Answer:
Sara must climb 10609 stairs to burn this food.
Explanation:
Chemical energy content in Sara's body = 788 kCal x 4.184 = 3296.99 kJ = 3296990 J
her body is only 25% effective in converting chemical to mechanical energy, therefore,
available mechanical energy = 25% of 3296990 = 0.25 x 3296990 = 824247.5 J --this is equivalent to the work she must do to burn this stored energy
flight of stairs = 12 cm = 0.12 m
Sara's mass m = 66 kg
Sara's weight = mg = mass x acceleration due to gravity 9.81 m/s^2
weight W = 66 x 9.81 = 647.46 N
work done = force x distance climbed d
putting in the necessary values, we have
824247.5 = 647.46 x d
d = 1273.05 m
the number of stairs Sara must climb is equal to the distance climbed divided by the height of a single stairs
number of stairs = 1273.05/0.12 = 10608.75 ≅ 10609 stairs
A circuit contains two elements, but it is not known if they are L, R, or C. The current in this circuit when connected to a 120-V 60 Hz source is 5.3 A and lags the voltage by 65∘.
Part A. What are the two elements?
Part B. What are their values?
Express your answer using two significant figures
Answer:
the two elements are resistor R and inductor L
answers in two significant figures
R = 9.6Ω
L = 54mH
Explanation:
) An electron moving along the x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, what is the direction of the magnetic field in this region
Answer:
- z direction
Explanation:
To find the direction of the magnetic field, you take into account that the magnetic force over a charge, is given by the following cross product:
[tex]\vec{F_B}=q\vec{v}\ X\ \vec{B}[/tex] (1)
F_B: magnetic force
q: charge of the particle
v: velocity of the charge
B: magnetic field
In this case you have that the electron is moving along x-axis. You can consider this direction as the ^i direction. The electron experiences a magnetic deflection in the -y direction, that is, in the -^j direction.
By the cross products between unit vectors, you have that:
-^j = ^i X ^k
That is, the cross product between two vectors, one in the +x direction, and another one in the +z direction, generates a vector in the -y direction. However, it is necessary to take into account that the negative charge of the electron change the sign of the result of the cross product, which demands that the second vector is in the -z direction. That is:
-^i X -k^ = ^i X ^k = - ^j
Hence, the direction of the magnetic field is in the -z direction
A stationary 6-kg shell explodes into three pieces. One 4.0 kg piece moves horizontally along the negative x-axis. The other two fragments, each 1.0 kg, move in directions that make 60o angle above and below the positive x-axis and their speeds are 60 m/s each. What is the velocity of the 4.0-kg fragment
Answer:
-15 m/s
Explanation:
The computation of the velocity of the 4.0 kg fragment is shown below:
For this question, we use the correlation of the momentum along with horizontal x axis
Given that
Weight of stationary shell = 6 kg
Other two fragments each = 1.0 kg
Angle = 60
Speed = 60 m/s
Based on the above information, the velocity = v is
[tex]1\times 60 \times cos\ 60 + 1\times 60 \times cos\ 60 - 4\ v = 0[/tex]
[tex]\frac{60}{2} + \frac{60}{2} - 4\ v = 0[/tex]
[tex]v = \frac{60}{4}[/tex]
= -15 m/s
The motion of spinning a hula hoop around one's hips can be modeled as a hoop rotating around an axis not through the center, but offset from the center by an amount h, where h is less than R, the radius of the hoop. Suppose Maria spins a hula hoop with a mass of 0.73 kg and a radius of 0.60 m around her waist. The rotation axis is perpendicular to the plane of the hoop, but approximately 0.38 m from the center of the hoop.(a) What is the rotational inertia of the hoop in this case? ________ kg m^2 (b) If the hula hoop is rotating with an angular speed of 14.1 rad/s, what is its rotational kinetic energy?
Answer and Explanation:
Based on the given information, the formula and the computation is given below:
a. The rotational inertia of the hoop is shown below:
[tex]I_H = I_R + Mh^2[/tex]
[tex]= MR^2 + Mh^2[/tex]
[tex]= 0.73 \times (0.60^2 + 0.38^2)[/tex]
= 0.73 × (0.36 + 0.1444)
= 0.368 [tex]kg\ mg^2[/tex]
b. Now the rotational kinetic energy is
[tex]= Half \times Inertia \times omega^2[/tex]
[tex]= 0.5 \times 0.368 \times 14.1^2[/tex]
= 36.58 J
We simply applied the above formula for rotational inertia and rotational kinetic energy in order to reach with the correct answer
Alan leaves Los Angeles at 8:00 a.m. to drive to San Francisco, 400 mi away. He travels at a steady 45.0 mph . Beth leaves Los Angeles at 9:00 a.m. and drives a steady 55.0 mph .
a. Who gets to San Francisco first?
b. How long does the first to arrive have to wait for the second?
Answer:
a.Beth
b.2232 s
Explanation:
We are given that
Distance,d=400 mi
Speed of Alan,v=45 mph
Speed of Beth,v'=55 mph
a.Time =[tex]\frac{distance}{speed}[/tex]
Using the formula
Time taken by Alan=[tex]\frac{400}{45}=8.89 hr[/tex]
Time taken by Beth=[tex]\frac{400}{55}=7.27hr[/tex]
Alan will reach San Francisco at 4:53 PM
Beth will reach San Francisco at 4:16 PM
Beth will reach before Alan.
b.Difference between time=8.89-7.27=1.62 hr
t=1.62 hr
1.62-1=0.62 hr
0.62 hr=[tex]0.62\times 60\times 60=2232 s[/tex]
Hence, Beth has to wait 2232 s for Alan to arrive .
What is a possible state for an object in the absence of a net force?
There is only one possible state: constant uniform motion. That means constant speed in a straight line.
(If the constant speed happens to be zero, this description also covers the case where the object isn't moving. That special case is called "at rest".)
Answer:
at restzero accelerationconstant speedHope this helps
Imagine you made a little pendulum by tying a 6.0 gram washer to a thin piece of thread and dangled it from your hand as you were given a test drive in a sports car. If the driver floored the engine and the car went from 0 to 60 mph down a straight level road in 3.2 seconds, how far off vertical would the washer hang as the car accelerated
Answer:
The displacement of the pendulum from equilibrium position is 0.855 of the length of the pendulum.
Explanation:
A pendulum's motion can be modelled as a simple harmonic motion and the differential equation describing simple harmonic motion is given as
(d²x/dt²) + (k/m)x = 0
(d²x/dt²) = acceleration of the pendulum
x = displacement of the pendulum from equilibrium position
k = spring constant of the pendulum = (mg/L) = (0.006×9.8/L) = (0.0588/L)
L = length of the pendulum
m = mass of the pendulum = 6 g = 0.006 kg
(k/m) = (g/L) = (9.8/L)
To calculate the pendulum's acceleration
Initial velocity of the car = 0 mph = 0 m/s
Final velocity of the car = 60 mph = 26.822 m/s
Time taken for velocity change = 3.2 s
Acceleration = (Δv/Δt) = (26.822/3.2) = 8.382 m/s²
(d²x/dt²) + (k/m)x = 0
Becomes
8.383 + (9.8/L)x = 0
(9.8x/L) = -8.382
9.8x = -8.382L
x = -(8.382/9.8) = -0.855 L (the minus sign shows the direction of movement of the pendulum)
Ignoring the sign and only focusing on the magnitude of the pendulum's displacement from equilibrium position, the displacement of the pendulum from equilibrium position is 0.855 of the length of the pendulum.
Hope this Helps!!!
Assume you have a rocket in Earth orbit and want to go to Mars. The required change in velocity is ΔV≈9.6km/s . There are two options for the propulsion system --- chemical and electric --- each with a different specific impulse. Recall that the relationship between specific impulse and exhaust velocity is: Vex=g0Isp Using the Ideal Rocket Equation and setting g0=9.81m/s2 , calculate the propellant fraction required to achieve the necessary ΔV for each of propulsion system. Part 1: Cryogenic Chemical Propulsion First, consider a cryogenic chemical propulsion system with Isp≈450s . Enter the required propellant fraction as a proportion with at least 2 decimal places (i.e., enter 0.25 to represent 25%): incorrect Part 2: Electric Propulsion Next, consider an electric propulsion system with Isp≈2000s . Enter the required propellant fraction as a proportion with at least 2 decimal places (i.e., enter 0.25 to represent 25%):
Answer: Part 1: Propellant Fraction (MR) = 8.76
Part 2: Propellant Fraction (MR) = 1.63
Explanation: The Ideal Rocket Equation is given by:
Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]
Where:
[tex]v_{ex}[/tex] is relationship between exhaust velocity and specific impulse
[tex]\frac{m_{f}}{m_{e}}[/tex] is the porpellant fraction, also written as MR.
The relationship [tex]v_{ex}[/tex] is: [tex]v_{ex} = g_{0}.Isp[/tex]
To determine the fraction:
Δv = [tex]v_{ex}.ln(\frac{m_{f}}{m_{e}} )[/tex]
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
Knowing that change in velocity is Δv = 9.6km/s and [tex]g_{0}[/tex] = 9.81m/s²
Note: Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.
Part 1: Isp = 450s
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
ln(MR) = [tex]\frac{9.6.10^{3}}{9.81.450}[/tex]
ln (MR) = 2.17
MR = [tex]e^{2.17}[/tex]
MR = 8.76
Part 2: Isp = 2000s
[tex]ln(MR) = \frac{v}{v_{ex}}[/tex]
ln (MR) = [tex]\frac{9.6.10^{3}}{9.81.2.10^{3}}[/tex]
ln (MR) = 0.49
MR = [tex]e^{0.49}[/tex]
MR = 1.63
As a system expands, it absorbs 52.5 J of energy in the form of heat from the surroundings. The piston is working against a pressure of 0.500 atm. The final volume of the system is 58.0 L. What was the initial volume of the system if the internal energy of the system decreased by 102.5
Answer:
Vi = 0.055 m³ = 55 L
Explanation:
From first Law of Thermodynamics, we know that:
ΔQ = ΔU + W
where,
ΔQ = Heat absorbed by the system = 52.5 J
ΔU = Change in Internal Energy = -102.5 J (negative sign shows decrease in internal energy of the system)
W = Work Done in Expansion by the system = ?
Therefore,
52.5 J = - 102.5 J + W
W = 52.5 J + 102.5 J
W = 155 J
Now, the work done in a constant pressure condition is given by:
W = PΔV
W = P(Vf - Vi)
where,
P = Constant Pressure = (0.5 atm)(101325 Pa/1 atm) = 50662.5 Pa
Vf = Final Volume of System = (58 L)(0.001 m³/1 L) = 0.058 m³
Vi = Initial Volume of System = ?
Therefore,
155 J = (50662.5 Pa)(0.058 m³ - Vi)
Vi = 0.058 m³ - 155 J/50662.5 Pa
Vi = 0.058 m³ - 0.003 m³
Vi = 0.055 m³ = 55 L
The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 4.50 m/s at point A and 5.00 m/s at point C. The cart takes 4.00 s to go from point A to point C, and the cart takes 1.60 s to go from point B to point C. What is the cart's speed at point B
Answer:
Vi = 4.8 m/s
Explanation:
First we need to find the magnitude of constant tangential acceleration. For that purpose we use the following formula between points A and C:
a = (Vf - Vi)/t
where,
a = constant tangential acceleration from A to C = ?
Vf = Final Velocity at C = 5 m/s
Vi = Initial Velocity at A = 4.5 m/s
t = time taken to move from A to C = 4 s
Therefore,
a = (5 m/s - 4.5 m/s)/4 s
a = 0.125 m/s²
Now, applying the same equation between points B and C:
a = (Vf - Vi)/t
where,
a = constant tangential acceleration from A to B = 0.125 m/s²
Vf = Final Velocity at C = 5 m/s
Vi = Initial Velocity at B = ?
t = time taken to move from B to C = 1.6 s
Therefore,
0.125 m/s² = (5 m/s - Vi)/1.6 s
Vi = 5 m/s - (0.125 m/s²)(1.6 s)
Vi = 4.8 m/s
Two cars are traveling around identical circular racetracks. Car A travels at a constant speed of 20 m/s. Car B starts at rest and speeds up with constant tangential acceleration until its speed is 40 m/s. When car B has the same (tangential) velocity as car A, it is always true that:
a. it has the same linear (tangential) acceleration as car A.
b. it has the same centripetal acceleration as car A.
c. it has traveled farther than car A since starting.
d. it has the same total acceleration as car A.
e. it is passing car A.
Answer:
b. it has the same centripetal acceleration as car A.
Explanation:
According to the question, the data provided is as follows
Constant speed of car A = 20 m/s
Constant tangential acceleration until its speed is 40 m/s
Based on the above information, the true statement is the same centripetal acceleration as car A because
As we know that
Centripetal acceleration is
[tex]= \frac{V^2}{r}[/tex]
where,
[tex]V^2[/tex] = velocity
r = radius of the path
Now if both car A and car B moving in the same or identical circular path having the same velocity so in this case there is the same centripetal acceleration for that particular time
hence, the second option is correct
An object suspended from a spring vibrates with simple harmonic motion. Part A At an instant when the displacement of the object is equal to one-fourth the amplitude, what fraction of the total energy of the system is kinetic
Complete Question
An object suspended from a spring vibrates with simple harmonic motion.
a. At an instant when the displacement of the object is equal to one-half the amplitude, what fraction of the total energy of the system is kinetic?
b. At an instant when the displacement of the object is equal to one-half the amplitude, what fraction of the total energy of the system is potential?
Answer:
a
The fraction of the total energy of the system is kinetic energy [tex]\frac{KE}{T} = \frac{3}{4}[/tex]
b
The fraction of the total energy of the system is potential energy [tex]\frac{PE}{T} = \frac{1}{4}[/tex]
Explanation:
From the question we are told that
The displacement of the system is [tex]e = \frac{a}{2}[/tex]
where a is the amplitude
Let denote the potential energy as PE which is mathematically represented as
[tex]PE = \frac{1}{2} * k* x^2[/tex]
=> [tex]PE = \frac{1}{2} * k* [\frac{a}{2} ]^2[/tex]
[tex]PE = k* [\frac{a^2}{8} ][/tex]
Let denote the total energy as T which is mathematically represented as
[tex]T = \frac{1}{2} * k * a^2[/tex]
Let denote the kinetic energy as KE which is mathematically represented as
[tex]KE = T -PE[/tex]
=> [tex]KE =k [ \frac{a^2}{2} - \frac{a^2}{8} ][/tex]
=> [tex]KE =k [ \frac{3}{8} a^2 ][/tex]
Now the fraction of the total energy that is kinetic energy is
[tex]\frac{KE}{T} = \frac{ \frac{3ka^2}{8} }{\frac{ka^2}{2} }[/tex]
[tex]\frac{KE}{T} = \frac{3}{4}[/tex]
Now the fraction of the total energy that is potential energy is
[tex]\frac{PE}{T} = \frac{\frac{k a^2}{8} }{\frac{k a^2}{2} }[/tex]
[tex]\frac{PE}{T} = \frac{1}{4}[/tex]
Which circuits are parallel circuits?
Answer:
The bottom two lines.
Explanation:
They need their own line of voltage quantity. A parallel circuit has the definition of 'two or more paths for current to flow through.' The voltage does stay the same in each line.
Suppose that a circular parallel-plate capacitor has radius R0 = 3.0 cm and plate separation d = 5.0 mm. A sinusoidal potential difference V = V0 sin(2πft) is applied across the plates, where V0 = 150 V and f = 60 Hz. In the region between the plates, find the magnitude of the induced magnetic field versus R and t, where R is the radial distance from the capacitor’s central axis. Assume that B = B0(R) is the amplitude of the magnetic field.
Answer:
[tex]B=\frac{\mu_o \epsilon_o R_o^2V_o}{2rd}cos(\omega t)[/tex]
Explanation:
By the information of the statement you have that the sinusoidal potential difference is given by:
[tex]V=V_osin(\omega t)=V_osin(2\pi ft)=150sin(2\pi (60)t)[/tex] (1)
In order to calculate the induced magnetic field in between the plates, you first take into account the following formula, which is the Ampere-Maxwell law:
[tex]\int B\cdot ds=\mu_o \epsilon_o\frac{d\Phi_E}{dt}+\mu_oI_c[/tex] (2)
B: induced magnetic field
μo: magnetic permeability of vacuum = 4π*10^-7 A/T
εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2
Ic: conduction current
ФE: electric flux
There is no conduction current in between the plates, then Ic = 0A
Next, you calculate dФE/dt, as follow:
The electric field, and the electric flux, are:
[tex]E=\frac{V}{d}=\frac{V_osin(\omega t)}{d}\\\\\Phi_E=EA[/tex]
d: separation between plates = 5.0mm = 5.0*10^-3 m
A: area of the circular plates = [tex]\pi R_o^2[/tex]
Ro: radius of the circular capacitor = 3.0cm = 0.03m
Thus, dФE/dt is:
[tex]\frac{d\Phi_E}{dt}=\frac{d(EA)}{dt}=\pi R_o^2\frac{d}{dt}[\frac{V_osin(\omega t)}{d}]\\\\\frac{d\Phi_E}{dt}=\frac{\pi \omega R_o^2 V_o}{d}cos(\omega t)[/tex] (3)
The induced magnetic field is calculated by taking into account that the integral of the equation (2) is:
[tex]\int B \cdot ds=B\int ds=B(2\pi r)[/tex] (4)
Next, you replace the results of (3) and (4) into the equation (2) and you solve for B:
[tex]B(2\pi r)=\mu_o \epsilon_o (\frac{\pi \omega R_o^2 V_o}{d}cos(\omega t))\\\\B=\frac{\mu_o \epsilon_o R_o^2V_o}{2rd}cos(\omega t)[/tex] (5)
The last expression is de induced magnetic field in between the plates in terms of t and r
Another way of expressing the formula (5) is as follow:
[tex]B=B_ocos(\omega t)\\\\B_o=\frac{\mu_o \epsilon_o R_o^2V_o}{2rd}[/tex]
A 50-loop circular coil has a radius of 3 cm. It is oriented so that the field lines of a magnetic field are perpendicular to the coil. Suppose that the magnetic field is varied so that B increases from 0.10 T to 0.35 T in 2 ms. Find the induced emf in the coil.
Answer:
-17.8 V
Explanation:
The induced emf in a coil is given as:
[tex]E = \frac{-NdB\pi r^2}{dt}[/tex]
where N = number of loops
dB = change in magnetic field
r = radius of coil
dt = elapsed time
From the question:
N = 50
dB = final magnetic field - initial magnetic field
dB = 0.35 - 0.10 = 0.25 T
r = 3 cm
dt = 2 ms = 0.002 secs
Therefore, the induced emf is:
[tex]E = \frac{-50 * 0.25 * \pi * 0.03^2}{0.002} \\E = -17.8 V[/tex]
Note: The negative sign implies that the EMf acts in an opposite direction to the change in magnetic flux.
Suppose that the emf from a rod moving in a magnetic field was used to supply the current to illuminate a light bulb in a circuit, and the force needed to keep the rod moving is . What happens to the force needed drive the motion of the rod if the light bulb is removed
Explanation:
The rod moving in a magnetic field and induces an emf which is used to illuminate the bulb. But if the bulb is removed form the circuit, the circuit is opened.
For an open circuit, no current is passes through the moving rod. If there is no current along the rod, then no magnetic field developed around the rod. Because moving charges nothing but current produces the magnetic field around the rod.
The formula for the magnetic force on the rod is,
[tex]F_{\mathrm{B}} &=I(l \times B)[/tex]
[tex]=I l B \sin \theta[/tex]
The current along the rod is zero because the bulb is removed and the magnetic field around the rod is zero because no current is passes through the rod.
Then calculate the magnetic force on the rod as follows:
\[tex]F_{\mathrm{B}} &=I l B \sin \theta[/tex]
[tex]=(0)(l)(0) \sin \theta =0 \mathrm{N}[/tex]
Thus, no force is needed because there is no longer magnetic field developed around the rod.
A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 gram mass be attached so that the meterstick will be balanced in rotational equilibrium
Answer:
The 40g mass will be attached at 69 cm
Explanation:
First, make a sketch of the meterstick with the masses placed on it;
--------------------------------------------------------------------------
↓ Δ ↓
20 g.................50 cm.................40g
38 cm y cm
Apply principle of moment;
sum of clockwise moment = sum of anticlockwise moment
40y = 20 (38)
40y = 760
y = 760 / 40
y = 19 cm
Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm
12cm 50 cm 69cm
--------------------------------------------------------------------------
↓ Δ ↓
20 g.................50 cm.................40g
38 cm 19 cm
The drawing shows a top view of a hockey puck as it slides across frictionless ice. Three forces act on the puck, and it is in equilibrium. The force F is applied at the center and has a magnitude of 32 N. The force F1 is applied at the top edge, and F2 is applied half way between the center and the bottom edge. Find the magnitude of F1 and F2.
Answer:
The values of the forces are
[tex]F_1 = 10.6 \ N[/tex] , [tex]F_2 = 21.33 \ N[/tex]
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The magnitude of F is [tex]F = 32 \ N[/tex]
Generally at equilibrium the torque is mathematically evaluated as
[tex]\sum \tau = 0[/tex]
From the diagram we have
[tex]r * F_1 - [\frac{r}{2} ] F_2 + 0 F = 0[/tex]
=> [tex]F_1 = 0.5 F_2[/tex]
Generally at equilibrium the Force is mathematically evaluated as
[tex]\sum F = 0[/tex]
From the diagram
[tex]F - F_ 1 - F_2 = 0[/tex]
substituting values
[tex]32 - (0.5F_2 ) - F_2 = 0[/tex]
[tex]F_2 = 21.33 \ N[/tex]
So
[tex]F_1 = 0.5 * 21.33[/tex]
[tex]F_1 = 10.6 \ N[/tex]
QUESTION ONE
(a) Zindhile and Phindile are rowing a boat across a river which is 40m wide. They row in a direction
perpendicular to the bank. However, the river is flowing downstream and by the time they reach the other
side, they end up 30m downstream from their starting point. Over what distance did the boat travel?
If we approximate the rack to be completely flat and the racecar is travelling a constant 30.5 m/s around the turn, what forces are responsible for the centripetal acceleration?
We have,
The race car is travelling a constant 30.5 m/s around the turn. When the car is travelling around a turn, the centripetal force acts on it. The centripetal force is balanced by the force of friction between the car and the surface. Hence, frictional force is responsible for the centripetal acceleration.
g Using the available equations in your lab manual , what is the acceleration of an object when it hits the ground if that object is thrown from a building 100 meters above the ground, with an initial velocity of 10 meters per second (10M/S), and a flight time of 5 seconds (assume no friction)
Answer:
The acceleration of an object is 4 m/s².
Explanation:
The object is thrown from a building 100 meters above the ground, with an initial velocity of 10 meters per second and a flight time of 5 seconds.
It is required to find the acceleration of the object. Let a is the acceleration. Using second equation of kinematics to find it as :
[tex]h=ut+\dfrac{1}{2}at^2\\\\100=10(5)+\dfrac{1}{2}a(5)^2\\\\100=50+\dfrac{1}{2}\times 25a\\\\50=\dfrac{25a}{2}\\\\a=\dfrac{100}{25}=4\ m/s^2[/tex]
So, the acceleration of an object is 4 m/s².
A charged Adam or particle is called a
Answer:
A charged atom or particle is called an ion :)
What is the frequency a stationary observer hears when a train approaches her with a speed of 30 m/s. The frequency of the train horn is 0.600 kHz and the speed of sound is 340 m/s. A) 551 Hz B) 600 Hz C) 653 Hz D) 658 Hz
Answer:
The frequency a stationary observer hears when a train approaches her with a speed of 30 m/s is 658 Hz (option D)
Explanation:
The Doppler effect is defined as the change in the apparent frequency of a wave produced by the relative movement of the source with respect to its observer. In other words, this effect is the change in the perceived frequency of any wave motion as the emitter and receiver, or observer, move relative to each other.
The following expression is considered the general case of the Doppler effect:
[tex]f'=f*\frac{v+-vR}{v-+vE}[/tex]
Where:
f ’, f: Frequency perceived by the receiver and frequency emitted by the emitter respectively. Its unit of measurement in the International System (S.I.) is hertz (Hz), which is the inverse unit of the second (1 Hz = 1 s-1)
v: propagation speed of the wave in the medium. It is constant and depends on the characteristics of the medium.
vR, vE: Receiver and emitter speed respectively. Your unit of measure in the S.I. is the m / s
±, ∓:
The + sign is used:
In the numerator if the receiver approaches the sender In the denominator if the sender moves away from the receiverThe sign - is used:
In the numerator if the receiver moves away from the sender In the denominator if the sender approaches the receiverIn this case:
f= 0.600 kHz= 600 Hz (being 1 kHz= 1000 Hz)v= 340 m/svR= 0 m/s because the observer is in a stationary state.vE= 30 m/s The emitter (the train) approaches the receiver, so the sign in the denominator is negative.Then:
[tex]f'=600 Hz*\frac{340 \frac{m}{s} +-0 \frac{m}{s} }{340 \frac{m}{s} -30 \frac{m}{s} }[/tex]
Solving:
[tex]f'=600 Hz*\frac{340 \frac{m}{s} }{340 \frac{m}{s} -30 \frac{m}{s} }[/tex]
[tex]f'=600 Hz*\frac{340 \frac{m}{s} }{310 \frac{m}{s} }[/tex]
f'= 658 Hz
The frequency a stationary observer hears when a train approaches her with a speed of 30 m/s is 658 Hz (option D)
what is the most likely elevation of point x?
A. 150 ft
B. 200 ft
C. 125 ft
D. 250 ft
In a contest, two tractors pull two identical blocks of stone thesame distance over identical surfaces. However, block A is moving twice as fast as block B when it crosses the finish line. Which statement is correct?a) Block A has twiceas much kinetic energy as block B.b) Block B has losttwice as much kinetic energy to friction as block A.c) Block B has losttwice as much kinetic energy as block A.d) Both blocks havehad equal losses of energy to friction.e) No energy is lostto friction because the ground has no displacement.
Answer:
d) Both blocks have had equal losses of energy to friction
Explanation:
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces
Moreover, the block A is twice as fast than block B at the time of crossing the finish line
So based on the above information, it contains the losses of identical friction
And we also know that
Friction energy loss is
[tex]= \mu \times m \times g \times D[/tex]
It would be the same for both the blocks
hence, the option d is correct
The correct answer will be both blocks have had equal losses of energy to friction.
What is friction?Friction is defined as when any object is slides on a surface by means of any external force then the force in the opposite direction generated between the surface and the body restrict the motion of the body this force is called as the friction.
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces.
Moreover, the block A is twice as fast as block B at the time of crossing the finish line.
So based on the above information, it contains the losses of identical friction.
And we also know that
Friction energy loss is
[tex]E_f=\mu m g D[/tex]
It would be the same for both the blocks
Hence both blocks have had equal losses of energy to friction.
To know more about friction, follow
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A compact disc rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s.
(a) What is the angular acceleration of the disc, assuming the angular acceleration is uniform?
(b) Through what angle does the disc turn while coming up to speed?
(c) If the radius of the disc is 4.45 cm, find the tangential speed of a microbe riding on the rim of the disc when t=0.892 s.
(d) What is the magnitude of the tangential acceleration at the given time?
Answer:
(a) α = 35.20 rad/s^2
(b) θ = 802°
(c) v = 139.73 cm/s
(d) a = 156.64 cm/s^2
Explanation:
(a) To find the angular acceleration of the disc you use the following formula:
[tex]\alpha=\frac{\omega-\omega_o}{t}[/tex] (1)
w: angular speed of the disc = 31.4 rad/s
wo: initial angular speed = 0 rad/s
t: time = 0.892s
You replace the values of the parameters in the equation (1):
[tex]\alpha=\frac{31.4rad/s-0rad/s}{0.892s}=35.20\frac{rad}{s^2}[/tex]
The angular acceleration of the disc, for the given time, is 35.20rad/s^2
(b) To calculate the angle describe by the disc in such a time you use:
[tex]\theta=\frac{1}{2}\alpha t^2[/tex] (2)
[tex]\theta=\frac{1}{2}(35.20rad/s^2)(0.892s)^2=14.00rad[/tex]
In degrees you have:
[tex]\theta=14.00rad*\frac{180\°}{\pi \ rad}=802\°[/tex]
The angle described by the disc is 802°
(c) To calculate the tangential speed of the microbe for t=0.892s, you use the following formula:
[tex]v=\omega r[/tex] (3)
w: angular speed for t = 0.892s = 31.4rad/s
r: radius of the disc = 4.45cm
[tex]v=(31.4rad/s)(4.45cm)=139.73\frac{cm}{s}[/tex]
The tangential speed is 139.73 cm/s
(d) The tangential acceleration is calculated by using the following formula:
[tex]a=\alpha r[/tex]
α: angular acceleration for t=0.892s
[tex]a=(35.20rad/s^2)(4.45cm)=156.64\frac{cm}{s^2}[/tex]
The tangential acceleration is 156.64cm/s^2
A disk between vertebrae in the spine is subjected to a shearing force of 640 N. Find its shear deformation taking it to have the shear modulus of 1.00 109 N/m2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.30 cm in diameter.
Answer:
3.08*10^-6 m
Explanation:
Given that
Total shearing force, F = 640 N
Shear modulus, S = 1*10^9 N/m²
Height of the cylinder, L = 0.7 cm
Diameter of the cylinder, d = 4.3 cm
The solution is attached below.
We have our shear deformation to be 3.08*10^-6 m
A proton of mass m is at rest when it is suddenly struck head-on by an alpha particle (which consistsof 2 protons and 2 neutrons) moving at speed v. If the collision is perfectly elastic, what speed will the alpha particle have after the collision
Answer:
3/5 v
Explanation:
The computation of speed will the alpha particle have after the collision is shown below:-
In a perfectly elastic the kinetic energy and collision the momentum are considered.
The velocity of the particles defines the below equation:
[tex]VA_f=(\frac{m_A-m_B}{m_A+m_B})VA_i+(\frac{2m_B}{m_A+m_B})VB_i[/tex]
As we know that
[tex]VA_i=v[/tex]
[tex]\\VB_i=0[/tex]
Here, we consider A is the alpha particle and B is the proton and now by the above values we can solve the equation which is below:-
[tex]VA_f=(\frac{4m-m}{4m+m})v[/tex]
[tex]\\VA_f=\frac{3m}{5m}v[/tex]
[tex]\\VA_f=\frac{3}{5}v[/tex]
Therefore the correct answer is [tex]\frac{3}{5}v[/tex]
An object moving in the +x direction experiences an acceleration of +2.0 m/s2. This means the object (a) travels 2.0 m in every second. (b) is traveling at 2.0 m/s. (c) is decreasing its velocity by 2.0 m/s every second. (d) is increasing its velocity by 2.0 m/s every second.
Answer:
(d) is increasing its velocity by 2.0 m/s every second.
Explanation:
Acceleration is the change in velocity over time.
An acceleration of 2.0 m/s/s means the object's velocity is increasing 2.0 m/s per second.
This question is related to the concept of acceleration.
The correct answer is "(d) is increasing its velocity by 2.0 m/s every second".
The acceleration of an object is defined as the rate of change of its velocity over a given interval of time. Mathematically, acceleration is defined as the change in velocity of an object divided by the time interval taken for that change in velocity.
[tex]Acceleration=\frac{Change\ in\ Velocity}{Time}\\\\[/tex]
The Positive sign of the acceleration means the change in velocity is positive and the velocity is increasing, while the negative sign indicates a decrease in velocity.
Hence, an acceleration of 2 m/s² means that the velocity of the object is increasing by 2 m/s every second.
Attached picture describes acceleration.
Learn more about acceleration here:
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Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12.0-V battery, the current from the battery is 1.78 A. When the resistors are connected in parallel to the battery, the total current from the battery is 11.3 A. Determine the two resistances.
Answer:
R(smaller) = 1.3 Ω and R(larger) = 5.4 Ω
Explanation:
Ohm's Law states that:
V = IR
R = V/I
where,
R = Resistance
V = Potential Difference
I = Current
Therefore, for series connection:
Rs = Vs/Is
where,
Rs = Resistance when connected in series = R(smaller) + R(larger)
Vs = Potential Difference when connected in series = 12 V
Is = Current when connected in series = 1.78 A
Therefore,
R(smaller) + R(larger) = 12 V/1.78 A
R(smaller) + R(larger) = 6.74 Ω --------------- equation 1
R(smaller) = 6.74 Ω - R(larger) --------------- equation 2
Therefore, for series connection:
Rp = Vp/Ip
where,
Rp = Resistance when connected in parallel = [1/R(smaller) + 1/R(larger)]⁻¹
Rp = [{R(smaller) + R(larger)}/{R(smaller).R(larger)]⁻¹
Rp = R(smaller).R(larger)/[R(smaller) + R(larger)]
Vp = Potential Difference when connected in parallel = 12 V
Ip = Current when connected in parallel = 11.3 A
Therefore,
R(smaller).R(larger)/[R(smaller) + R(larger)] = 12 V/11.3 A
using equation 1 and equation 2, we get:
[6.74 Ω - R(larger)].R(larger)/6.74 Ω = 1.06 Ω
6.74 R(larger) - R(larger)² = (6.74)(1.06)
R(larger)² - 6.74 R(larger) + 7.16 = 0
solving this quadratic equation we get:
R(larger) = 5.4 Ω (OR) R(larger) = 1.3 Ω
using these values in equation 2, we get:
R(smaller) = 1.3 Ω (OR) R(smaller) = 5.4 Ω
Since, it is given in the question that R(smaller)<R(larger).
Therefore, the correct answers will be:
R(smaller) = 1.3 Ω and R(larger) = 5.4 Ω
