minimum would be safe and the on at angle greater than the diffraction first minimum would be slightly affected
The minimum is as a result of destructive interference
And for single-slit this is mathematically represented as
[tex]D sin \ \theta =m \lambda[/tex]
where D is the slit with
[tex]\theta[/tex] is the angle relative to the original direction of the wave
m is the order of the minimum j
[tex]\lambda[/tex] is the wavelength
Now since in the question we are told to obtain the largest angle at which the boat would be safe
And the both is safe at the angle equal to the second minimum then
The the angle is evaluated as
[tex]\theta = sin ^{-1}[\frac{m\lambda}{D} ][/tex]
Since for second minimum m= 2
The equation becomes
[tex]\theta = \frac{2 \lambda}{D}[/tex]
A negative charge of - 0.0005 C exerts an attractive force of 19.0 N on a second charge that is 25 m away. What is the magnitude of the second charge?
Answer:
0.002638 C or 2.6388*10^-3 C
Explanation:
Given that
Quantity of the charge, q = -0.0005 C
Force on the charge of magnitude, F1 = 19 N
Distance from the second charge, r = 25 m
Magnitude of force of the second charge, q2 = ? N
F = (kq1q2) / r², where
k = 9*10^9
19 = (9*10^9 * 0.0005 * q2) / 25²
19 * 625 = 4.5*10^6 * q2
q2 = 11875 / 4.5*10^6
q2 = 0.002638 C or 2.6388*10^-3 C
Thus, the magnitude of the second charge is 0.002638 C or 2.6388*10^-3 C
A battery with emf e m f is connected in series with an inductance L and a resistance R. (a) Assuming the current has reached steady state when it is at 99% of its maximum value, how long does it take to reach steady state, assuming the initial current is zero? (Use any variable or symbol stated above as necessary. To represent e m f, use E.) t99% = (b) If an emergency power circuit needs to reach steady state within 1.2 ms of turning on and the circuit has a total resistance of 72 Ω, what values of the total inductance of the circuit are needed to satisfy the requirement? (Give the maximum value.) H
Answer:
a) [tex]t = 4.6\tau[/tex]
b) [tex]L = 0.0187 \: H[/tex]
Explanation:
The current flowing in a R-L series circuit is given by
[tex]I = I_{0} (1 - e^{\frac{-t}{\tau} })[/tex]
Where τ is the time constant and is given by
[tex]\tau = \frac{L}{R}[/tex]
Where L is the inductance and R is the resistance
Assuming the current has reached steady state when it is at 99% of its maximum value,
[tex]0.99I_{0} = I_{0} (1 - e^{\frac{-t}{\tau} })\\0.99 = (1 - e^{\frac{-t}{\tau} })\\1 - 0.99 = e^{\frac{-t}{\tau}}\\ln(0.01) = ln(e^{\frac{-t}{\tau}})\\-4.6 = \frac{-t}{\tau}\\t = 4.6\tau\\[/tex]
Therefore, it would take t = 4.6τ to reach the steady state.
(b) If an emergency power circuit needs to reach steady state within 1.2 ms of turning on and the circuit has a total resistance of 72 Ω, what values of the total inductance of the circuit are needed to satisfy the requirement?
[tex]t = 4.6\frac{L}{R}\\t = 4.6\frac{L}{R}\\0.0012 = 4.6\frac{L}{72}\\0.0864 = 4.6 L\\L = 0.0864/4.6\\L = 0.0187 \: H[/tex]
Therefore, an inductance of 0.0187 H is needed to satisfy the requirement.
A torpedo is to be designed to be 3 m long with a diameter of 0.5 m. Treating the torpedo as a cylinder, it is to be made to have a velocity of 10 m/s in sea water. Sea water has a dynamic viscosity of 0.00097 Ns/m and a density of 1023 kg/m3 . A 1:15 scale model is going to be tested in air. What velocity will be needed for the model and prototype to be similar
The velocity that will be needed for the model and prototype to be similar is 108.97m/s
Explanation:
length of Torpedo = 3m
diameter, [tex]d_{1} = 0.5m[/tex]
velocity of sea water, [tex]v_{1}[/tex]= 10m/s
dynamic viscosity of sea water, η[tex]_{1}[/tex] = 0.00097 Ns/m²
density of sea water, ρ[tex]_{1}[/tex] = 1023 kg/m³
Scale model = 1:15
[tex]\frac{d_{1} }{d_{2} }[/tex] = [tex]\frac{1}{15}[/tex]
Cross multiplying: d[tex]_{2}[/tex] = [tex]15d_{1} }[/tex] = 15 ×0.5 = 7.5m
Let:
velocity of air, [tex]v_{2}[/tex]
viscosity of air, η[tex]_{2}[/tex] = 0.000186Ns/m²
density of air, ρ[tex]_{2}[/tex] = 1.2 kg/m³
For the model and the prototype groups to be equal, Non-dimensional groups should be equal.
Reynold's number: (ρ[tex]_{2}[/tex] ×[tex]v_{2}[/tex] ×d[tex]_{2}[/tex])/η[tex]_{2}[/tex] = (ρ[tex]_{1}[/tex] ×[tex]v_{1}[/tex] ×d[tex]_{1}[/tex])/η[tex]_{1}[/tex]
[tex]v_{2}[/tex] = η[tex]_{2}[/tex]/(ρ[tex]_{2}[/tex] ×d[tex]_{2}[/tex]) × (ρ[tex]_{1}[/tex] ×[tex]v_{1}[/tex] ×d[tex]_{1}[/tex])/η[tex]_{1}[/tex]
[tex]v_{2}[/tex] = [tex]\frac{0.000186}{1.2* 7.5}[/tex]×[tex]\frac{1023 *10*0.5}{0.00097}[/tex] , note: * means multiplication
[tex]v_{2}[/tex] = 108.97m/s
velocity that will be needed for the model and prototype to be similar = 108.97m/s
A long straight wire carries a current of 40 A to the right. An electron, traveling at 2.7 10 7 m/s, is 5.3 cm from the wire. What force, magnitude and direction, acts on the electron if the electron velocity is directed toward the wire
Answer:
[tex]6.53\times10^-^1^7N[/tex]
Explanation:
The magnet of the magnetic field is 53 cm = 0.53m from wire is
[tex]B = \frac{\mu_0 I}{2\pi d}[/tex]
[tex]= \frac{(4\pi \times 10^-^7)(40)}{2 \pi (0.53)} \\\\= \frac{5.0265\times 10^-^5}{3.33} \\\\= 1.5095 \times 10^-^5[/tex]
the magnetic force exerted by the wire on the electron is
[tex]F = Bqv \sin \theta\\\\= 1.5095 \times 10^-^5 \times1.602\times10^-^1^9\times2.7\times10^7\\\\= 6.53\times10^-^1^7N[/tex]
From the right hand rule the direction of the force is parallel to the current (since the particle is electron)
Answer: f = 6.52*10^-16 N
Explanation:
if we assume that the force is directed at the y positive direction, then
B = μi / 2πr, where
μ = 4π*10^-7
B= (4π*10^-7 * 40) / 2 * π * 5.3*10^-2
B = 5.027*10^-5 / 0.333
B = 1.51*10^-4 T
Since v and B are perpendicular, then,
F = qvB
F = 1.6*10^-19 * 2.7*10^7 * 1.51*10^-4
F = 2.416*10^-23 * 2.7*10^7
F = 6.52*10^-16 N
Therefore, the magnitude of the force is, F = 6.52*10^-16 N and it moves in the i negative direction
Octane (C8H18) is burned with dry air. The volumetric analysis of the products on a dry basis is as below. (Fig. 15–12) Determine (a) the air–fuel ratio, (b) the percentage of theoretical air used, and (c) the amount of H2O that condenses as the products are cooled to 25°C at 100 kPa
Answer:
a. Air fuel Ratio = 19.76 kg air/kg fuel
b. % Theoretical air used = 131%
c. Amount of H2O that condenses as the products are cooled to 25°C at 100kPa = 6.59 kmol
Explanation:
1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position will the speed of the mass be 25% of its maximum speed?
Answer:
The value of the distance is [tex]\bf{14.52~cm}[/tex].
Explanation:
The velocity of a particle(v) executing SHM is
[tex]v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)[/tex]
where, [tex]\omega[/tex] is the angular frequency, [tex]A[/tex] is the amplitude of the oscillation and [tex]x[/tex] is the displacement of the particle at any instant of time.
The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., [tex]x = 0[/tex].
The maximum velocity([tex]\bf{v_{m}}[/tex]) is
[tex]v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Divide equation (1) by equation(2).
[tex]\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)[/tex]
Given, [tex]v = 0.25 v_{m}[/tex] and [tex]A = 15~cm[/tex]. Substitute these values in equation (3).
[tex]&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm[/tex]
The human ear is sometimes called a Fourier analyzer. What does this mean and why is it an apt description? 1. Our ears can sort out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone. 2. Our ears measure the intensity of sound, which is just what a Fourier analyzer does. 3. Our ears can measure the speed of sound. 4. Our ears have nothing to do with a Fourier analyzer
Answer:
1. Our ears can sort out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone.
Explanation:
A complex tone is a sound wave that consist of two or more forms of audible sound frequencies. Sound wave is a mechanical wave that is longitudinal, and could be represented by a sine wave because of it sinusoidal manner of propagation.
A Fourier analyzer can be used to differentiate individual sine waves from a combination of two or more of it; which is as the same function performed by human ear. To the human ear, a sound wave that consist of more than one sine wave will have perceptible harmonics which would be distorted and turn to a noise.
Thus, the human ear makes it possible to hear the pure tones that make up a complex tone.
Answer:
1. A Fourier analyzer sorts out the individual sine waves from a mixture of two or more sine waves, so we hear the pure tones that make up a complex tone.
Explanation:
Fourier analysis is a technique that is used to determine which sine waves constitute a given signal, i.e. to deconstruct the signal into its individual sine waves. It is the process of decomposing a periodic function into its constituent sine or cosine waves.
What goes on inside our ears is a mathematical process called a Fourier transform. In the ear, there's a combination of different waves, Fourier analysis identifies contributions at different frequencies, allowing us to reconstruct the individual signals that go into it.
A complex tone perceived by the air is is an individual sine wave that the ear, by acting as a Fourier analyzer, decomposes to serious of sine waves that we hear as pure tones.
maria was riding her bike at a velocity of 3 m/s to the north. Her velocity changed to 11 m/s to the north. What was her change in velocity
Maria's change in velocity while riding her bike is 8 meters per second to the north.
Since the motion is along the same direction (to the north), we do not need to consider the direction as negative or positive. Here's the calculation:
Change in velocity = Final velocity - Initial velocity
= 11 m/s - 3 m/s
= 8 m/s.
So, Maria's change in velocity is 8 meters per second to the north.
A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequency of the sound is 5920 Hz. In water sound propagates at a speed of 1485 m/s. The sound pulse is then reflected back from the bottom of the ocean and it is detected by the vessel 5.63 s after it was emitted. How deep is the ocean just below the vessel? Submit Answer Tries 0/12 What is the wavelength of this sound wave?
Answer:
d = 4180.3m
wavelengt of sound is 0.251m
Explanation:
Given that
frequency of the sound is 5920 Hz
v=1485m/s
t=5.63s
let d represent distance from the vessel to the ocean bottom.
an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.
[tex]velocity=\frac{distance}{time}\\\\ v=\frac{2d}{t} \\\\vt=2d\\\\d=\frac{vt}{2}[/tex]
[tex]d=\frac{1485*5.63}{2}\\d= 4180.3m[/tex]
wavelengt of sound is [tex]\lambda[/tex] = v/f
= (1485)/(5920)
= 0.251 m
State all four factors and explain how those factors affect the resistance of a wire.
Answer:
There are four factors affecting resistance which are Temperature,Length of wire,Area of the cross section of wire and nature of the material.When there is current in a conductive material,The free electrons move through the material and occasionally collide with atoms.
Explanation:
I HOPE ITS HELPFUL FOR YOUAnswer: its length, material, temperature, and cross section area which can also be considered as diameter.
Explanation:
You leave the doctor's office after your annual checkup and recall that you weighed 688 N in her office. You then get into an elevator that, conveniently, has a scale. Find the magnitude and direction of the elevator's acceleration if the scale readsQ: Find the magnitude of the elevator's acceleration if the scale reads 726 NQ: Find the direction of the elevator's acceleration if the scale reads 726 NQ: Find the magnitude of the elevator's acceleration if the scale reads 598 NQ: Find the direction of the elevator's acceleration if the scale reads 598 N
Answer:
Explanation:
weight of the person = 688 N
a) reading while in the elevator using the scale was 726 N
since when the elevator is going upward the floor of the elevator and the the scale pushes against the person leading to the person experiences a normal force greater than the weight as a result of the acceleration of the elevator and also
F = ma and W, weight = mg
mass of the body = weight / g = 688 / 9.8 = 70.20 kg
net force on the body = force of normal - weight of the body = 726 - 688 = 38 N
ma = 38 N
70.20 kg × a ( acceleration) = 38 N
a = 38 / 70.20 = 0.54 m/s² and the elevator is moving upward
b) net force, ma = force of normal - weight of the body = 598 - 688 = -90 N
a = -90 / 70.20 = -1.282 m/s² and the elevator is coming downward
A thin uniform film of refractive index 1.750 is placed on a sheet of glass with a refractive index 1.50. At room temperature ( 18.8 ∘C), this film is just thick enough for light with a wavelength 580.9 nm reflected off the top of the film to be canceled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 170 ∘C, you find that the film cancels reflected light with a wavelength 588.2 nm .
Complete Question
A thin uniform film of refractive index 1.750 is placed on a sheet of glass with a refractive index 1.50. At room temperature ( 18.8 ∘C), this film is just thick enough for light with a wavelength 580.9 nm reflected off the top of the film to be canceled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 170 ∘C, you find that the film cancels reflected light with a wavelength 588.2 nm .
What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.) Express your answer using two significant figures.
Answer:
the coefficient of linear expansion of the film is [tex]\alpha = 7.93 *10^{-5} / ^oC[/tex]
Explanation:
From the question we are told that
The refractive index of the film is [tex]n_f = 1.750[/tex]
The refractive index of the glass is [tex]n_g = 1. 50[/tex]
The wavelength of light reflected at 18°C is [tex]\lambda _r = 580.9nm = 580.9*10^{-9}m[/tex]
The wavelength of light reflected at 170°C is [tex]\lambda_h = 588.2 nm = 588.2 * 10^{-9}m[/tex]
For destructive interference the condition is
[tex]2t = \frac{m \lambda }{n_f}[/tex]
Where m is the order of interference
t is the thickness
For the smallest thickness is when m= 1 and this is represented as
[tex]t = \frac{\lambda }{2n_f }[/tex]
At 18°C the thickness would be
[tex]t_{r} = \frac{580.9 *10^{-9}}{2 * 1.750}[/tex]
[tex]t_{r} = 166nm[/tex]\
At 170° the thickness is
[tex]t_h = \frac{588.2 *10^{-9}}{2 * 1.750}[/tex]
[tex]t_h = 168 nm[/tex]
The coefficient of linear expansion f the film is mathematically represented as
[tex]\alpha = \frac{t_h - t_r}{t_r \Delta T}[/tex]
Substituting value
[tex]\alpha = \frac{168 *10^{-9} - 166 *10^{-9} }{166*10^{-9} * (170 -18)}[/tex]
[tex]\alpha = 7.93 *10^{-5} / ^oC[/tex]
Answer:
α=8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]
Explanation:
Complete question : What is the coefficient of linear expansion of the film?
SOLUTION:
There is a net ( λ/2 ) phase change due to reflection for this film, therefore, destructive interference is given by
2t = m( λ/n) where n=1.750
for smallest non-zero thickness
t= λ/2n
At 18.8°C, [tex]t_{o[/tex]=580.9 x [tex]10^{-9}[/tex]/(2 x 1.750)
[tex]t_{o[/tex]= 165.9nm
At 170°C, t= 588.2x [tex]10^{-9}[/tex]/(2x1.750)
t=168nm
t=[tex]t_{o[/tex](1 + αΔT)
=>α= (t-[tex]t_{o[/tex])/ ([tex]t_{o[/tex]ΔT) [ΔT= 170-18.8 =151.2°C]
α= (168 x [tex]10^{-9}[/tex] - 165.9 x [tex]10^{-9}[/tex])/ (165.9 x [tex]10^{-9}[/tex] x 151.2)
α= 2.1 x [tex]10^{-9}[/tex]/ 2.508 x [tex]10^{-5}[/tex]
α=8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]
Therefore, the coefficient of linear expansion of the film is 8.37 x [tex]10^{-5}[/tex] °[tex]C^{-1}[/tex]
1.- An elevator is being lowered at a constant speed by a steel cable attached to an electric motor. Which statement is correct? A. The cable does positive work on the elevator, and the elevator does positive work on the cable. B. The cable does positive work on the elevator, and the elevator does negative work on the cable. C. The cable does negative work on the elevator, and the elevator does positive work on the cable. D. The cable does negative work on the elevator, and the elevator does negative work on the cable
Answer:
the correct one is C
Explanation:
For this exercise we must use the work definition
W = F. s
Where the bold characters indicate vectors and the point is the scalar producer
W = F s cos θ
Where θ is the angles between force and displacement.
Let us support this in our case. The cable creates an upward tension and with the elevator going down the angle between them is 180º so the work of the cable on the elevator is negative.
The evade has a downward force, its weight so the force goes down and the displacement goes down, as both are in the same direction the work is positive
When examining the statements the correct one is C
Final answer:
The correct statement for an elevator being lowered at constant speed is that the steel cable does negative work on the elevator, and the elevator does negative work on the cable, illustrating the principle that work can be negative if force and displacement are in opposite directions.
Explanation:
The question pertains to the work done by an elevator cable while lowering an elevator at constant speed. According to the principles of work and energy in physics, work done is defined as the force applied in the direction of motion times the distance moved. If an elevator is being lowered at a constant speed, the steel cable exerts an upward force to counteract gravity but the elevator moves downward. Therefore, the displacement of the elevator is in the opposite direction to the force exerted by the cable, resulting in the cable doing negative work on the elevator. Conversely, because the elevator is moving downwards (in the direction opposite to the force exerted by the cable), we can interpret this as the elevator doing negative work on the cable as well, due to the concept that positive work adds energy to a system while negative work removes it.
Thus, the correct statement is: D. The cable does negative work on the elevator, and the elevator does negative work on the cable. This illustrates the application of the definition of work in physics, particularly in scenarios involving opposite directions of force and motion.
A 60.0kg skier starts from rest at the top of a ski slope of height 62.0 . Part A If frictional forces do −1.06×10 4 of work on her as she descends, how fast is she going at the bottom of the slope?
Answer:
v = 29.35 m /s
Explanation:
potential energy at the height of 62m
= m g h , m is mass , g is acceleration due to gravity and h is height
= 60 x 9.8 x 62
= 36456 J
negative work done by friction = -10600 J
energy at the bottom = 36456 - 10600 = 25856 J
This energy will be in the form of kinetic energy . If v be velocity at the bottom
1/2 m v² = 25856
1/2 x 60 x v² = 25856
v = 29.35 m /s
A particle with charge 1.60×10−19 C is placed on the x axis in a region where the electric potential due to other charges increases in the +x direction but does not change in the y or z direction.The particle, initially at rest is acted on by the electricforce and moves from point a to point b along the x axis increasingits kinetic energy by 8.00 x 10 -19J In what direction and through what potential difference does the particlemove
If the particle moves from point b to point c in the ydirection what is the change in its potential energy
Answer:
Explanation:
According to energy conservation principle;
[tex]\delta \ K.E = \delta \ P.E[/tex]
[tex]8.00*10^{-19} \ J = (1.6*10^{-19} \ C ) |V_b-V_a|[/tex]
[tex]|V_b-V_a|= \frac {8.00*10^{-19} }{(1.6*10^{-19} ) }[/tex]
[tex]|V_b-V_a|= 5 \ V[/tex]
The potential is positive hence the electric fielf=d is negative along the x-axis.
We can then say that the movement of the particle goes to the left through a potential difference of [tex]|V_b-V_a|= 5 \ V[/tex].
There will be no significant change in the y-direction of the potential energy when the particle moves from point b to point c in the y-direction.
A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50km/s in the +x-direction experiences a force of 2.06�10-16N in the +y-direction, and an electron moving at 4.40km/s in the -z-direction experiences a force of 8.40�10-16N in the +y-direction.
Part A
What is the magnitude of the magnetic field?
B = T
Part B
What is the direction of the magnetic field? (in the xz-plane)
theta = from the -z-direction
Part C
What is the magnitude of the magnetic force on an electron moving in the -y-direction at 3.70km/s ?
F = N
Part D
What is the direction of this the magnetic force? (in the xz-plane)
theta = from the -x-direction
Answer:
Explanation:
velocity of proton v = 1.5 x 10³ i m /s
charge on proton e = 1.6 x 10⁻¹⁹ C
Let the magnetic field be B = Bx i + Bz k
force on charged particle ( proton )
F = e ( v x B )
2.06 x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ [ 1.5 x 10³ i x ( Bx i + Bz k) ]
2.06 x10⁻¹⁶ j = - 1.6 x 10⁻¹⁹ x 1.5 x 10³ Bz j) ]
2.06 x10⁻¹⁶ = - 1.6 x 10⁻¹⁹ x 1.5 x 10³ Bz
Bz = - .8583
force on charged particle ( electron )
F = e ( v x B )
8.40 x10⁻¹⁶ j = -1.6 x 10⁻¹⁹ [ - 4.4 x 10³ k x ( Bx i + Bz k) ]
8.4 x10⁻¹⁶ j = 1.6 x 10⁻¹⁹ x 4.4 x 10³ Bx j ]
- 8.4 x10⁻¹⁶ = 1.6 x 10⁻¹⁹ x 4.4 x 10³ Bx
Bx = - 1.19
Magnetic field = - 1.19 i - .8583 k
magnitude = √ (1.19² + .8583²)
= 1.467 T
If it is making angle θ with x - axis in x -z plane
Tanθ = (.8583 / 1.19 )
36⁰ .
C )
v = - 3.7 x 10³j m /s
e = - 1.6 x 10⁻¹⁶ C
Force = F = e ( v x B )
= -1.6 x 10⁻¹⁹ [ -3.7 x 10³ j x ( Bx i + Bz k) ]
= - 1.6 x 10⁻¹⁹ x 3.7 x 10³ Bx k -1.6 x 10⁻¹⁹ x 3.7 x 10³Bzi ]
= 5.08 i - 7.04 k
Tanθ = 54 ° .
Determine the absolute pressure on the bottom of a swimming pool 30.0 m by 8.9 m whose uniform depth is 1.9 m . Express your answer using two significant figures. P = nothing N/m2 Request Answer Part B Determine the total force on the bottom of a swimming pool. Express your answer using two significant figures. F = nothing N Request Answer Part C What will be the pressure against the side of the pool near the bottom? Express your answer using two significant figures. P = nothing N/m2 Request Answer Provide Feedback
Answer:
A)P = 1.2 × 10⁵Pa
B)F = 3.2 × 10⁷N
C) P = 1.2 × 10⁵Pa
Explanation:
Part A)
The relative pressure at the bottom of a column of fluid is given by
[tex]p_r = \rho g h[/tex]
where
[tex]\rho[/tex] is the fluid density
g is the gravitational acceleration
h is the height of the column of fluid
At the bottom of the swimming pool, h=1.9 m, and the water density is
[tex]\rho[/tex] = 1000 kg/m^3, therefore the relative pressure is
[tex]p_r = (1000 kg/m^3)(9.81 m/s^2)(1.9 m)=1.86 \cdot 10^4 Pa[/tex]
To find the absolute pressure, we must add to this the atmospheric pressure, [tex]p_a[/tex] :
[tex]p= p_r + p_a\\= 1.86 \cdot 10^4 Pa + 1.01 \cdot 10^5 Pa \\=1.2 \times 10^5 Pa[/tex]
part B
Total force acting on the bottom
force = pressure * area
area of pool = 30.0 m × 8.9 m
= 267m²
Force F =
1.2 × 10⁵ * 267m² N
= 32040000 N
F = 3.2 × 10⁷N
Part C
The pressure acting on the side wall will be
now the pressure at the side of the pool at the bottom is simply equal to absolute pressure as they are at same level
P = 1.2 × 10⁵
A wire of length L is wound into a square coil with 117 turns and used in a generator that operates at 60.0 Hz and 120 V rms value in a 0.023-T magnetic field. What is the length L of the wire used to construct the coil?
Answer: The length L of the wire used to construct the coil is 191.4m
Explanation: Please see the attachments below
An AC generator has an output rms voltage of 100.0 V at a frequency of 42.0 Hz. If the generator is connected across a 45.0-mH inductor, find the following. (a) inductive reactance Ω (b) rms current A (c) maximum current in the circuit
Answer:
(a) 11.8692 ohm
(b) 12.447 A
(c) 17.6 A
Explanation:
a) inductive reactance Z = L Ω
= L x 2π x F
= 45.0 x 10⁻³ x 2(3.14) x 42
= 11.8692 ohm
b) rms current
= 100 / 8.034
= 12.447 A
c) maximum current in the circuit
= I eff x rac2
= 12.447 x 1.414
= 17.6 A
A hawk flew 600 meters in 60 seconds. A sparrow flew 400 meters in 30 seconds. Which bird flew faster? How fast did each bird fly?
Answer:
A sparrow flew faster. the sparrow flew 10 meters per second. The sparow flew 13.(3) meters per second
Explanation:
A sparrow flew faster than the hawk as it completes more distance in 60 seconds than that of hawk which is about 1200 meters. Speed is the distance travelled per unit time.
What is Speed?Speed is the measure of the distance travelled by an object per unit time taken. Speed is a vector quantity. It has both magnitude and direction.
Speed of an object can be calculated as: Distance travelled divided by time taken.
Speed of hawk is 600 meters/ 60 seconds
Speed of hawk = 10 m/s
Speed of sparrow is 400 meters/ 30 seconds
Speed of Sparrow = 13.33 m/s
Learn more about Speed here:
https://brainly.com/question/28224010
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A student has 67-cm-long arms. What is the minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any? The distance from the handle to the bottom of the bucket is 35 cm .
Answer:
29.61 rpm.
Explanation:
Given,
student arm length, l = 67 cm
distance of the bucket, r = 35 m
Minimum angular speed of the bucket so, the water not fall can be calculated by equating centrifugal force with weight.
Now,
[tex]mg = m r \omega^2[/tex]
[tex]\omega = \sqrt{\dfrac{g}{R}}[/tex]
R = 67 + 35 = 102 cm = 1.02 m
[tex]\omega = \sqrt{\dfrac{9.81}{1.02}}[/tex]
[tex]\omega = 3.101\ rad/s[/tex]
[tex]\omega = \dfrac{3.101}{2\pi} = 0.494\ rev/s[/tex]
[tex]\omega = 0.494 \times 60 = 29.61\ rpm[/tex]
minimum angular velocity is equal to 29.61 rpm.
Answer:
29.6 rpm
Explanation:
length of arm = 67 cm
distance of handle to the bottom = 35 cm
radius of rotation, R = 67 + 35 = 102 cm = 1.02 m
The centripetal force acting on the bucket is balanced by the weight of the bucket.
mRω² = mg
R x ω² = g
[tex]\omega = \sqrt\frac{g}{R}[/tex]
[tex]\omega = \sqrt\frac{9.8}{1.02}[/tex]
ω = 3.1 rad/s
Let f is the frequency in rps
ω = 2 x 3.14 x f
3.1 = 2 x 3.14 xf
f = 0.495 rps
f = 29.6 rpm
Two manned satellites approaching one another at a relative speed of 0.550 m/s intend to dock. The first has a mass of 2.50 ✕ 103 kg, and the second a mass of 7.50 ✕ 103kg. If the two satellites collide elastically rather than dock, what is their final relative velocity? Adopt the reference frame in which the second satellite is initially at rest and assume that the positive direction is directed from the second satellite towards the first satellite.
...............m/s
Answer: Their final relative velocity is -0.412 m/s.
Explanation:
According to the law of conservation,
[tex]m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v[/tex]
Putting the given values into the above formula as follows.
[tex]m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v[/tex]
[tex]2.50 \times 10^{3} kg \times 0 m/s + 7.50 \times 10^{3} kg \times -0.550 m/s = (2.50 \times 10^{3} kg + 7.50 \times 10^{3} kg)v[/tex]
[tex]-4.12 \times 10^{3} kg m/s = (10^{4} kg) v[/tex]
v = [tex]\frac{-4.12 \times 10^{3} kg m/s}{10^{4} kg}[/tex]
= -0.412 m/s
Thus, we can conclude that their final relative velocity is -0.412 m/s.
A person is diving in a lake in the depth of h = 15 m. The density of the water is rho = 1.0 x103 kg/m3. The pressure of the atmosphere is P0 = 1.0 x 105 Pa. The surface area of the top of the person's head is A = 0.036 m2.
(a) Express the absolute pressure at the depth of h , Po, Q in terms of Poe, and h
Answer:
Pabs = 247150 [Pa]
Explanation:
The pressure in the depth h can be calculated by the following expression.
Pabs = Po + (rho * g * h)
Where:
g = gravity = 9.81[m/s^2]
rho = density = 1000 [kg/m^3]
h = depth = 15 [m]
Po = 100000 [Pa]
Pabs = 100000 + (1000*9.81*15)
Pabs = 247150 [Pa]
A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 20.0 m/s2, what is the radius of the planet?
Answer:
Explanation:
Expression for times period of a satellite can be given as follows
Time period T = 1.8 x 60 x 60
= 6480
T² = [tex]\frac{4\times \pi^2\times r^3}{GM}[/tex] where T is time period , r is radius of orbit , G is gravitational constant and M is mass of the satellite.
6480² = 4 x 3.14² x 7.5³ x 10¹⁸ / GM
GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²
= 3.96 X 10¹⁴
Expression for acceleration due to gravity
g = GM / R² where R is radius of satellite
20 = 3.96 X 10¹⁴ / R²
R² = 3.96 X 10¹⁴ / 20
= 1.98 x 10¹³ m
R= 4.45 x 10⁶ m