Answer:
0.179 N
Explanation:
What is the magnitude of the horizontal part of the contact force on the shaker by the turntable?
The horizontal part of the constant force of the turntable on the shaker is the centripetal force of the turntable on the shaker, F.
So, F = mv²/r where m = mass of shaker = 79 g = 0.079 kg, v = speed of shaker = 0.7222 m/s and r = radius of turntable = 0.23 m
So, substituting the values of the variables into the equation, we have
F = mv²/r
F = 0.079 kg (0.7222 m/s)²/0.23 m
F = 0.0412 kgm/s² ÷ 0.23 m
F = 0.179 kgm/s²
F = 0.179 N
Plzzz answer this question correctly
Answer:
changing the direction in which a force is exerted
What is the correct coefficient for 2H2 + O2 →2H2O
Explanation:
2forH2,1for02,and2forH20
Cole drives to school from home, starting from rest and accelerating for 10 minutes as he travels 6.0 km to school.
1) What is Cole's acceleration?
2) What is his velocity when he reaches school?
Explanation:
this is the answer for your question. if you have any doubt.
you can send your doubt to:6369514784(what's app)
Use Newton's laws of motion to explain why it is important that baseballs and softballs each have a small acceptable range of masses.
Explanation:
new non law neutron means neutral then it's important that baseball and softball features small respectable range of masses soft it means that when a ball hits anything hard it comes back by the Newton Law if the baseball is big and the small boy small and then if the contract with each other they ignore triple so when a ball hits the wall if the comeback because of the Mutants and when a big ball if we throw it to the wall it doesn't come that it comes back but in a very low way because it contains less neutrons in it if it is helpful please share with me
As mentioned in the text, the tangent line to a smooth curve r(t) = ƒ(t)i + g(t)j + h(t)k at t = t0 is the line that passes through the point (ƒ(t0), g(t0), h(t0)) parallel to v(t0), the curve’s velocity vector at t0. In Exercises 23–26, find parametric equations for the line that is tangent to the given curve at the given parameter value t = t0.
Answer:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
Explanation:
Given
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex] at [tex]t = 0[/tex]
Point: [tex](f(t0), g(t0), h(t0))[/tex]
[tex]r(t) = ln\ t_i + \frac{t-1}{t+2}j + t\ ln\ tk[/tex], [tex]t0 = 1[/tex] -- Missing Information
Required
Determine the parametric equations
[tex]r(t) = ln\ ti + \frac{t-1}{t+2}j + t\ ln\ tk[/tex]
Differentiate with respect to t
[tex]r'(t) = \frac{1}{t}i +\frac{3}{(t+2)^2}j + (ln\ t + 1)k[/tex]
Let t = 1 (i.e [tex]t0 = 1[/tex])
[tex]r'(1) = \frac{1}{1}i +\frac{3}{(1+2)^2}j + (ln\ 1 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{3^2}j + (0 + 1)k[/tex]
[tex]r'(1) = i +\frac{3}{9}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + (1)k[/tex]
[tex]r'(1) = i +\frac{1}{3}j + k[/tex]
To solve for x, y and z, we make use of:
[tex]r(t) = f(t)i + g(t)j + h(t)k[/tex]
This implies that:
[tex]r'(1)t = xi + yj + zk[/tex]
So, we have:
[tex]xi + yj + zk = (i +\frac{1}{3}j + k)t[/tex]
[tex]xi + yj + zk = it +\frac{1}{3}jt + kt[/tex]
By comparison:
[tex]xi = it[/tex]
Divide by i
[tex]x = t[/tex]
[tex]yj = \frac{1}{3}jt[/tex]
Divide by j
[tex]y = \frac{1}{3}t[/tex]
[tex]zk = kt[/tex]
Divide by k
[tex]z = t[/tex]
Hence, the parametric equations are:
[tex]x = t[/tex]
[tex]y = \frac{1}{3}t[/tex]
[tex]z =t[/tex]
Q4. (a) An acre-foot is the volume of water that would cover 1 acre of flat land to a depth of 1
foot. How many gallons are in 1 acre-foot?
Answer:
326,000
Explanation:
One acre-foot equals about 326,000 gallons, or enough water to cover an acre of land, about the size of a football field, one foot deep. An average California household uses between one-half and one acre-foot of water per year for indoor and outdoor use.
a body of mass 20kg initially at rest is subjected to a force of 40N for 1sec calculate the change in kinetic energy showing the solution
Answer:
Change in KE is 40 J
Explanation:
Recall that the impulse exerted on an object equal the change of momentum of the object (ΔP), which in time is defined as the product of the force exerted on it times the time the force was acting:
Change in momentum is: ΔP = F * Δt
In our case,
ΔP = 40 N * 1 sec = 40 N s
Since the object was initially at rest, its initial momentum was zero, and the final momentum should then be 40 N s.
So, the initial KE was 0, and the final (KEf) can be calculated using:
KEf = 1 /(2 m) Pf^2 = 1 / (40) 40^2 = 40 J
So, the change in kinetic energy is:
KEf - KEi = 40 J - 0 j = 40 J
This is the build up of substance such as pesticides in an organism and occurs when an organism absorb a substance at a rate faster than that at which the substance is lost
Answer:
which the substance is lost by catabolism and excretion.
Explanation:
what happens when a wave passes through a medium ?
Answer:
When waves travel from one medium to another the frequency never changes. As waves travel into the denser medium, they slow down and wavelength decreases. Part of the wave travels faster for longer causing the wave to turn. The wave is slower but the wavelength is shorter meaning frequency remains the same.
Explanation:
While talking to a friend, a construction worker momentarily set her cell phone down on one end of an iron rail of length 7.50 m. At that moment, a second worker dropped a wrench so that it hit the other end of the rail. The person on the phone detected two pulses of sound, one that traveled through the air and a longitudinal wave that traveled through the rail. (Assume the speed of sound in iron is 5,950 m/s and the speed of sound in air is 343 m/s).
A) Which pulse reaches the cell phone first?
B) Find the separation in time (in s) between the arrivals of the two pulses.
Answer:
A)
The impulse that reaches the cell phone first is the Longitudinal wave
B) 0.0206 seconds
Explanation:
length of Iron rail = 7.5 m
speed of sound in Iron = 5950 m/s
speed of sound in Air = 343 m/s
A) Determine which pulse reaches the cell phone first
The impulse that reaches the cell phone first is the Longitudinal wave
Time for longitudinal pulse to be detected = 7.5 / 5950 = 0.00126 s
Time for pulse through air to be detected = 7.5 / 343 = 0.02186 s
B) separation in time between the arrivals of the two pulses
ΔT = 0.02186 - 0.00126 = 0.0206 seconds
In an experiment similar to the one pictured below, an electron is projected horizontally at a speed vi into a uniform electric field pointing up. The magnitude of the total vertical deflection, ye, of the electron is measured to be 1 mm. The same experiment is repeated with a proton (whose mass is 1840 times that of the electron) that is also projected horizontally at a speed vi into the same uniform electric field. What is the magnitude of the total vertical deflection, yp, for the proton
I think you need Graph to figure it out
Using Newton's second law and kinematic projectile motion we can find the proton deflection y = 5.43 10⁻⁷ m, in the opposite direction to the electron deflection.
given parameters
The deflection of the electorn y₁ = 1 mm = 0.001 m The initial velocity of the electron and proton v_i The mass of the proton m_p = 1840 meto find
deflection of the protonFor this exercise we will use Newton's second law where the force is electric
F = ma
F = q E
where F is the force, q the charge, E the electric field, m the mass and the acceleration of the particle
q E = m a
a = q / m E
This acceleration is the direction of the electric field that is perpendicular to the initial velocity (v_i)
Having the acceleration we can use the kinematics relations
If we make the direction of the initial velocity coincide with the x-axis
v_i = cte
v_i = x / t
t = x/ v_i
on the y-axis is in the direction of the electric field
y = v_{iy} t + ½ a t²
on this axis the initial velocity is zero
y = [tex]\frac{1}{2} (\frac{q}{m} E) \ t^2[/tex]
subtitute
y = (1)
Electron motion.
Let us propose the expression for the electron situation, the length of the displacement must be the same for electron and proton, suppose that it is x = L
In this case the charge q = -e and the mass m = m_e
its substitute in equation 1
y₁ = [tex]\frac{1}{2} \ ( \frac{-e}{m_e} E) \ \frac{x^2}{v_i^2}[/tex]
where y₁, is the lectron deflection.
Proton motion
Between the proton and the electron we have some relationships
q_p = -e
m_ = 1840 m_e
we substitute in the equation 1
y₂ = ½ e / 1840 me E x² / vi²
y₂ =
y₂ = - y₁ / 1840
y₂ = - 0.001 / 1840
y₂ = - 5.43 10⁻⁷ m
The negative sign indicates that the deflection of the proton is in the opposite direction to the deflection of the electron.
In conclusion they use Newton's second law and kinematics we can find the proton deflection is y = 5.43 10⁻⁷ m
learn more about electric charge movement here: https://brainly.com/question/19315467
Which of the following is the BEST explanation for why oceans have two different types of currents?
Answer:
sddww
Explanation:
szsswa
Displacement vector A points due east and has a magnitude of 1.9 km. Displacement vector B points due north and has a magnitude of 2.08 km. Displacement vector C points due west and has a magnitude of 2.4 km. Displacement vector D points due south and has a magnitude of 2.8 km. Find the magnitude and direction (relative to due east) of the resultant vector A + B + C + D
Answer:
Explanation:
We shall represent all the four displacement in vector form in terms of unit vector i and j where i represents unit vector towards east , j represents unit vector towards north .
Displacement of A
D₁ = 1.9 i
Displacement of B
D₂ = 2.08 j
Displacement of C
D₃ = - 2.4 i
Displacement of D
D₄ = -2.8 j
Resultant displacement
= 1.9 i + 2.08 j - 2.4 i - 2.8 j
= - 0.5 i - 0.72 j
magnitude of resultant vector
= √ ( .5² + .72² )
=√ ( .25 + .5184 )
= √ .7684
= .876 km
Both i and j are negative of resultant displacement
hence its direction is towards south of west . Angle with west is Ф .
TanФ = .5184 / .25 = 2.0736
Ф = 64.25° .
From east direction is = 180 + 64.25 = 244.25° .
What are regular and irregular reflection of light? plz help its
urgent..
Explanation:
Regular reflection: It is the reflection from a smooth surface such that the light rays are evenly parallel to each other and an image is formed. ... Irregular reflection: It is the diffused reflection from uneven surface such that the light rays are not parallel to each other and do not form an image.
A loaded wagon of mass 10,000 kg moving with a speed of 15 m/s strikes a stationary wagon of the same mass making a perfect inelastic collision. What will be the speed of coupled wagons after collision?
Answer:
7.5 m/s
Explanation:
Unfortunately, I don't have an explanation but I guessed the correct answer.
Two students are on a balcony a distance h above the street. One student throws a ball vertically downward at a speed vi; at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of vi, g, h, and t. (Take upward to be the positive direction.)
(a) What is the time interval between when the first ball strikes the ground and the second ball strikes the ground?
?t = ______
(b) Find the velocity of each ball as it strikes the ground.
For the ball thrown upward vf = ______
For the ball thrown downward vf = ______
(c) How far apart are the balls at a time t after they are thrown and before they strike the ground?
d = _______
Answer:
Explanation:
a )
Time for first ball to reach top position
v = u - gt
0 = vi - gt
t = vi / g
Time to reach balcony while going downwards
= vi /g
Total time = 2 vi / g
Time to go down further to the ground = t₁
Total time = 2 vi / g + t₁
Time for the other ball to go to the ground = t₁
Time difference = ( 2 vi / g + t₁ ) - t₁
= 2vi / g .
( b )
v² = u² + 2gh
For both the throw ,
final displacement = h , initial velocity downwards = vi
( For the first ball also , when it go down while passing the balcony , it acquires the same velocity vi but its direction is downwards.)
vf² = vi² + 2gh
vf = √ ( vi² + 2gh )
(c )
displacement of first ball after time t
s₁ = - vi t + 1/2 g t² [ As initial velocity is upwards , vi is negative ]
displacement of second ball after time t
s₂ = vi t + 1/2 g t²
Difference = d = s₂ - s₁
= vi t + 1/2 g t² - ( - vi t + 1/2 g t² )
d = 2 vi t .
A point charge, Q1 = -4.2 μC, is located at the origin. A rod of length L = 0.35 m is located along the x-axis with the near side a distance d = 0.45 m from the origin. A charge Q2 = 10.4 μC is uniformly spread over the length of the rod.Part (a) Consider a thin slice of the rod, of thickness dx, located a distance x away from the origin. What is the direction of the force on the charge located at the origin due to the charge on this thin slice of the rod? Part (b) Write an expression for the magnitude of the force on the point charge, |dF|, due to the thin slice of the rod. Give your answer in terms of the variables Q1, Q2, L, x, dx, and the Coulomb constant, k. Part (c) Integrate the force from each slice over the length of the rod, and write an expression for the magnitude of the electric force on the charge at the origin. Part (d) Calculate the magnitude of the force |F|, in newtons, that the rod exerts on the point charge at the origin.
Answer:
a) attractiva, b) dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex], c) F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex], d) F = -1.09 N
Explanation:
a) q1 is negative and the charge of the bar is positive therefore the force is attractive
b) For this exercise we use Coulomb's law, where we assume a card dQ₂ at a distance x
dF = [tex]k \frac{Q_1 \ dQ_2}{dx}[/tex]
where k is a constant, Q₁ the charge at the origin, x the distance
c) To find the total force we must integrate from the beginning of the bar at x = d to the end point of the bar x = d + L
∫ dF = [tex]k \ Q_1 \int\limits^{d+L}_d {\frac{1}{x^2} } \, dQ_2[/tex]
as they indicate that the load on the bar is uniformly distributed, we use the concept of linear density
λ = dQ₂ / dx
DQ₂ = λ dx
we substitute
F = [tex]k \ Q_1 \lambda \int\limits^{d+L}_d \, \frac{dx}{x^2}[/tex]
F = k Q1 λ ([tex]-\frac{1}{x}[/tex])
we evaluate the integral
F = k Q₁ λ [tex](- \frac{1}{d+L} + \frac{1}{d} )[/tex]
F = k Q₁ λ [tex]( \frac{L}{d \ (d+L)})[/tex]
we change the linear density by its value
λ = Q2 / L
F = [tex]k Q_1 \frac{Q_2}{d \ (d+L)}[/tex]
d) we calculate the magnitude of F
F =9 10⁹ (-4.2 10⁻⁶) [tex]\frac{10.4 10x^{-6} }{0.45 ( 0.45 +0.35)}[/tex]
F = -1.09 N
the sign indicates that the force is attractive
Answer:
a)Toward the rod
b)|dF| = k|Q1|Q2(dx/L)/x^2
c)|F| = k|Q1|Q2/(d(d+L))
d)Plug in for answer c and solve
Explanation:
A)
Q1 is negative and Q2 is positive so it is an attractive force to where the rod is located.
B)
The formula for Force due to electric charges is F=kQ1Q2/r^2
In this case, Q2 is distrusted through the length of the rod as opposed to a single point charge. As such Q2 is actually Q2*dx/L as dx is a small portion of the full length, L.
The radius between Q1 and Q2 depends on the section of the rod taken so r will be the variable x distance from Q1.
The force is only from a small portion of the rod so more accurately, we are finding |dF| as opposed to the full force, F, caused by the whole rod.
The final formula is |dF| = k|Q1|Q2(dx/L)/x^2
C)
Integrating with respect to the only changing variable, x, which spans the length of the rod, from radius = d to d+L we get this:
F = integral from d to d+L of k|Q1|Q2(dx/L)/x^2
factor out constants
F = kQ1Q2/L * integral d to d+L(1/x^2)dx
F = kQ1Q2/L * (-1/x)| from d to d+L
F = kQ1Q2/L * (-1/d+L - -1/d)
F = kQ1Q2/L * (-d/(d(d+L)) + (d+L)/(d(d+L))
F = kQ1Q2/L * (L)/(d(d+L))
F = kQ1Q2/(d(d+L))
D)
Plug in the given values into c and you have your answer.
To fully describe velocity you must have a _____
A. Magnitude and unit
B. Speed and unit
C. Average speed and position
D. Magnitude and direction
20. For each improvement in glider design, engineers follow
O A. the written instructions that are provided in the hang glider build kit.
O B. an iterative process of testing, modifying, retesting, and modifying again.
O C. a complicated process of checks and balances while obtaining financing.
O D. a mathematical process, rejecting designs that don't follow blueprint dimensions.
Turn In
Astronauts aboard the ISS move at about 8000 m/s, relative to us when we look upward.How long does an astronaut need to stay aboard the space station to be a full second youngerthan people on the ground? Please show and explain how you would set-up the problem,before you actually try to solve it. If you cannot solve it exactly, please try to offer an estimate.(5 pts)
Answer:
#_time = 7.5 10⁴ s
Explanation:
In order for the astronaut to be younger than the people on earth, it follows that the speed of light has a constant speed in vacuum (c = 3 108 m / s), therefore with the expressions of special relativity we have.
t = [tex]\frac{t_p}{ \sqrt{1- (v/c)^2} }[/tex]
where t_p is the person's own time in an immobile reference frame,
[tex]t_{p} = t \sqrt{1 - (\frac{v}{c})^2 }[/tex]
let's calculate
we assume that the speed of the space station is constant
[tex]t_p = 1 \sqrt{1 - \frac{8 \ 10^3}{3 \ 10^8} }[/tex]
[tex]t_p = 1 \sqrt{1- 2.6666 \ 10^{-5}}[/tex]
t_ = 0.99998666657 s
therefore the time change is
Δt = t - t_p
Δt = 1 - 0.9998666657
Δt = 1.3333 10⁻⁵ s
this is the delay in each second, therefore we can use a direct rule of proportions. If Δt was delayed every second, how much second (#_time) is needed for a total delay of Δt = 1 s
#_time = 1 / Δt
#_time =[tex]\frac{1}{1.3333 \ 10^{-5}}[/tex]
#_time = 7.5 10⁴ s
A cyclist cover 6km in 20minutes. His speed is
Answer:
The speed of a cyclist is 0.3 km/min.
Explanation:
Given
The distance d = 6km Time t = 20 minutesTo determine
We need to determine the speed of a cyclist.
In order to determine the speed of a cyclist, all we need to do is to divide the distance covered by a cyclist by the time taken to cover the distance.
We know the formula involving speed, time, and distance
[tex]s=\frac{d}{t}[/tex]
where
s = speedd = distance coveredt = time takensubstitute d = 6, and t = 20 in the formula
[tex]s=\frac{d}{t}[/tex]
[tex]s=\frac{6}{20}[/tex]
Cancel the common factor 2
[tex]s=\frac{3}{10}[/tex]
[tex]s=0.3[/tex] km/min
Thus, the speed of a cyclist is 0.3 km/min.
In a certain region of space the electric potential increases uniformly from east to west and does not vary in any other direction. The electric field:Group of answer choicespoints east and varies with positionpoints east and does not vary with positionpoints west and varies with positionpoints west and does not vary with positionpoints north and does not vary with position
Answer:
Explanation:
The relation between electric field and potential difference is as follows
E = - dV / dr
That means if dV is positive , E is negative . In other words , if potential increases , E is negative or in opposite direction in which potential increases .
Here the electric potential increases uniformly from east to west , that means electric field is from west to east . Since potential is uniformly increasing that means
dV / dr = constant
E = constant
Electric field is constant .
So the option which is correct is
" points east and does not vary with position " .
1
What kind of adaptation is a long neck on a tortoise? *
(10 Points)
O
A. Structural
B. Behavioral
a
C. Functional
a
D. Physiological
Answer:
The answer is ......... structural adaptation
Explanation:
because structural adaptations is a physical thinng in their body so its A please give me brainliest
A toy car can go 5 mph. How long would it take to go 12 miles?
How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?
Light of wavelength 425.0 nm in air falls at normal incidence on an oil film that is 850.0 nm thick. The oil is floating on a water layer 1500 nm thick. The refractive index of water is 1.33, and that of the oil is 1.40. The number of wavelengths of light that fit in the oil film is closest to:
Answer:
in oil film λ = 303.57 10⁻⁹ m
in the water film λ = 319.55 10⁻⁹ m
Explanation:
When electromagnetic radiation reaches a material, its propagation is by a process that we call absorption and reflection,
when light reaches a surface it has a mass much greater than the mass of the photons (m = 0), therefore there is an elastic collision where the frequency does not change, due to the speed of light in the material medium changes, therefore the only possibility is that the wavelength in the material changes, to maintain the relationship
v = λ f
in the void we have
c = λ₀ f
we divide the two expression
c / v = λ₀ / λ
the refractive index is
n = c / v
n = λ₀ /λ
λ = λ₀ / n
let's calculate
in oil film
λ = 425 10⁻⁹ / 1.40
λ = 303.57 10⁻⁹ m
in the water film
λ = 425 10⁻⁹ / 1.33
λ = 319.55 10⁻⁹
those wavelengths are in the ultraviolet
In the human arm, the forearm and hand pivot about the elbow joint. Consider a simplified model in which the biceps muscle is attached to the forearm 3.80 cm from the elbow joint. Assume that the person's hand and forearm together weigh 15.0 N and that their center of gravity is 15.0 cm from the elbow (not quite halfway to the hand). The forearm is held horizontally at a right angle to the upper arm, with the biceps muscle exerting its force perpendicular to the forearm.
A. Find the force exerted by the biceps when the hand is empty.
B. Now the person holds a 80.0-N weight in his hand, with the forearm still horizontal. Assume that the center of gravity of this weight is 33.0 cm from the elbow. Find the force now exerted by the biceps.
C. Explain why the biceps muscle needs to be very strong.
D. Under the conditions of part B, find the magnitude of the force that the elbow joint exerts on the forearm.
E. Under the conditions of part B, find the direction of the force that the elbow joint exerts on the forearm.
F. While holding the 80.0-N weight, the person raises his forearm until it is at an angle of 53.0∘ above the horizontal. If the biceps muscle continues to exert its force perpendicular to the forearm, what is this force when the forearm is in this position?
G. Has the force increased or decreased from its value in part B? Explain why this is so, and test your answer by actually doing this with your own arm.
Answer:
Answer is explained in the explanation section below.
Explanation:
Part A)
From conserve moment of force, we have:
F1d1 = F2d2
F1 x (3.80 x [tex]10^{-2}[/tex] m) = 15N x (15 x [tex]10^{-2}[/tex] m)
F1 = [tex]\frac{15 . 15 . 10^{-2} }{3.80 . 10^{-2} }[/tex]
F1 = 59.2 N
Force exerted by the biceps when the hand is empty.
Part B)
The 80 N weight acts at 33 cm and 15 N at 15 cm, then the center of mass is:
x = [tex]\frac{m1x1 + m2x2}{m1+m2}[/tex]
x = [tex]\frac{\frac{80}{9.8} (33 .10^{-2}) + \frac{15}{9.8}(15.10^{-2} }{\frac{80}{9.8} + \frac{15}{9.8} }[/tex]
x = 30.16 x [tex]10^{-2}[/tex] m
Total Weight is:
F = 80N + 15N = 95N
From the conserve moment of force, we have:
F ( 3.8 x [tex]10^{-2}[/tex] ) = 95N (30.16 x [tex]10^{-2}[/tex])
F = 754 N
Part C:
From the above two examples solved, the force exerted by the biceps is higher than downward force, due to this muscle need to be very strong.
Part D)
The force exerted by elbow on the forearm is:
The force exerted by the elbow and biceps are in upward direction and total weight is in downward direction. So, the balancing force in vertical direction is:
F2 + 754N = 95N
F2 = 95N -754N
F2 = -659N
Negative sign shows the force is in downward direction.
Part E)
The bicep muscle acts perpendicular to the forearm, so it is lever arm stays the same. but those of the other two forces decreases as the arm is raised. There tension in the biceps muscle decreases.
Part F)
Angle = 53 degrees.
So,
Force = FcosФ
Force = 754 cos 53
Force = 453.76 N
Part G)
The value of force has gone downwards. It has decreased from that of part B.
The blood pressure at your heart is approximately 100 mm Hg. As blood is pumped from the left ventricle of your heart, it flows through the aorta, a single large vessel with a diameter of about 2.5 cm. The speed of blood flow in the aorta is about 60 cm/s. Any change in pressure as blood flows in the aorta is due to the change in height: the vessel is large enough that viscous drag is not a major factor into successively smaller and smaller blood vessels until it reaches the capillaries. Blood flows in the capillaries at the much lower speed of approximately 0.7 mm/s. The diameter of capillaries and other small blood vessels is so small that viscous drag is a major factor..Because the flow speed in your capillaries is much less than in the aorta, the total cross-section area of the capillaries considered together must be much larger than that of the aorta. Given the flow speeds noted, the total area of the capillaries considered together is equivalent to the cross-section area of a single vessel of approximately what diameter?
a. 25 cm
b. 50 cm
c. 75 cm
d. 100 cm
Answer:
The correct option is c. 75 for this question
Explanation:
The correct option is c. 75 for this question:
Let's see how.
Continuity Equation is given as:
AcVc = AaVa
Where,
Aa = Area of Aorta
Ac = Area of the capillary
Va = Fluid speed in Aorta
Vc = Fluid speed in Capillary
So,
Assuming the fluid is the ideal one/
[tex]\pi[/tex]/4 [tex]Dc^{2}[/tex] Vc= [tex]\pi[/tex]/4 [tex]Da^{2}[/tex] Va
[tex]Dc^{2}[/tex] Vc= [tex]Da^{2}[/tex] Va
Dc = Da x [tex]\sqrt{\frac{Va}{Vc} }[/tex]
Dc = 2.5 cm x [tex]\sqrt{\frac{60 cm}{0.07 cm } }[/tex]
Dc = 73.192 cm
Dc = 75 approximately
Hence, the diameter of the capillary = 75 cm approximately
A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20oC. Determine the volume of the second tank and the final equilibrium pressure of air.
Answer:
- the volume of the second tank is 1.77 m³
- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Explanation:
Given that;
[tex]V_{A}[/tex] = 1 m³
[tex]T_{A}[/tex] = 10°C = 283 K
[tex]P_{A}[/tex] = 350 kPa
[tex]m_{B}[/tex] = 3 kg
[tex]T_{B}[/tex] = 35°C = 308 K
[tex]P_{B}[/tex] = 150 kPa
Now, lets apply the ideal gas equation;
[tex]P_{B}[/tex] [tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex]
[tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex] / [tex]P_{B}[/tex]
The gas constant of air R = 0.287 kPa⋅m³/kg⋅K
we substitute
[tex]V_{B}[/tex] = ( 3 × 0.287 × 308) / 150
[tex]V_{B}[/tex] = 265.188 / 150
[tex]V_{B}[/tex] = 1.77 m³
Therefore, the volume of the second tank is 1.77 m³
Also, [tex]m_{A}[/tex] = [tex]P_{A}[/tex][tex]V_{A}[/tex] / R[tex]T_{A}[/tex] = (350 × 1)/(0.287 × 283) = 350 / 81.221
[tex]m_{A}[/tex] = 4.309 kg
Total mass, [tex]m_{f}[/tex] = [tex]m_{A}[/tex] + [tex]m_{B}[/tex] = 4.309 + 3 = 7.309 kg
Total volume [tex]V_{f}[/tex] = [tex]V_{A}[/tex] + [tex]V_{B}[/tex] = 1 + 1.77 = 2.77 m³
Now, from ideal gas equation;
[tex]P_{f}[/tex] = [tex]m_{f}[/tex]R[tex]T_{f}[/tex] / [tex]V_{f}[/tex]
given that; final temperature [tex]T_{f}[/tex] = 20°C = 293 K
we substitute
[tex]P_{f}[/tex] = ( 7.309 × 0.287 × 293) / 2.77
[tex]P_{f}[/tex] = 614.6211119 / 2.77
[tex]P_{f}[/tex] = 221.88 kPa ≈ 222 kPa
Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
The new springs will be identical to the original springs, except the force constant will be 5655.00 N/m smaller. When James removes the original springs, he discovers that the length of each spring expands from 8.55 cm (its length when installed) to 12.00 cm (its length with no load placed on it). If the mass of the car body is 1355.00 kg, by how much will the body be lowered with the new springs installed, compared to its original height
Answer:
Explanation:
For original spring , compression in spring due to a load of 1355 kg is
x = 12 - 8.55 = 3.45 cm = .0345 m
spring constant = W / x
= 1355 x 9.8 / .0345
= 384898.55 N /m
Spring constant of new spring
k = 384898.55 - 5655 = 379243.55 N /m
New compression for new spring
= W / k
= 1355 x 9.8 / 379243.55
= .035 m
= 3.50 cm
Difference of compression = 3.50 - 3.45
= .05 cm .
In later case , car will be more lowered by .05 cm .