Answer:
The maximum emf generated in the coil is 60527.49 V
Explanation:
Given;
area of coil, A = 0.320 m²
angular frequency, f = 100 rev/s
magnetic field, B = 0.43 T
number of turns, N = 700 turns
The maximum emf generated in the coil is calculated as,
E = NBAω
where;
ω is the angular speed = 2πf
E = NBA(2πf)
Substitute in the given values and solve for E
E = 700 x 0.43 x 0.32 x 2π x 100
E = 60527.49 V
Therefore, the maximum emf generated in the coil is 60527.49 V
A record player rotates a record at 45 revolutions per minute. When the record player is switched off, it makes 4.0 complete turns at a constant angular acceleration before coming to rest. What was the magnitude of the angular acceleration (in rads/s2) of the record as it slowed down
Answer:
The angular acceleration is [tex]\alpha = 0.4418 \ rad /s^2[/tex]
Explanation:
From the question we are told that
The angular speed is [tex]w_f = 45 \ rev / minutes = \frac{45 * 2 * \pi }{60 }= 4.713 \ rad/s[/tex]
The angular displacement is [tex]\theta =4 \ rev = 4 * 2 * \pi = 25.14 \ rad[/tex]
From the first equation of motion we can define the movement of the record as
[tex]w_f ^2 = w_o ^2 + 2 * \alpha * \theta[/tex]
Given that the record started from rest [tex]w_o = 0[/tex]
So
[tex]4.713^2 = 2 * \alpha * 25.14[/tex]
[tex]\alpha = 0.4418 \ rad /s^2[/tex]
A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is the velocity of the exiting water? Ignore all orificelosses.
Answer:
4.75 m/s
Explanation:
The computation of the velocity of the existing water is shown below:
Data provided in the question
Tall = 2 m
Inside diameter tank = 2m
Hole opened = 10 cm
Bottom of the tank = 0.75 m
Based on the above information, first we have to determine the height which is
= 2 - 0.75 - 0.10
= 2 - 0.85
= 1.15 m
We assume the following things
1. Compressible flow
2. Stream line followed
Now applied the Bernoulli equation to section 1 and 2
So we get
[tex]\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2[/tex]
where,
P_1 = P_2 = hydrostatic
z_1 = 0
z_2 = h
Now
[tex]\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\ \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}[/tex]
= 4.7476 m/sec
= 4.75 m/s
Two point charges attract each other with an electric force of magnitude F. If one charge is reduced to one-third its original value and the distance between the charges is doubled, what is the resulting magnitude of the electric force between them
Answer:
F' = F/12
Therefore, the electrostatic force is reduced to one-twelve of its original value.
Explanation:
The electrostatic force of attraction or repulsion between to charges is given by Coulomb's Law:
F = kq₁q₂/r² ---------- equation 1
where,
F = Electrostatic Force
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of 2nd charge
r = distance between charges
Now, if we double the distance between charges and reduce one charge to one-third value, then the force will become:
F' = kq₁'q₂'/r'²
where,
q₁' = (1/3)q₁
q₂' = q₂
r' = 2r
Therefore,
F' = k(1/3 q₁)(q₂)/(2r)²
F' = (1/12)kq₁q₂/r²
using equation 1:
F' = F/12
Therefore, the electrostatic force is reduced to one-twelve of its original value.
Two capacitors, CA and CB, are such that CA > CB. These are connected with a battery in various ways: each individually, series, and parallel. Rank these four cases according to the total amount of charge, greatest first.
a. (CA) > (C) > (CA and CB in series) > (CA and Co in parallel)
b. (CA)>(Cb)> (CA and CB in parallel) > (CA and CB in series)
c. (CA and CB in series) > (CA) > (CB) > (CA and CB in parallel)
d. (CA and Cg in parallel) > (CA) > (CB) > (CA and Cg in series)
Answer:
b. (CA)>(Cb)> (CA and CB in parallel) > (CA and CB in series)
Explanation:
This is because capacitors in series is the sum of the reciprocal of the capacitances while that of parallel is the sum of the individual capacitances
Light with a frequency of 5.70×10^14 Hz travels in a block of glass that has an index of refraction of 1.56. What is the wavelength of the light in the glass?
✔First you have to calculate the light's speed in the glass,
You know that in the air and in the void (where the refraction index n is zero) the light's speed C corresponds to 3,0 x 10^8 m/s
So We have :
V = C/n
V = 3,0 x 10^8/1,56 V ≈ 1,92 x 10^8 m/s✔ Now, you know the light's speed in glass, and you know that : the wavelength λ is the quotient of light's speed V on its frequency ν, so :
λ = V/ ν
λ = 1,82 x 10^8/5,70 x 10^14 λ ≈ 3.40 x 10^-7 m λ ≈ 340 nmUnpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?
Answer:
I₂ = 0.25 I₀
Explanation:
To know the light transmitted by a filter we must use the law of Malus
I = I₀ cos² θ
In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.
I₁ = I₀ cos² 45
I₁ = I₀ 0.5
this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂
I₂ = I₁ cos² 45
I₂ = (I₀ 0.5) 0.5
I₂ = 0.25 I₀
this is the intensity of the light transmitted by the set of polarizers
Consider the following three objects, each of the same mass and radius:
(1) a solid sphere
(2) a solid disk
(3) a hoop
All three are released from rest at the top of an inclined plane. The three objects proceed down the incline undergoing rolling motion without slipping. Use work-kinetic energy theorem to determine which object will reach the bottom of the incline first.
a) 1, 2, 3
b) 2, 3, 1
c) 3, 1, 2
d) 3, 2, 1
e) All three reach the bottom at the same time.
Answer:
Explanation:a 1
Which of the following represents a concave mirror? +f,-f,-di,+di
Answer:
fully describes the concave mirror is + f
Explanation:
A concave mirror is a mirror where light rays are reflected reaching a point where the image is formed, therefore this mirror has a positive focal length, the amount that fully describes the concave mirror is + f
This allows defining a sign convention, for concave mirror + f, the distance to the object is + d0 and the distance to the image is + di
Answer:
+f
Explanation:
because you have to be really dumb to get an -f
Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 times larger than Earth's and a radius that is 1.61 times larger than Earth's. Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 times larger than Earth's and a radius that is 1.61 times larger than Earth's. Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 times larger than Earth's and a radius that is 1.61 times larger than Earth's. Calculate the acceleration of gravity on the surface of Kepler-62e.
Answer:
g' = 13.5 m/s²
Explanation:
The acceleration due to gravity on surface of earth is given by the formula:
g = GMe/Re² --------------- euation 1
where,
g = acceleration due to gravity on surface of earth
G = Universal Gravitational Constant
Me = Mass of Earth
Re = Radius of Earth
Now, the the acceleration due to gravity on the surface of Kepler-62e is:
g' = GM'/R'² --------------- euation 1
where,
g' = acceleration due to gravity on surface of Kepler-62e
G = Universal Gravitational Constant
M' = Mass of Kepler-62e = 3.57 Me
R' = Radius of Kepler-62e = 1.61 Re
Therefore,
g' = G(3.57 Me)/(1.61 Re)²
g' = 1.38 GMe/Re²
using equation 1:
g' = 1.38 g
where,
g = 9.8 m/s²
Therefore,
g' = 1.38(9.8 m/s²)
g' = 13.5 m/s²
The sound intensity at a distance 2.00 m from a sound source is 5.00 Find the total sound energy emitted by the source in each second.
Answer:
P = 251, 3 W
Explanation:
The intensity is defined as the power emitted per unit area
I = P / A
Since sound is distributed in all directions spherical shape, the area of a sphere is
A = 4π r²
let's clear the power and replace
P = I A
P = I (4π r²)
let's calculate
P = 5.00 (4π 2²)
P = 251, 3 W
Since you analyzed the charging of a capacitor for a DC charging voltage, how is it possible that you
Answer:
I = E/R e^{-t/RC}
Explanation:
In a capacitor charging circuit you must have a DC power source, the capacitor, a resistor, and a switch. When closing the circuit,
E -q / c-IR = 0
we replace the current by its expression and divide by the resistance
I = dq / dt
dq / dt = E / R -q / RC
dq / dt = (CE -q) / RC
we solve the equation
dq / (Ce-q) = -dt / RC
we integrate and evaluate for the charge between 0 and q and for the time 0 and t
ln (q-CE / -CE) = -1 /RC (t -0)
eliminate the logarithm
q - CE = CE [tex]e^{-t/RC}[/tex]
q = CE (1 + 1/RC e^{-t/RC} )
In general the teams measure the current therefore we take the derivative to find the current
i = CE (e^{-t/RC} / RC)
I = E/R e^{-t/RC}
This expression is the one that describes the charge of a condensate in a DC circuit
Diamagnetic materialsA) have small negative values of magnetic susceptibility.B) are those in which the magnetic moments of all electrons in each atom cancel.C) experience a small induced magnetic moment when placed in an external magnetic field.D) exhibit the property of diamagnetism independently of temperature.E)are described by all
Answer:
C) experience a small induced magnetic moment when placed in an external magnetic field.
Explanation:
Diamagnetics materials are those that experience a small induced magnetic moment when placed in an external magnetic field. These materials, such as bismuth, copper, silver and lead, have elementary magnets in their compositions. When they are exposed to an external magnetic cap, these elemental magnets tend to follow an orientation contrary to the external magnetic field. As a result, a magnetic field is created in the opposite direction to the external magnetic field.
water and air are both fluids. why is it easier to lift a rock in water rather thatn lifting a rock in air? a the force of gravity. b the bouyant force is greater on the rock in water. c the bouyant force is greater on the rock in air. d the force of gravity on the rock is less in water.
Answer:
The answer is option b.the buoyant force is greater on the rock in water.
If an X-ray beam of wavelength 1.4 × 10-10 m makes an angle of 30° with a set of planes in a crystal causing first order constructive interference, what is the plane spacing?
Answer:
Plane spacing, [tex]d=1.4\times 10^{-10}\ m[/tex]
Explanation:
It is given that,
Wavelength of x-ray, [tex]\lambda=1.4\times 10^{-10}\ m[/tex]
Angle the x-ray made with a set of planes in a crystal causing first order constructive interference is 30 degrees
We need to find the plane spacing. It is based on Bragg's law such that,
[tex]2d\sin\theta=n\lambda[/tex]
d is plane spacing
n = 1 here
[tex]d=\dfrac{\lambda}{2\sin\theta}\\\\d=\dfrac{1.4\times 10^{-10}}{2\times \sin (30)}\\\\d=1.4\times 10^{-10}\ m[/tex]
So, the plane spacing is [tex]1.4\times 10^{-10}\ m[/tex].
The inner and outer surface temperature of a glass window 10 mm thick are 25 and 5 degree-C, respectively. What is the heat loss through a 1 m x 3 m window
Answer:
The heat loss is [tex]H = 8400\ W[/tex]
Explanation:
From the question we are told that
The thickness is [tex]t = 10 \ mm = 0.01 \ m[/tex]
The inner temperature is [tex]T_i = 25 ^oC[/tex]
The outer temperature is [tex]T_o = 5 ^oC[/tex]
The length of the window is L = 1 m
The width of the window is w = 3 m
Generally the heat loss is mathematically represented as
[tex]H = \frac{k * A * \Delta T}{t}[/tex]
Where k is the thermal conductivity of glass with value [tex]k = 1.4\ W/m \cdot K[/tex]
and A is the area of the window with value
[tex]A = 1 * 3[/tex]
[tex]A = 3 \ m^2[/tex]
substituting values
[tex]H = \frac{1.4 * 3 * (23-5)}{0.01}[/tex]
[tex]H = 8400\ W[/tex]
A stone is thrown vertically upward with a speed of 29.0 m/s and when it reaches a height of 13 m, the velocity is 24.2 m/s.
Using the formula x = v0 • t + ½ • a • t2, find the time it takes to reach this height? Why do you get two values for time? Explain.
Answer:
the value of t = 0.49 seconds shows that its upward journey
and
at t = 5.43 seconds shows in downward journey
Explanation:
Given:
initial speed, u = 29 m/s
acceleration due to gravity, g = - 9.8 m/s^2
h = 13 m
Let it is moving with velocity v at a height of 13 m.
Use third equation of motion
v² = u² + 2gh
By substituting the values
v² = 29² - (2 * 9.8 * 13)
v = sqrt 585.94
v = 24.2 m/s
Let it takes time t to reach at height 13 m
Use second equation of motion
s = u * t + 1/2 * g * t²
13 = 29t - 4.9t²
4.9t² - 29t + 13 = 0
using quadratic equation to solve time
29 ± [tex]\sqrt{29^2 - 4 * 4.9 * 13}\\[/tex]
t = ------------------------------------
2 * 4.9
t = 5.43 second or t = 0.49 second
Therefore...
the value of t = 0.49 seconds shows that its upward journey
and
at t = 5.43 seconds shows in downward journey
On a certain planet a body is thrown vertically upwards with an initial speed of 40 m / s. If the maximum height was 100 m, the acceleration due to gravity is
a) 15 m / s 2
b) 12.5 m / s 2
c) 8 m / s 2
d) 10 m / s 2
Answer:
C) 8 m/s²
Explanation:
Given:
v₀ = 40 m/s
v = 0 m/s
Δy = 100 m
Find: a
v² = v₀² + 2aΔy
(0 m/s)² = (40 m/s)² + 2a (100 m)
a = -8 m/s²
A sharp edged orifice with a 50 mm diameter opening in the vertical side of a large tank discharges under a head of 5m. If the coefficient of contraction is 0.62 and the coefficientof velocity is 0.98, what is thedischarge?
Answer:
0.24
Explanation:
See attached file
During a visit to the beach, you get in a small rubber raft and paddle out beyond the surf zone. Tiring, you stop and take a rest. Describe the movement of your raft during your rest. How does this movement differ, if at all, from what you would have experienced if you had stopped paddling while in the surf zone?
Answer:
The main difference in these two movements is that the first is a pure swing movement and the followed form a wave travels from the beach
Explanation:
The movement in the two parts is very different, when the surf zone has passed it is in a deeper part of the water where the seabed does not rise much, therefore due to the movement of the waves there is an upward oscillatory movement and descending, in this movement there is no horizontal displacement.
When it is within the southern zone, there is a rapid rise of the sea floor, which generates a horizontal movement, having a traveling wave, therefore your movement is more complicated, you can have some oscillating movement on the axis and, but in addition to this you have a horizontal movement that reaches you towards the beach, forming a Traveling wave.
The main difference in these two movements is that the first is a pure swing movement and the followed form a wave travels from the beach
A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the negative plate with a speed of 45000 m/s.
What will be the final speed of an electron released from rest at the negative plate?
Express your answer to two significant figures and include the appropriate units
V=_________
Answer:
2.1x10^6m/s
Explanation:
One electron has a charge of –1.602e-19 C
mass of electron is 9.1e-31 kg
mass of proton is 1.6726e−27 kg
mass ratio is 1.6726e−27 / 9.1e-31 = 1838
The force is constant, F
distance is constant, d
a = F/m
a increases by a factor 1838, as m decreases by that factor
a = a₀1838
v₀² = 2a₀d
v² = 2a₀d1838
v²/v₀² = 2a₀d1838 / 2a₀d = 1838
v² = 1838v₀² = 1838(45000)²
v = 45000√1838 = 2.1e6 m/s
3) If a ball launched at an angle of 10.0 degrees above horizontal from an initial height of 1.50 meters has a final horizontal displacement of 3.00 meters, what is its launch velocity
Answer:
35.6 m
Explanation:
The given ball possesses a projectile motion from its initial height. So, the required launch velocity of the ball is 6.55 m/s.
What is launch velocity?The horizontal component of velocity during the projection of an object is known as launch velocity. It is obtained when the horizontal range is known.
Given data -
The angle of projection is, [tex]\theta = 10.0 {^\circ}[/tex].
The initial height of the projection is, h = 1.50 m.
The horizontal displacement is, R = 3.00 m.
The mathematical expression for the horizontal displacement (Range) of the projectile is given as,
[tex]R = \dfrac{u^{2} \times sin2 \theta}{g}[/tex]
here,
u is the launch velocity.
g is the gravitational acceleration.
Solving as,
[tex]u =\sqrt{\dfrac{R \times g}{sin2 \theta}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin(2 \times 10)}}\\\\\\u =\sqrt{\dfrac{1.50 \times 9.8}{sin20^\circ}}\\\\\\u=6.55 \;\rm m/s[/tex]
Thus, we can conclude that the required launch velocity of the ball is 6.55 m/s.
Learn more about the projectile motion here:
https://brainly.com/question/11049671
Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris wheel rotates once every 24 s.What is the apparent weight of a 40 kg passenger at the lowest point of the circle?What is the apparent weight of a 40 kg passenger at the highest point of the circle?
Answer:
a
[tex]F_A =425.42 \ N[/tex]
b
[tex]F_A_H = 358.58 \ N[/tex]
Explanation:
From the question we are told that
The diameter of the Ferris wheel is [tex]d = 80 \ ft = \frac{80}{3.281} = 24.383[/tex]
The period of the Ferris wheel is [tex]T = 24 \ s[/tex]
The mass of the passenger is [tex]m_g = 40 \ kg[/tex]
The apparent weight of the passenger at the lowest point is mathematically represented as
[tex]F_A_L = F_c + W[/tex]
Where [tex]F_c[/tex] is the centripetal force on the passenger, which is mathematically represented as
[tex]F_c =m * r * w^2[/tex]
Where [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{2* \pi }{T}[/tex]
substituting values
[tex]w = \frac{2* 3.142 }{24}[/tex]
[tex]w = 0.2618 \ rad/s[/tex]
and r is the radius which is evaluated as [tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{24.383}{2}[/tex]
[tex]r = 12.19 \ ft[/tex]
So
[tex]F_c = 40 * 12.19* (0.2618)^2[/tex]
[tex]F_c = 33.42 \ N[/tex]
W is the weight which is mathematically represented as
[tex]W = 40 * 9.8[/tex]
[tex]W = 392 \ N[/tex]
So
[tex]F_A = 33.42 + 392[/tex]
[tex]F_A =425.42 \ N[/tex]
The apparent weight of the passenger at the highest point is mathematically represented as
[tex]F_A_H = W- F_c[/tex]
substituting values
[tex]F_A_H = 392 - 33.42[/tex]
[tex]F_A_H = 358.58 \ N[/tex]
A hoop, a solid disk, and a solid sphere, all with the same mass and the same radius, are set rolling without slipping up an incline, all with the same initial kinetic energy.
Which goes furthest the incline?
a. The hoop
b. The disk
c. The sphere
d. They all roll to the same height
Answer:
The sphere
Explanation:
Because it has a smaller inertia (I) value in the explanation in the attached file
A car travels at 100 km / h, collides head-on against a pole. Assuming the vehicle stopped at 2.2 seconds after impact, calculate the magnitude of the deceleration suffered by the driver.
Answer:
12.6 m/s²
Explanation:
First, convert to m/s.
100 km/h × (1000 m/km) × (1 hr / 3600 s) = 27.8 m/s
a = Δv / Δt
a = (0 m/s − 27.8 m/s) / 2.2 s
a = -12.6 m/s²
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the wavelength of the illuminating light is doubled
Answer:
Doubling the wavelength of the diffracting doubles the angle of diffraction. So, the width of the central bright spot pattern formed on the screen will also be doubled.
Explanation:
For a single slit diffraction, the path length difference is related to the wavelength of the light leaving the slit onto the screen by
D sin θ = mλ
where D sin θ is the path length of the waves, each.
mλ is the wavelength of the wavelet
where m is the the order of each minimum
m = m = 1,−1,2,−2,3, . . .
The wavelength of each wavelet is always a multiple of the wavelength of the light source, and from the equation, we can see that the angle of diffraction depend on the wavelength of the light. From this we can see that increasing the wavelength of the light increases the angle of diffraction, and hence we can say that doubling the wavelength will double the diffraction angle. Also, the width of the central bright spot of the screen will spread or increase with the angle of diffraction, so doubling the wavelength doubles the central bright spot on the screen.
An electronic stove is rated 1000w, 240v. explain this statement. please I'm in need
Answer and Explanation:
When an electronic appliance such as electric heater, electric stove e.t.c is rated, the rating actually specifies the ideal working properties of the appliance. For example if it is rated 200W, 220V, it shows that the power the appliance will consume at a voltage of 220V is 200W.
Therefore, for the electronic stove mentioned with a rating of 1000W, 240V, the stove will consume or draw a power of 1000 watts at a voltage of 240volts.
Ratings can also help determine some other properties of the appliance such as current consumption and resistance in the appliance. For the given electronic stove, the current consumed can be found by using the following relation:
P = IV -------------(i)
Where;
P = Power rating = 1000W
I = Current used
V = Voltage rating = 240V
Substituting these values into equation (i) gives;
1000 = I x 240
I = [tex]\frac{1000}{240}[/tex] = 4.17A
Therefore, the current used by the stove is 4.17A
To get the resistance R of the stove, we can use the relation;
P = [tex]\frac{V^2}{R}[/tex]
R = [tex]\frac{V^2}{P}[/tex]
R = [tex]\frac{240^2}{1000}[/tex]
R = 57.6Ω
Therefore, the resistance of the stove is 57.6Ω
g A tube open at both ends, resonated at it's fundamental frequency, to a sound wave traveling at 330m/s. If the length of the tube is 4cm, find the frequency of the sound wave.
Answer:
frequency =4125Hz
Explanation:
L = 4cm = 0.04m
f =v/2L
f = 330/2 x 0.04
f = 4125Hz
If a system has 4.50×102 kcal of work done to it, and releases 5.00×102 kJ of heat into its surroundings, what is the change in internal energy (ΔE or Δ????) of the system?
The change in internal energy (ΔE) of the system is equal to -18823 Kilojoules.
Given the following data:
Quantity of heat = [tex]5.00 \times 10^2 \;kJ[/tex]Work done = [tex]4.50 \times 10^2 \;kcal[/tex]Conversion:
1 kcal = 4.184 kJ
[tex]4.50 \times 10^2 \;kcal[/tex] = [tex]4.50 \times 10^2 \times 4.184 = 18828 \; kJ[/tex]
To determine the change in internal energy (ΔE) of the system, we would apply the first law of thermodynamics.
Mathematically, the first law of thermodynamics is given by the formula:
[tex]\Delta E = Q - W[/tex]
Where;
[tex]\Delta E[/tex] is the change in internal energy.Q is the quantity of heat released.W is the work done.Substituting the given parameters into the formula, we have;
[tex]\Delta E = 5 - 18828\\\\\Delta E = -18823[/tex]
Change in internal energy, E = -18823 Kilojoules
Read more: https://brainly.com/question/20599052
In general, how do highland climates compare with nearby areas at lower elevations? They are cooler and drier. They are cooler and wetter. They are warmer and drier. They are warmer and wetter.
Answer:
They are cooler and wetter
Explanation:
Highland areas have lower temperatures compared to low lying areas. The climate is more wetter because of more rainfalls compared to low lying areas and the wind carries moist air over the highlands.
Blood is pumped from the heart at a rate of 5.0 L/min into the aorta (of radius 1.0 cm). Determine the speed of blood through the aorta in units of m/s
Answer:
The speed of blood through the aorta is 0.265 m/s
Explanation:
Given;
volumetric flow rate, Q = 5.0L/min = 0.005 m³/min x 1min/60s = 8.333 x 10⁻⁵ m³/s
radius of the aorta, r = 1.0 cm = 0.01 m
Area of the aorta = πr²
Area of the aorta = π(0.01)² = 3.142 x 10⁻⁴ m²
Volumetric flow rate is given by;
Q = Av
where;
v is the speed of blood through the aorta
v = Q /A
v = (8.333 x 10⁻⁵ ) / (3.142 x 10⁻⁴)
v = 0.265 m/s
Therefore, the speed of blood through the aorta is 0.265 m/s
The blood's speed through the aorta will be "0.265 m/s". To understand the calculation, check below.
Blood and AortaAccording to the question,
Volumetric flow rate, Q = 5.0 L/min or,
= 0.005 × [tex]\frac{1 \ min}{60 \ s}[/tex]
= 8.333 × 10⁻⁵ m³/s
Aorta's radius, r = 1.0 cm
We know the formula,
→ Q = A × v
or,
Speed, v = [tex]\frac{Q}{A}[/tex]
By substituting the values,
= [tex]\frac{8.333\times 10^{-5}}{3.142\times 10^{-4}}[/tex]
= 0.265 m/s
Thus the above answer is correct.
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