Answer:
Explanation:
Refraction is defined as the bending of light rays as an incident ray pass from one medium to another. If the incident ray is passing from the media with low refractive index to a greater refractive index, the refracted ray tends to bend away from the normal.
Refractive index is the ratio of the sin of angle of incidence to the sine of angle of refraction.
n = sin i/sin r
For us to have a greater index of refraction, the denominator must be lesser than the numerator. This means that the angle of refraction must be smaller and if the angle of refraction must get smaller, this means that the refracted ray must bend towards the normal
Need help understanding this. If anyone help, that would be greatly appreciated!
Answer:
8.33` m/s^2 and 8333.3 N
Explanation:
a) acceleration:
ā=v^2/r
ā=(15m/s)^2/27m
ā=225/27 m/s^2
ā=8.333 m/s^2
force:
F=mā. where the is equal to v^2/r
F=1000kg*8.3 m/s^2
F=8333.3 N
Answer:
8.33` m/s^2 and 8333.3 N
Explanation:
g How many rpm would a 25 m diameter Ferris wheel need to travel if a 75 kg person were to experience an effective weight of 810 N at the lower-most point of the ride
Answer:
2.52 rpm
Explanation:
given that
diameter of the wheel, d = 25 m
Mass of the person, m = 75 kg
Weight experienced, N = 810 N
Since diameter is 25, radius then is 25/2 = 12.5 m
We all know that,
v = rw
Also, the passengers weight is equal to the centripetal acceleration, and thus
mg = mv²/r
Substitute for v, we have
mg = m/r * (rw)²
mg = mr²w²/r
g = rw²
If we make w the subject of formula, we have
w² = g/r
w = √(g/r)
mg = 810
75 * g = 810
g = 810 / 75
g = 1.08 m/s²
w = √(g/r)
w = √(1.08 / 12.5)
w = √0.0864
w = 0.294 rad/s
Since the question asked us in rpm, we convert to rpm
0.294 * (60 / 2π)
2.52 revolution per minute.
A typical electric oven has two separate heating elements: one on top and one on the bottom. The bottom element is used for baking while the top element is used to broil foods. When only the bottom element is active and glowing red hot, what heat transfer mechanisms carry most of the heat to the food in the oven?
Answer:
Convection and Radiation mechanisms carry most of the heat
Explanation:
This is because Convection proceeds strongy as heated air rises from the hot element while Radiation is also strong, although the material of the cooking pots will how effective it is.
a positively charged ion, due to a cosmic ray, is headed through earth's atmosphere toward the center of Earth. Due to Earth's magnetic field, the ion will be delfected:
Answer:
East direction
Explanation:
Given that
Charge on the particle is positive.
Moving towards the center of earth .
We know that N(north ) pole in magnetic fields work as source of magnetic lines and S(South ) pole works and sink for magnetic lines.
Therefore due to the earth magnetic fields , the positive ions will deflect towards East direction.
Thus the answer will be East direction.
An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.285 T. If the kinetic energy of the electron is 2.10 10-19 J, find the speed of the electron and the radius of the circular path. (a) the speed of the electron
Answer:
The speed of the electron is 6.79 x 10⁵ m/s
The radius of the circular path is 1.357 x 10⁻⁵ m
Explanation:
Given;
magnetic field, B = 0.285 T
energy of electron, E = 2.10 x 10⁻¹⁹ J
The kinetic energy of the electron is calculated as;
[tex]K.E = \frac{1}{2} m_eV^2[/tex]
Where;
[tex]m_e[/tex] is the mass of electron = 9.11 x 10⁻³¹ kg
V is the speed of the electron
[tex]K.E = \frac{1}{2} m_eV^2\\\\V^2 = \frac{2.K.E}{m_e} \\\\V = \sqrt{\frac{2K.E}{m_e} } \\\\V = \sqrt{\frac{2*(2.1*10^{-19})}{9.11*10^{-31}} }\\\\V = 6.79 *10^{5} \ m/s[/tex]
The radius of the circular path is given by;
[tex]R = \frac{M_eV}{qB}[/tex]
where;
q is the charge of the electron = 1.6 x 10⁻¹⁹ C
[tex]R = \frac{M_eV}{qB} \\\\R = \frac{9.11 *10^{-31}*6.79 *10^{5}}{1.6*10^{-19}*0.285} \\\\R = 1.357 *10^{-5} \ m[/tex]
which example describes a nonrenewable resource?
A. everyone in our neighborhood uses solar panels to generate electricity to run their pool pumps.
B. once up and running, the power plant will convert the energy from tides and waves into electricity.
C. there is a long stretch of land in the desert with many windmills that are able to generate enough electricity to run the town.
D. there are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity.
Answer:
D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity
Explanation:
Solar panels are a renewable resource because the sun will not run out. The power plant uses water, so it is also a renewable resource. Windmills use wind, and wind will not run out so it is a renewable resource. However, natural gas and oil are not renewable resources because they will run out one day.
What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an angle of 62.5°?
Answer:
The distance is [tex]d = 1.747 *10^{-6} \ m[/tex]
Explanation:
From the question we are told that
The order of maximum diffraction is m = 2
The wavelength is [tex]\lambda = 775 nm = 775 * 10^{-9} \ m[/tex]
The angle is [tex]\theta = 62.5^o[/tex]
Generally the condition for constructive interference for diffraction grating is mathematically represented as
[tex]dsin \theta = m * \lambda[/tex]
where d is the distance between the lines on a diffraction grating
So
[tex]d = \frac{m * \lambda }{sin (\theta )}[/tex]
substituting values
[tex]d = \frac{2 * 775 *1^{-9} }{sin ( 62.5 )}[/tex]
[tex]d = 1.747 *10^{-6} \ m[/tex]
Suppose you wish to make a solenoid whose self-inductance is 1.8 mH. The inductor is to have a cross-sectional area of 1.6 x 10-3 m2 and a length of 0.066 m. How many turns of wire are needed
Answer:
The number of turns of the wire needed is 243 turns
Explanation:
Given;
self inductance of the solenoid, L = 1.8 mH
cross sectional area of the inductor, A = 1.6 x 10⁻³ m²
length of the inductor, l = 0.066 m
The self inductance of long solenoid is given by;
L = μ₀n²Al
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ H/m
n is number of turns per length
A is the area of the solenoid
l is length of the solenoid
[tex]n = \sqrt{\frac{L}{\mu_o Al} } \\\\n = \sqrt{\frac{1.8*10^{-3}}{(4\pi*10^{-7}) (1.6*10^{-3})(0.066)} } \\\\n = \sqrt{13562583.184} \\\\n = 3682.74 \ turns/m[/tex]
The number of turns is given by;
N = nL
N = (3682.74)(0.066)
N = 243 turns
Therefore, the number of turns of the wire needed is 243 turns
A 0.210-kg metal rod carrying a current of 11.0 A glides on two horizontal rails 0.490 m apart. If the coefficient of kinetic friction between the rod and rails is 0.200, what vertical magnetic field is required to keep the rod moving at a constant speed?
Answer:
The magnetic field is [tex]B = 0.0764 \ T[/tex]
Explanation:
From the question we are told that
The mass of the metal is [tex]m = 0.210 \ kg[/tex]
The current is [tex]I = 11.0 \ A[/tex]
The distance between the rail(length of the rod ) is [tex]d = 0.490 \ m[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.200[/tex]
Generally the magnetic force is mathematically represented as
[tex]F_b = B * I * d[/tex]
Given that the rod is moving at a constant velocity, it
=> [tex]F_b = F_k[/tex]
Where [tex]F_k[/tex] is the kinetic frictional force which is mathematically represented as
[tex]F_k = \mu_k * m * g[/tex]
So
[tex]B * I * d = \mu_k * m * g[/tex]
=> [tex]B = \frac{\mu_k * m * g}{I * d }[/tex]
substituting values
=> [tex]B = \frac{0.200 * 0.210 * 9.8 }{ 11 * 0.490 }[/tex]
=> [tex]B = 0.0764 \ T[/tex]
Ohm’s Law
pls answer this photos
Answer:
Trial 1: 2 Volts, 0 %
Trial 2: 2.8 Volts, 0%
Trial 3: 4 Volts, 0 %
Explanation:
Th experimental values are given in the table, while the theoretical value can be found by using Ohm/s Law:
V = IR
TRIAL 1:
V = IR
V = (0.1 A)(20 Ω)
V = 2 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2 - 2)/2| x 100%
% Difference = 0 %
TRIAL 2:
V = IR
V = (0.14 A)(20 Ω)
V = 2.8 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(2.8 - 2.8)/2.8| x 100%
% Difference = 0 %
TRIAL 3:
V = IR
V = (0.2 A)(20 Ω)
V = 4 volts
% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%
% Difference = |(4 - 4)/4| x 100%
% Difference = 0 %
A converging lens of focal length 7.40 cm is 18.0 cm to the left of a diverging lens of focal length -7.00 cm . A coin is placed 12.0 cm to the left of the converging lens.
A) Find the location of the coin's final image relative to the diverging lens.
B) Find the magnification of the coin's final image.
Answer:
Explanation:
The set up is a compound microscope. The converging lens is the objective lens while the diverging lens is the eyepiece lens.
In compound microscopes, the distance between the two lenses is expressed as L = v0+ue
v0 is the image distance of the objective lens and ue is the object distance of the eye piece lens.
Befre we can get the location of the coin's final image relative to the diverging lens (ve), we need to get ue first.
Given L = 18.0cm
Using the lens formula to get v0 where u0 = 12.0cm and f0 = 7.40cm
1/f0 = 1/u0+1/v0
f0 and u0 are the focal length and object distance of the converging lens (objective lens)respectively.
1/v0 = 1/7.4-1/12
1/v0 = 0.1351-0.0833
1/v0 = 0.0518
v0 = 1/0.2184
v0 = 19.31cm
Note that v0 = ue = 19.31cm
To get ve, we will use the lens formula 1/fe = 1/ue+1/ve
1/ve = 1/fe-1/ue
Given ue = 19.31cm and fe = -7.00cm
1/ve = -1/7.0-1/19.31
1/ve = -0.1429-0.0518
1/ve = -0.1947
ve = 1/-0.1947
ve = -5.14cm
Hence, the location of the coin's final image relative to the diverging lens is 5.14cm to the lens
b) Magnification of the final image M = ve/ue
M = 5.14/19.31
M = 0.27
Magnification of the final image is 0.27
You are in the frozen food section of the grocery store and you notice that your hand gets cold when you place it on the glass windows of the display cases. Your friend says this is because coolness is transferred from the display case to your hand. What do you think?
Answer:
I think my friend got it all wrong, as coolness can not be transferred but heat was actually transferred between my hand and the glass windows
Explanation:
In thermodynamics, coolness can not be transferred, only heat can be transferred
Here is how the mechanism of why i felt cold works, my body gave out heat, hence there was heat transfer from a region of high to a low heat region, equilibrium was reached and I started feeling the coolness in my hands.
Light emitted by element X passes through a diffraction grating that has 1200 slits/mm. The interference pattern is observed on a screen 77.0 cm behind the grating. First-order maxima are observed at distances of 58.0 cm , 65.4 cm , and 94.5 cm from the central maximum. What are the wavelengths of light emitted by element X?
Answer:
500 nm, 530 nm, 650 nm
Explanation:
Let's say that there is diffraction grating observed with a slit spacing of s. Respectively we must determine the angle θ which will help us determine the 3 wavelengths ( λ ) of the light emitted by element X. This can be done applying the following formulas,
s( sin θ ) = m [tex]*[/tex] λ, such that y = L( tan θ ) - where y = positioning, or the distance of the first - order maxima, and L = constant, of 77 cm
Now the grating has a slit spacing of -
s = 1 / N = 1 / 1200 = 0.833 [tex]*[/tex] 10⁻³ mm
The diffraction angles of the " positionings " should thus be -
θ = tan⁻¹ [tex]*[/tex] ( 0.58 / 0.77 ) = 37°,
θ = tan⁻¹ [tex]*[/tex] ( 0.654 / 0.77 ) = 40°,
θ = tan⁻¹ [tex]*[/tex] ( 0.945 / 0.77 ) = 51°
The wavelengths of these three bright fringes should thus be calculated through the formula : λ = s( sin θ ) -
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 37° ) = ( 500 [tex]*[/tex] 10⁻⁹ m )
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 40° ) = ( 530 [tex]*[/tex] 10⁻⁹ m )
λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 51° ) = ( 650 [tex]*[/tex] 10⁻⁹ m )
Wavelengths : 500 nm, 530 nm, 650 nm
This question will be solved using the "grating equation".
The wavelengths of the light emitted by element X are:
"1. 6.654 x 10⁻⁷ m = 665.4 nm
2. 6.349 x 10⁻⁷ m = 634.9 nm
3. 5.262 x 10⁻⁷ m = 526.2 nm"
The diffraction grating equation is given as follows:
[tex]m\lambda = d Sin\ \theta[/tex]
where,
m = order of maxima = 1
λ = wavelength of light = ?
d = grating element = [tex]\frac{1}{no.\ of\ slits\ per\ unit\ length} = \frac{1}{1200\ slits/mm}[/tex]
d = (8.33 x 10⁻⁴ mm/slit)(1 m/ 1000 mm) = 8.33 x 10⁻⁷ m/slit
θ = angle of diffraction = [tex]tan^{-1}(\frac{L}{y})[/tex]
where,
L = distance of grating from the screen = 77 cm
y = distance of maxima from central maxima
Hence, the general equation after substituting constant values becomes:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{y}))[/tex]
FOR y = 58 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{58\ cm}))[/tex]
λ = 6.654 x 10⁻⁷ m = 665.4 nm
FOR y = 65.4 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{65.4\ cm}))[/tex]
λ = 6.349 x 10⁻⁷ m = 634.9 nm
FOR y = 94.5 cm:
[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{94.5\ cm}))[/tex]
λ = 5.262 x 10⁻⁷ m = 526.2 nm
The attached picture shows the arrangement of the light rays in a diffraction grating.
Learn more about diffraction grating here:
https://brainly.com/question/17012571?referrer=searchResults
1. Notice that the voltmeter moves in response to the coil entering or leaving the magnetic gap.
2. Let's apply Faraday's Law to this situation. Faraday's Law says that the induced voltage (or emf )in a loop of wire caused by a changing magnetic field is:
€ = 1
Where is the magnetic flux which is
Q = BA
In this case, the flux density B is not changing. Instead, the changing flux is due to the motion of the coil as it enters or leaves the magnetic gap:
do = BdA
Given that the area immersed in the gap is changing as the coil enters the gap, what is the correct expression of Faraday's Law for this situation?
Answer:
Explanation:
let the coil of length l and breathe b entering the magnetic field B with speed v.
So, the magnetic flux through the coil is
Ф = B(l×b)
length × breathe = area
Ф = BA
dФ = BdA
therefore induced emf is given as
ε = [tex]Bl(\frac{db}{dt})[/tex]
note: [tex]\frac{db}{dt} = v[/tex]
ε[tex]= Blv[/tex]
attached is the diagram for the solution
Suppose a 225 kg motorcycle is heading toward a hill at a speed of 29 m/s. The two wheels weigh 12 kg each and are each annular rings with an inner radius of 0.280 m and an outer radius of 0.330 m. How high can it coast up the hill, if you neglect friction in m?
a) m = 180 kg
b) v = 29 m/s
c) h = 32 m
Answer:
It can coast uphill 6.2m
Explanation:
See attached file pls
Wind gusts create ripples on the ocean that have a wavelength of 5.00 cm and propagate at 2.00 m/s. What is their frequency
Answer:
f = 40Hz
Explanation:
v=f x wavelength
f =v / wavelength
f = 2/5 x 10-²= 40 Hz
f = 40Hz
f = 40Hz
What is frequency?In physics, the term frequency refers to the number of waves that pass a fixed point in unit time.
It also describes the number of cycles or vibrations undergone during one unit of time by a body in periodic motion.
Wavelength (λ) - The wavelength of light is defined as the distance between the crests or troughs of a wave motion.
The wave equation: v = fλ
As per question,
Wavelength = 5.00 cm
v = 2.00 m/s.
v=fλ
f =v / λ
f = 2/5 x 10⁻² = 40 Hz
f = 40Hz
Therefore,
The frequency is 40Hz.
Learn more about frequency here:https://brainly.com/question/14316711
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A sinusoidal wave travels along a string. The time for a particular point to move from maximum displacement to zero is 0.17 s. What are the (a) period and (b) frequency? (c) The wavelength is 1.5 m; what is the wave speed?
Answer:
31
Explanation:
You would like to store 8.1 J of energy in the magnetic field of a solenoid. The solenoid has 620 circular turns of diameter 6.6 cm distributed uniformly along its 33 cm length.
A. How much current is needed?
_____________ A
B. What is the magnitude of the magnetic field inside the solenoid?
________________T
C. What is the energy density (energy/volume) inside the solenoid?
________________ kJ/m^3
Answer:
(a) The current needed is 56.92 A
(b) The magnitude of the magnetic field inside the solenoid is 0.134 T
(c) The energy density inside the solenoid is 7.144 kJ/m³
Explanation:
Given;
energy stored in the magnetic field of solenoid, E = 8.1 J
number of turns of the solenoid, N = 620 turns
diameter of the solenoid, D = 6.6 cm = 0.066 m
radius of the solenoid, r = D/2 = 0.033 m
length of the solenoid, L = 33 cm = 0.33 m
Inductance of the solenoid is given as;
[tex]L= \frac{\mu_o N^2 A}{l}[/tex]
where;
A is the area of the solenoid = πr² = π (0.033)² = 0.00342 m²
μ₀ is permeability of free space = 4π x 10⁻⁷ H/m
[tex]L= \frac{4\pi*10^{-7} *620^2 *0.00342}{0.33} \\\\L = 0.005 \ H[/tex]
(A). How much current needed
Energy stored in magnetic field of solenoid is given as;
[tex]E = \frac{1}{2} LI^2\\\\[/tex]
Where;
I is the current in the solenoid
[tex]E = \frac{1}{2} LI^2\\\\I^2 = \frac{2E}{L}\\\\I = \sqrt{\frac{2*8.1}{0.005}}\\\\ I = 56.92 \ A[/tex]
(B) The magnitude of the magnetic field inside the solenoid
B = μ₀nI
where;
n is number of turns per unit length
B = μ₀(N/L)I
B = (4π x 10⁻⁷)(620/0.33)(56.92)
B = 0.134 T
(C) The energy density (energy/volume) inside the solenoid
[tex]U_B = \frac{B^2}{2\mu_0} \\\\U_B = \frac{(0.134)^2}{2*4\pi*10^{-7}} \\\\U_B = 7143.54 \ J/m^3\\\\U_B = 7.144 \ kJ/m^3[/tex]
A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 390 kV. The secondary of this transformer is being replaced so that its output can be 515 kV for more efficient cross-country transmission on upgraded transmission lines. (a) What is the ratio of turns in the new secondary compared with turns in the old secondary
Answer:
1.32 is the turns ratio
Explanation:
Note that The transformer steps up the voltage from 12000 V to 390000V
12000 V is the primary and in the secondary it is 390000 V in old transformer
If n₁ be no of turns in primary coil and n₂ be no of turns in secondary coils
the formula is
n₂ / n₁ = voltage in secondry / voltage in primary
n₂ / n₁ = 390000 / 12000
ratio of turns in old transformer is 32.5
ratio of turns in new transformer
n₃ / n₁ = 515 / 12 ( n₃ is no of turns in the secondary of new transformer )
= 42.9
T he ratio of turns in the new secondary compared with the old secondary
n₃ / n₂ = 42.9 / 32.5
= 1.32
If the car decelerates uniformly along the curved road from 27 m/s m/s at A to 13 m/s m/s at C, determine the acceleration of the car at B
Answer:
0.9m/s²
Explanation:
See attached files
A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the friction force between the car and the track, calculate how much the car travels in 10 s
Answer:
50 m
Explanation:
F = ma
10 N = (10 kg) a
a = 1 m/s²
Given:
v₀ = 0 m/s
a = 1 m/s²
t = 10 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²
Δx = 50 m
Light from an argon laser strikes a diffraction grating that has 4,917 lines per cm. The first-order principal maxima are separated by 0.4 m on a wall 1.62 m from the grating. What is the wavelength of the laser light in nm
Answer:
Wavelength is 4.8x10^-7m
Explanation:
See attached file
I need help with this question?
Answer:
You got it right, didn't you?
c) vector C
Explanation:
opposite to vector V
Answer:
A, vector B
Explanation:
A negative vector is a vector which points in the opposite direction, even though it’s in a different quadranot it’s still the opposite direction.
Solve 3* +5-220t = 0
Answer:
t = 27.5
Explanation:
[tex]3 + 5 -220t = 0[/tex]
Well to solve for t we need to combine like terms and seperate t.
So 3+5= 8
8 - 220t = 0
We do +220 to both sides
8 = 220t
And now we divide 220 by 8 which is 27.5
Hence, t = 27.5
What portion of the difference in the angular speed before and after you increased the mass can be accounted for by frictional losses
Answer:
As the mass increases, the moment of inertia(I) increases, therefore, the angular momentum(L) increases too.
Explanation:
friction can be defined as resistance in motion of bodies in relative to one another
momentum is the product of mass and velocity
torque is the time rate of change in momentum
τ = [tex]\frac{dL}{dt}[/tex]
where L = Iω = mvr
I = moment of inertia
ω= angular frequency
if there is no external force(torque) acting on the system, then
[tex]\frac{dL}{dt}[/tex] = 0
dL = 0 = constant
moment of inertia I depends on the distribution of mass on the axis of rotation.
as the mass increases, the angular momentum(L) increases
angular frequency, ω, remains constant
A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done by gravity during the roll, in Joules
Answer:
The work done by gravity during the roll is 490.6 J
Explanation:
The work (W) is:
[tex] W = F*d [/tex]
Where:
F: is the force
d: is the displacement = 20 m
The force is equal to the weight (W) in the x component:
[tex]F = W_{x} = mgsin(\theta)[/tex]
Where:
m: is the mass of the bowling ball = 5 kg
g: is the gravity = 9.81 m/s²
θ: is the degree angle to the horizontal = 30°
[tex]F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N[/tex]
Now, we can find the work:
[tex]W = F*d = 24.53 N*20 m = 490.6 J[/tex]
Therefore, the work done by gravity during the roll is 490.6 J.
I hope it helps you!
Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.
Answer:
Explanation:
radius of aorta = 1.5 cm
cross sectional area = π r²
= 3.14 x 1.5²
= 7.065 cm²
volume of blood flowing out per second out of heart
= a x v , a is cross sectional area , v is velocity of flow
= 7.065 x 11.2
= 79.128 cm³
heart beat per second = 67 / 60
= 1.116666
If V be the volume of heart
1.116666 V = 79.128
V = 70.86 cm³.
A person bends over to grab a 20 kg object. The back muscle responsible for supporting his upper body weight and the object is located 2/3 of the way up his back (where it attaches to the spine) and makes an angle of 12 degrees with the spine. His upper body weighs 36 kg. What is the tension in the back muscle
Answer:
T = 2689.6N
Explanation:
Considering the situation, one can say that torque due to tension in the spine is counter balanced by the torque due to weight of upper part of the body and the weight of the object. Hence, the tension force is acting at an angle of 12 degree
while both weight are acting perpendicular to the length. Hence we have :
Torque ( clockwise) = Torque ( anticlockwise)
m1g (L/2)+ m2g(L) = Tsin 12(2L/3)........1
Where m1 = 36kg
m2 = 20kg
g = 9.81m/s^2
Theta = 12
Substituting into equation 1
36(9.81) * (L/2)+20(9.81)(L) = Tsin12(2L/3)
353.16L/2+196.2L = T ×0.2079(2L/3)
176.58L+196.2L = T × 0.1386L
372.78L = 0.1386LT
T = 372.78L/0.1386L
T = 2689.6N
Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?
Answer:
The frequency is [tex]F = 325 Hz[/tex]
Explanation:
From the question we are told that
The frequency for the first note is [tex]F_1 = 330 Hz[/tex]
The beat frequency of the first note is [tex]f_b = 5 \ Hz[/tex]
The frequency for the second note is [tex]F_2 = 350 \ H_z[/tex]
The beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]
Generally beat frequency is mathematically represented as
[tex]F_{beat} = | F_a - F_b |[/tex]
Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source
Now in the case of this question
For the first note
[tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]
Where F is the frequency of the string note
For the second note
[tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]
Adding equation 1 from 2
[tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]
[tex]f_b + f_a = F_1 + F_2 -2F[/tex]
substituting values
[tex]5 +25 = 330 + 350 -2F[/tex]
=> [tex]F = 325 Hz[/tex]
Equal charges, one at rest, the other having a velocity of 104 m/s, are released in a uniform magnetic field. Which charge has the largest force exerted on it by the magnetic field
Answer:
case 1 of physics is the answer