Answer:
The rms voltage (in V) measured across the secondary coil is 459.62 V
Explanation:
Given;
number of turns in the primary coil, Np = 375 turns
number of turns in the secondary coil, Ns = 1875 turns
peak voltage across the primary coil, Ep = 130 V
peak voltage across the secondary coil, Es = ?
[tex]\frac{N_P}{N_s} = \frac{E_p}{E_s} \\\\E_s = \frac{N_sE_p}{N_p} \\\\E_s = \frac{1875*130}{375} \\\\E_s = 650 \ V[/tex]
The rms voltage (in V) measured across the secondary coil is calculated as;
[tex]V_{rms} = \frac{V_0}{\sqrt{2} } = \frac{E_s}{\sqrt{2} } \\\\V_{rms} = \frac{650}{\sqrt{2} } = 459.62 \ V[/tex]
Therefore, the rms voltage (in V) measured across the secondary coil is 459.62 V
A light wave will *Blank* if it enters a new medium perpendicular to the surface.
Answer:
A light wave will not stop if it enters a new medium perpendicular to the surface.
Explanation:
A light wave will not have any deviation if it enters a new medium perpendicular to the surface.
What is meant by refraction ?Refraction is defined as an optical phenomenon by which the direction of a light wave gets changed when it travels from one medium to another. This is because of the change in speed.
Here,
The light wave is entering a new medium such that it enters perpendicular to the surface. Angle of incidence is the angle between the incident ray and the line perpendicular to the surface at the point of incidence. Since, here the light ray is incident normal to the surface that means the angle of incidence is 0.
According to Snell's law,
sin i = μ sin r
where i is the angle of incidence, r is the angle of refraction and μ is the constant called refractive index.
As i = 0, sin i = 0
So, μ sin r = 0
Since μ is a constant, we can say that sin r = 0 or the angle of refraction,
r = 0
This means that there is no refraction and hence the light wave won't get deviated when it enters the medium normally.
Hence,
A light wave will not have any deviation if it enters a new medium perpendicular to the surface.
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A car of mass 410 kg travels around a flat, circular race track of radius 83.4 m. The coefficient of static friction between the wheels and the track is 0.286. The acceleration of gravity is 9.8 m/s 2 . What is the maximum speed v that the car can go without flying off the track
Answer:
The maximum speed v that the car can go without flying off the track = 15.29 m/s
Explanation:
let us first lay out the information clearly:
mass of car (m) = 410 kg
radius of race track (r) = 83.4 m
coefficient of friction (μ) = 0.286
acceleration due to gravity (g) = 9.8 m/s²
maximum speed = v m/s
For a body in a constant circular motion, the centripetal for (F) acting on the body is given by:
F = mass × ω
where:
F = maximum centripetal force = mass × μ × g
ω = angular acceleration = [tex]\frac{(velocity)^2}{radius}[/tex]
∴ F = mass × ω
m × μ × g = m × [tex]\frac{v^{2} }{r}[/tex]
410 × 0.286 × 9.8 = 410 × [tex]\frac{v^{2} }{83.4}[/tex]
since 410 is on both sides, they will cancel out:
0.286 × 9.8 = [tex]\frac{v^{2} }{83.4}[/tex]
2.8028 = [tex]\frac{v^{2} }{83.4}[/tex]
now, we cross-multiply the equation
2.8028 × 83.4 = [tex]v^{2}[/tex]
[tex]v^{2}[/tex] = 233.754
∴ v = √(233.754)
v = 15.29 m/s
Therefore, the maximum speed v that the car can go without flying off the track = 15.29 m/s
A carousel has a diameter of 6.0-m and completes one rotation every 1.7s. Find the centripetal acceleration of the traveler in m / s2.
Answer:
The centripetal acceleration of the traveler is [tex]40.98\ m/s^2[/tex]
Explanation:
It is given that, A carousel has a diameter of 6.0-m and completes one rotation every 1.7 s.
We need to find the centripetal acceleration of the traveler. It is given by the formula as follows :
[tex]a=\dfrac{v^2}{r}[/tex]
r is radius of carousel
[tex]v=\dfrac{2\pi r}{T}[/tex]
So,
[tex]a=\dfrac{4\pi ^2r}{T^2}[/tex]
Plugging all the values we get :
[tex]a=\dfrac{4\pi ^2\times 3}{(1.7)^2}\\\\a=40.98\ m/s^2[/tex]
So, the centripetal acceleration of the traveler is [tex]40.98\ m/s^2[/tex].
A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of 5.37x10-2 kg/s. The density of the gasoline is 739 kg/m3, and the radius of the fuel line is 3.37x10-3 m. What is the speed at which gasoline moves through the fuel line
Answer:
Speed v = 2.04 m/s
the speed at which gasoline moves through the fuel line is 2.04 m/s
Explanation:
Given;
Mass transfer rate m = 5.37x10^-2 kg/s.
Density d = 739 kg/m3
radius of pipe r = 3.37x10^-3 m
We know that;
Density = mass/volume
Volume = mass/density
Volumetric flow rate V = mass transfer rate/density
V = m/d
V = 5.37x10^-2 kg/s ÷ 739 kg/m3
V = 0.00007266576454 m^3/s
V = 7.267 × 10^-5 m^3/s
V = cross sectional area × speed
V = Av
Area A = πr^2
V = πr^2 × v
v = V/πr^2
Substituting the given values;
v = 7.267 × 10^-5 m^3/s/(π×(3.37x10^-3 m)^2))
v = 0.203678639672 × 10 m/s
v = 2.04 m/s
the speed at which gasoline moves through the fuel line is 2.04 m/s
A heavy copper ball of mass 2 kg is dropped from a fiftieth-floor apartment window. Another one with mass 1 kg is dropped immediately after 1 second. Air resistance is negligible. The difference between the speeds of the two balls:__________.
a. increases over time at first, but then stays constant.
b. decreases over time.
c. remains constant over time.
d. increases over time.
Answer:
C
Explanation:
Because everything on Earth falls at the same speed, the masses of the balls do not matter. Since the acceleration due to gravity is constant, their speeds will both be increasing at the same rate, and therefore the difference in speeds would remain constant until they hit the ground. Hope this helps!
A trap-jaw ant snaps its mandibles shut at very high speed, good for catching small prey. But these ants also slam their mandibles into the ground; the resulting force can launch the ant into the air for a quick escape. If a 12 mg ant hits the ground with an average force of 47 mN for a time of 0.13 ms, at what speed will it leave the ground?
Answer:
FInal speed (v) = 0.509 m/s (Approx)
Explanation:
Given:
Mass of ant (m) = 12 mg
Force (f) = 47 N
Time taken (t) = 0.13 ms
Find:
FInal speed (v) = ?
Computation:
Initial velocity (u) = 0
Impulse = change in momentum
Force × TIme = change in momentum
47 × 0.13 = mv - mu
6.11 = 12 (V)
FInal speed (v) = 0.509 m/s (Approx)
Our Sun shines bright with a luminosity of 3.828 x 1025 Watt. Her energies
responsible for many processes and the habitable temperatures on the earth that
make our life possible.
a) Calculate the amount of energy arriving on the Earth in a single day
b) To how many litres of heating oil (energy density 37.3 x 10^6 J/litre is the equivalent?
C) The Earth reflects 30% of this energy : Determine the temperature on Earth's sufact
d) what other factors should be considered to get an even more precisa temperature postiache
Note: The Earth's radius is 6370km; the Sun's sadius is 696 ×10^3km, I AU is 1.495 × 10^8km)
Answer:
a) E = 1.58 10²¹ J , b) Oil = 4,236 107 liter , e) T = 54.3 C
Explanation:
a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface
I = P / A
A = 4π r²
in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m
I = P / A
I = P / 4π r²
let's calculate
I = 3,828 10²⁵/4 pi (1.5 10¹¹)²
I = 1.3539 10²W / m² = 135.4 W / m2
the energy that reaches the disk of the Earth is
E = I A
the area of a disc
A = π r²
E = I π r²
where r is the radius of the Earth 6.37 10⁶ m
E = 135.4 π(6.37 10⁶)
E = 1,726 10¹⁶ W
This is the energy per unit of time that reaches Earth
t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s
E = 1,826 10¹⁶ 86400
E = 1.58 10²¹ J
b) for this part we can use a direct proportions rule
Oil = 1.58 10²¹ (1 / 37.3 10⁶)
Oil = 4,236 10⁷ liter
c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law
P = σ A e T⁴
T = [tex]\sqrt[4]{P/Ae}[/tex]
nos indicate the refect, therefore the amount of absorbencies
P_absorbed = 0.7 P
let's calculate
T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)
T = RER (8,676 106)
T = 54.3 C
b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere
small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct? A small car has a head-on collision with a large truck. Which of the following statements concerning the magnitude of the average force due to the collision is correct? It is impossible to tell since the velocities are not given. The truck experiences the greater average force. It is impossible to tell since the masses are not given. The small car and the truck experience the same average force. The small car experiences the greater average force.
Answer:
The correct option is D: "The small car and the truck experience the same average force."
Explanation:
The magnitude of the average force experienced by both bodies in motion is the same as explained by Newton's third law of motion. The force exerted by each body is equal and opposite in direction. The resulting acceleration experienced by each vehicle, however, will not be the same. It is greater for the small car.
A ship can float on water as long as it weighs less than water.
O A. True
O B. False
Answer:
It's true
Explanation:
Because the ship is mafe up of aluminium, which is a light metal.
Answer:
False
Explanation:
Took The Quiz
g 95 N force exerted at the end of a 0.50 m long torque wrench gives rise to a torque of 15 N • m. What is the angle (assumed to be less than 90°) between the wrench handle and the direction of the applied force?
Answer:
Angle = 18.41°
Explanation:
Torque = F•r•sin θ
where;
F = force
r = distance from the rotation point
θ = the angle between the force and the radius vector.
We are given;
Torque = 15 N.m
F = 95 N
r = 0.5 m
Thus, plugging in the relevant values ;
15 = 95 × 0.5 × sin θ
sin θ = 15/(95 × 0.5)
sin θ = 0.3158
θ = sin^(-1)0.3158
θ = 18.41°
A turntable rotates with a constant 1.85 rad/s2 clockwise angular acceleration. After 4.00 s it has rotated through a clockwise angle of 30.0 rad . Part A What was the angular velocity of the wheel at the beginning of the 4.00 s interval?
Answer: The angular velocity of the wheel at the beginning of the 4.00 s interval is 3.8 rad/s
Explanation: Please see the attachment below
The angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.
The given parameters:
Angular speed of the turn table = 1.85 rad/s²Time of motion, t = 4.0 sAngular displacement, θ = 30.0 radThe angular velocity of the wheel at the beginning of the 4.0 s time is calculated as follows;
[tex]\theta = \omega_i t + \frac{1}{2} \alpha t^2[/tex]
where;
[tex]\omega_i[/tex] is the initial angular velocity
[tex]30 = \omega_i (4) \ + \frac{1}{2}(1.85)(4)^2\\\\30 = 4\omega _i + 14.8\\\\4\omega _i = 30 - 14.8\\\\ 4\omega _i = 15.2\\\\\omega _i = \frac{15.2}{4} \\\\\omega _i = 3.8 \ rad/s[/tex]
Thus, the angular velocity of the wheel at the beginning of the 4.0 s time is 3.8 rad/s.
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You are comparing a reaction that produces a chemical change and one that produces a physical change. What evidence could you use to determine which type of change is occurring?
Answer: A chemical change results from a chemical reaction, while a physical change is when matter changes forms but not chemical identity. Examples of chemical changes are burning, cooking, rusting, and rotting. Examples of physical changes are boiling, melting, freezing, and shredding. Often, physical changes can be undone, if energy is input.
Explanation: hope this helps have a good day
Answer:
If the reaction is a chemical change, new substances with different properties and identities are formed. This may be indicated by the production of an odor, a change in color or energy, or the formation of a solid.
What is the answer for this question
An object with a mass of 1500 g (grams) accelerates 10.0 m/s2 when an
unknown force is applied to it. What is the amount of the force
Answer:
15N
Explanation:
F=ma
m=1500g = 1.5kg
a=10m/s2
1.5×10=15 N
Answer:15000gms^-2
Explanation:
F=m×a
m=1500g, a=10ms^-2
F=(1500×10)gms^-2
F=15000gms^-2
0.92 kg of R-134a fills a 0.14-m^3 weighted piston–cylinder device at a temperature of –26.4°C. The container is now heated until the temperature is 100°C. Determine the final volume of R-134a.
Answer:
The final volume of R-134a is 0.212m³Explanation:
Using one of the general gas equation to find the final volume of the R-134a.
According to pressure law; The volume of a given mas of gas is directly proportional to its temperature provided that the pressure remains constant.
VαT
V = kT
k = V/T
V1/T1 = V2/T2 = k
Given V1 = 0.14-m³ at T1 = –26.4°C = –26.4° + 273 = 246.6K
V2 = ? at T = 100°C = 100+273 = 373K
On substituting this values for T2;
0.14/246.6 = V2/373
373*0.14 = 246.6V2
V2 = 373*0.14 /246.6
V2 = 0.212m³
The final volume of R-134a is 0.212m³
In the Life Cycle of Stars diagram, what stage does letter J represent?
A.) white dwarf
B.) black dwarf
C.) black hole
D.) neutron star
Which letters in the Life Cycle of Stars diagram represent stars on the main sequence?
A.) F & I
B.) C & G
C.) A & E
D.) B & D
In the Life Cycle of Stars diagram, what stage does letter L represent?
A.) neutron star
B.) black hole
C.) white dwarf
D.) black dwarf
In the Life Cycle of Stars diagram, what stage does letter I represent?
A.) neutron star
B.) black dwarf
C.) black hole
D.) white dwarf
In the Life Cycle of Stars diagram, what does letter D represent?
A.) a high mass star
B.) a white dwarf
C.) a cool star
D.) a low mass star
In the Life Cycle of Stars diagram, what stage does letter C represent?
A.) nuclear fusion
B.) a supernova
C.) a planetary nebula
D.) protostar formation
Which letter in the Life Cycle of Stars diagram represents a star forming region of space?
A.) M
B.) H
C.) J
D.) G
Which letter in the Life Cycle of Stars diagram represents a planetary nebula?
Group of answer choices
A.) G
B.) H
C.) L
D.) M
ANSWER: num 1 is black hole
Photoelectric effect:
A. What is the maximum kinetic energy of electrons ejected from barium (W0=2.48eV) when illuminated by white light, lambda=410-750nm?
B. The work functions for sodium, cesium, copper, and iron are 2.3, 2.1, 4.7, and 4.5eV, respectively. Which of these metals will not emit electrons when visible light shines on it?
Answer:
A. K = 0.546 eV
B. cooper and iron will not emit electrons
Explanation:
A. This is a problem about photoelectric effect. Then you have the following equation:
[tex]K=h\nu-\Phi=h\frac{c}{\lambda} -\Phi[/tex] (1)
K: kinetic energy of the ejected electron
Ф: Work function of the metal = 2.48eV
h: Planck constant = 4.136*10^{-15} eV.s
λ: wavelength of light = 410nm - 750nm
c: speed of light = 3*10^8 m/s
As you can see in the equation (1), higher the wavelength, lower the kinetic energy. Then, the maximum kinetic energy is obtained with the lower wavelength (410nm). Thus, you replace the values of all variables :
[tex]K=(4.136*10^{-15}eV)\frac{3*10^8m/s}{410*10^{-9}m}-2.48eV\\\\K=0.546eV[/tex]
B. First you calculate the energy of the photon with wavelengths of 410nm and 750nm
[tex]E_1=(4.136*10^{-15}eV)\frac{3*10^{8}m/s}{410*10^{-9}m}=3.02eV\\\\E_2=(4.13610^{-15}eV)\frac{3*10^{8}m/s}{750*10^{-9}m}=1.6544eV[/tex]
You compare the energies E1 and E2 with the work functions of the metals and you can conclude:
sodium = 2.3eV < E1
cesium = 2.1 eV < E1
cooper = 4.7eV > E1 (this metal will not emit electrons)
iron = 4.5eV > E1 (this metal will not emit electrons)
You are watching an object that is moving in SHM. When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left. Part A How much farther from this point will the object move before it stops momentarily and then starts to move back to the left
Answer:
2.95m
Explanation:
The farthest distance the object can move is the radius of the circle of which the Simple harmonic motion is assumed to be a part
But V = w× r; where V is velocity,
w is angular velocity and r is radius.
Also,
a= w2r; where a is linear acceleration
but a = v× r ; by comparing both equations
Hence r = a/v =8.6/2.45 =3.51m
But the horizontal distance of the motion is given by:
X = rcosx ; where x is the angle
X is the distance covered.
We know that the maximum value of cos x is 1 which is 0°
When the object moves in a fashion directly parallel to an horizontal distance, maximum distance would be reached and hence:
X = r=3.51m
Meaning the object needs to travel 3.51-0.56=2.95m further.
Note: the acceleration of the motion is constant whether it is swinging towards the left or right.
When the object is displaced 0.560 m to the right of its equilibrium position, it has a velocity of 2.45 m/s to the right and an acceleration of 8.60 m/s2 to the left and the amplitude of motion A = 0.732 m.
What is Amplitude of motion?
The distance between the central and extreme points for a moving particle is known as the amplitude of motion.
The given data to find the amplitude of motion,
Object displaced = 0.560 m
Velocity = 2.45 m/s
Acceleration = 8.60 m/s²
Starting with sine:
x(t) = Asin(ωt)
so that t = 0, x = 0
x(t) = 0.56 m = Asin(ωt)
v(t) = x(t)'= 2.45 m/s = Aωcos(ωt)
a(t) = v(t)'= -8.60 m/s² = -Aω²sin(ωt)
x(t) / a(t) = Asin(ωt) / -Aω²sin(ωt)
0.56m / -8.60 m/s² = -1 / ω²
ω² = 15.3571 rad^2/s^2
ω = 3.91881 rad/s
x(t) / v(t) = Asin(ωt) / Aωcos(ωt)
0.560m / 2.45m/s = tan(3.91t) / 3.91rad/s
0.8937= tan(3.91t)
t = 0.176 s
x(0.176) = Asin(3.59×0.176)
0.65 m= Asin(0.631)
A = 0.732 m is the amplitude of motion.
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Zinc is added to a breaker containing hydrochloric acid and the beaker gets warm what type os reaction is this
Answer:
Exothermic
Explanation:
Depending on the unit you are in, the answer may vary.
This is an exothermic reaction because it produces heat (the beaker gets warm).
For the parallel plates mentioned above, the DC power supply is set to 31.5 Volts and the plate on the right is at x = 14 cm. What is the magnitude of the electric field at a point on the x-axis where x = 7.0 cm? Answer with a number in the format ### in Newtons per Coulombs.
Note: The complete question is attached as a file to this solution. The parallel plate mentioned can be seen in this picture attached.
Answer:
E = 225 N/C
Explanation:
Note: At any point on the parallel plates of a capacitor, the electric field is uniform and equal.
Therefore, Electric field at x = 14 cm equals the electric field at x = 7 cm
V(x) = 31.5 Volts
x = 14 cm = 0.14 m
The magnitude of the electric field at any point between the parallel plate of the capacitor is given by the equation:
E = V(x)/d
E(x = 0.14) = 31.5/0.14
E(x=0.14) = 225 N/C
E(x=0.14) = E(x=0.07) = 225 N/C
Determined to test the law of gravity for himself, a student walks off a skyscraper 180 m high, stopwatch in hand, and starts his free fall (zero initial velocity). Five seconds later, Superman arrives at the scene and dives off the roof to save the student.
a) Superman leaves the roof with an initial velocity that he produces by pushing himself downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. What must the value of the initial velocity be so that Superman catches the student just before they reach the ground?
b) On the same graph, sketch the positions of the student and of Superman as functions of time. Take Superman's initial speed to have the value calculated in part (a).
c) If the height of the skyscraper is less than some minimum value, even Superman can't reach the student before he hits the ground. What is this minimum height?
Answer:
a) v₀ = - 164.62 m / s , c) y = 122.5 m
Explanation:
We can solve this exercise using the free fall kinematic relations.
We put our reference system on the floor, so the height of the skyscraper is y₀ = 180m and the floor level is y = 0 m
For the boy
y = y₀ + v₀ t - ½ g t²
with free fall its initial speed is zero
y = ½ g t2
For superman
y = y₀ + v₀ (t-5) - ½ g (t-5)²
how superman grabs the lot just before hitting the ground
we look for the time it takes the boy down
t = √ (2 y₀ / g)
t = √ (2 180 / 9,8)
t = 6.06 s
in the equation for superman, we clear the volume and calculate
v₀ (t-5) = -y₀ + ½ g (t-5)²
v₀ (6.06 -5) = -180 + ½ 9.8 (6.06 -5)²
v₀ 1.06 = -174.49
v₀ = - 174.49 / 1.06
v₀ = - 164.62 m / s
the negative sign indicates that the initial speed is down
b) to graph the position of the two we use the table
t (s) Y_boy (m) Y_superman (m)
0 180 180
1 175.1 180
5 57.5 180
6 3.6 10.18
see attachment for the two curves
c) calculate the height that falls a lot in the 5 seconds (t = 5)
y = -1/2 g t²
y = ½ 9.8 5²
y = 122.5 m
for this height superman has not yet left the skyscraper, so the boy hits the ground
Coherent light that contains two wavelengths, 660 nm and 470 nm , passes through two narrow slits with a separation of 0.280 mm and an interference pattern is observed on a screen which is a distance 5.50 m from the slits.
Required:
What is the disatnce on the screen between the first order bright fringe for each wavelength?
Answer:
λ1 = 0.0129m = 1.29cm
λ2 = 0.00923m = 0.92 cm
Explanation:
To find the distance between the first order bright fringe and the central peak, can be calculated by using the following formula:
[tex]y_m=\frac{m\lambda D}{d}[/tex] (1)
m: order of the bright fringe = 1
λ: wavelength of the light = 660 nm, 470 nm
D: distance from the screen = 5.50 m
d: distance between slits = 0.280mm = 0.280 *10^⁻3 m
ym: height of the m-th fringe
You replace the values of the variables in the equation (1) for each wavelength:
For λ = 660 nm = 660*10^-9 m
[tex]y_1=\frac{(1)(660*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.0129m=1.29cm[/tex]
For λ = 470 nm = 470*10^-9 m
[tex]y_1=\frac{(1)(470*10^{-9}m)(5.50m)}{0.280*10^{-3}m}=0.00923m=0.92cm[/tex]
A particle with a charge of 5.1 μC is 3.02 cm from a particle with a charge of 2.51 μC . The potential energy of this two-particle system, relative to the potential energy at infinite separation, is
Answer:
U = 3.806 J
Explanation:
The potential energy between the two charges q1 and q2, is given by the following formula:
[tex]U=k\frac{q_1q_2}{r}[/tex] (1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
q1 = 5.1*10^-6 C
q2 = 2.51*10^-6 C
r: distance of separation between particles = 3.02cm = 3.02*10^-2 m
You replace the values of all parameters in the equation (1):
[tex]U=(8.98*10^9Nm^2/C^2)\frac{(5.1*10^{-6}C)(2.51*10^{-6}C)}{3.02*10^{-2}m}\\\\U=3.806J[/tex]
The potential energy of the two particle system is 3.806 J
A car travels around an oval racetrack at constant speed. The car is accelerating:________.
A) at all points except B and D.
B) at all points except A, B, C, and D.
C) everywhere, including points A, B, C, and D.
D) nowhere, because it is traveling at constant speed.
2) A moving object on which no forces are acting will continue to move with constant:_________
A) Acceleration
B) speed
C) both of theseD) none of these
Answer:
1A,2D,3B
Explanation:
hope this helps
Car A is traveling at twice the speed of car B. They both hit the brakes at the same time and decrease their velocities at the same rate. If car B travels a distance D before stopping, how far does car A travel before stopping?
A) 4D
B) 2D
C) D
D) D/2
E) D/4
Answer:
A) 4D
Explanation:
The distance traveled by the cars before coming to rest can be determined by 3rd equation of motion:
2as = Vf² - Vi²
s = (Vf² - Vi²)/2a
where,
s = distance traveled
Vf = Final Speed = 0 m/s
Vi = Initial Speed
a = deceleration rate
First, we consider Car B and we assign a subscript 2 for it:
Vf₂ = 0 m/s (As, car finally stops)
s₂ = D
a₂ = - a (due to deceleration)
D = (0² - Vi₂²) /(-2a)
D = Vi₂²/2a -------- equation (1)
Now, we consider Car A and we assign a subscript 1 for it:
Vf₁ = 0 m/s (As, car finally stops)
s₁ = ?
a₁ = - a (due to deceleration)
Vi₁ = 2 Vi₂ (Since, car A was initially traveling at twice speed of car B)
s₁ = (0² - Vi₁²) /(-2a)
s₁ = (2Vi₂)²/2a
s₁ = 4 (Vi₂²/2a)
using equation (1), we get:
s₁ = 4D
Therefore, the correct option is:
A) 4D
Organ pipe A, with both ends open, has a fundamental frequency of 475 Hz. The third harmonic of organ pipe B, with one end open, has the same frequency as the second harmonic of pipe A. Use 343 m/s for the speed of sound in air. How long are (a) pipe A and (b) pipe B?
Answer:
The length of organ pipe A is [tex]L = 0.3611 \ m[/tex]
The length of organ pipe B is [tex]L_b = 0.2708 \ m[/tex]
Explanation:
From the question we are told that
The fundamental frequency is [tex]f = 475 Hz[/tex]
The speed of sound is [tex]v_s = 343 \ m/s[/tex]
The fundamental frequency of the organ pipe A is mathematically represented as
[tex]f= \frac{v_s}{2 L}[/tex]
Where L is the length of organ pipe
Now making L the subject
[tex]L = \frac{v_s}{2f}[/tex]
substituting values
[tex]L = \frac{343}{2 *475}[/tex]
[tex]L = 0.3611 \ m[/tex]
The second harmonic frequency of the organ pipe A is mathematically represented as
[tex]f_2 = \frac{v_2}{L}[/tex]
The third harmonic frequency of the organ pipe B is mathematically represented as
[tex]f_3 = \frac{3 v_s}{4 L_b }[/tex]
So from the question
[tex]f_2 = f_3[/tex]
So
[tex]\frac{v_2}{L} = \frac{3 v_s}{4 L_b }[/tex]
Making [tex]L_b[/tex] the subject
[tex]L_b = \frac{3}{4} L[/tex]
substituting values
[tex]L_b = \frac{3}{4} (0.3611)[/tex]
[tex]L_b = 0.2708 \ m[/tex]
A metal ring 4.60 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.280 T/s.
A. What is the magnitude of the electric field induced in the ring?
B. In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?1. Counterclockwise2. Clockwise
Answer:
A. Ein = 8.05*10^-4 V/m
B. Clockwise sense
Explanation:
A. the magnitude of the electric field induced in the ring is obtaind by using the following formula:
[tex]\int E_{in} \cdot ds=-\frac{d\Phi_B}{dt}[/tex] (1)
Ein: induced electric field
ds: differential of a path of the ring
ФB: magnetic flux in the ring
The Ein vector is parallel to ds in the complete ring. Furthermore, the area of the ring is constant, hence, you have in the equation (1):
[tex]\int E_{in}ds=E_{in}(2\pi r)=-A\frac{dB}{dt}\\\\E_{in}=-\frac{A}{2\pi r}\frac{dB}{dt}[/tex] (2)
dB/dt = -0.280T/s (it is decreasing)
A: area of the ring = π(r/2)^2= (π/4) r^2
r: radius of the ring = 4.60/2 = 2.30 cm
Then, you replace the values of all variables in the equation (2):
[tex]E_{in}=-\frac{(\pi/4)r^2}{2\pi r}\frac{dB}{dt}=\frac{r}{8}\frac{dB}{dt}\\\\E_{in}=-\frac{0.0230m}{8}(-0.280T)=8.05*10^{-4}\frac{V}{m}[/tex]
hence, the induced electric field is 8.05*10^-4 V/m
B. The induced current in the ring produced a magnetic field that is opposite to the magnetic field of the magnet. The, in this case you have that the induced current is in a clockwise sense.
Which person will most likely hear the loudest sound?
A
B
C
D
Answer:
The youngest person
Explanation:
Hearing worsens with age
Please mark brainliest
Answer:
A
Explanation:
The person closest to the origin of the sound will most likely hear the loudest sound. ^^
Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the horizontal) under the following road conditions. You may assume that the weight of the car is evenly distributed on all four tires and that the static coefficient of friction is involved—that is, the tires are not allowed to slip during the deceleration.
The question is incomplete. Here is the complete question.
Calculate the maximum deceleration of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.
Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²
Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:
[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]
The [tex]W_x[/tex] is a x-component of force due to gravity (W) and, in this case, is given by: [tex]W_x[/tex] = W.sin(14)
W is described as: W = m.g
Force due to friction ([tex]f_s[/tex]) is given by: [tex]f_s[/tex] = μs.N
N is the normal force and, in the system, is equivalent of [tex]W_y[/tex], so:
[tex]W_y[/tex] = m.g.cos(14)
Therefore, the formula will be:
[tex]F_{net}[/tex] = - [tex]W_x[/tex] - [tex]f_s[/tex]
m.a = - (m.g.sin14) - (μs.mg.cos14)
a = - g (sin14 + μscos 14)
a) For dry concrete, μs = 1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 1.cos14)
a = - 11.05 m/s²
b) For wet concrete, μs = 0.7:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin 14 + 0.7.cos14)
a = - 10.64 m/s²
c) For ice, μs = 0.1:
a = - g (sin14 + μscos 14)
a = - 9.8 (sin14 + 0.1cos14)
a = - 9.84 m/s²
In this problem you will consider the balance of thermal energy radiated and absorbed by a person.Assume that the person is wearing only a skimpy bathing suit of negligible area. As a rough approximation, the area of a human body may be considered to be that of the sides of a cylinder of length L=2.0m and circumference C=0.8m.For the Stefan-Boltzmann constant use σ=5.67×10−8W/m2/K4.Part aIf the surface temperature of the skin is taken to be Tbody=30∘C, how much thermal power Prb does the body described in the introduction radiate?Take the emissivity to be e=0.6.Express the power radiated into the room by the body numerically, rounded to the nearest 10 W.part bFind Pnet, the net power radiated by the person when in a room with temperature Troom=20∘C
Answer:
The thermal power emitted by the body is [tex]P_t = 286.8 \ Wm^{-2}[/tex]
The net power radiated is [tex]P_{net} = 460 \ W[/tex]
Explanation:
From the question we are told that
The length of the assumed hum[tex]T_{room} = 20 ^oC[/tex]an body is L = 2.0 m
The circumference of the assumed human body is [tex]C = 0.8 \ m[/tex]
The Stefan-Boltzmann constant is [tex]\sigma = 5.67 * 10^{-8 } \ W\cdot m^{-2} \cdot K^{-4}.[/tex]
The temperature of skin [tex]T_{body} = 30^oC[/tex]
The temperature of the room is
The emissivity is e=0.6
The thermal power radiated by the body is mathematically represented as
[tex]P_t = e * \sigma * T_{body}^4[/tex]
substituting value
[tex]P_t = 0.6 * 5.67*10^{-8} * (303)^4[/tex]
[tex]P_t = 286.8 \ Wm^{-2}[/tex]
The net power radiated by the body is mathematically evaluated as
[tex]P_{net} = P_t * A[/tex]
Where A is the surface area of the body which is mathematically evaluated as
[tex]A = C* L[/tex]
substituting values
[tex]A = 0.8 * 2[/tex]
[tex]A = 1.6 m^2[/tex]
=> [tex]P_{net} = 286.8 * 1.6[/tex]
=> [tex]P_{net} = 460 \ W[/tex]
